Class 12

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 13 Probability Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1

Question 1.
Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P(E/F) and P(F/E).
Solution.
It is given that, P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2
⇒ P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{0.2}{0.3}=\frac{2}{3}\)

and P(F/E) = \(\frac{P(E \cap F)}{P(E)}=\frac{0.2}{0.6}=\frac{1}{3}\)

Question 2.
Compute P(A/B), if P(B) 0.5 and P(A ∩ B) = 0.32.
Solution.
It is given that, P(B) = 0.5 and P(A ∩ B)= 0.32
⇒ P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{0.32}{0.5}=\frac{16}{25}\).

Question 3.
If P(A) = 0.8, P(B)= 0.5 and P(B/A)= 0.4, find
(i) P(AB)
(ii) P(A/B)
(iii) P(A ∪ B)
Solution.
k is given that, P(A) = 0.8, P(B) 0.5 and P(B/A) = 0.4
(i) P(B|A) = 0.4
∴ \(\frac{P(A \cap B)}{P(A)}\) = 0.4

⇒ \(\frac{P(A \cap B)}{0.8}\) = 0.4

⇒ P(A ∩ B) = 0.32

(ii) P (A/B) = \(\frac{P(A \cap B)}{P(B)}\)
⇒ P(A/B) = \(\frac{0.32}{0.5}\) = 0.64

(iii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ P(A ∪ B) = 0.8 + 0.5 – 0.32 = 0.98

Question 4.
Evaluate P(A ∪ B), if 2P(A) = P(B) = \(\frac{5}{13}\) and P(A/B) = \(\frac{2}{5}\).
Solution.
It is given that, 2P(A) = P(B) = \(\frac{5}{13}\)

Question 5.
If P(A) = \(\frac{6}{11}\), P(B) = \(\frac{5}{11}\) and P(A uB) = \(\frac{7}{11}\), find
(i) P(A ∩ B)
(ii) P(A/B)
(iii) P(B/A)
Solution.

Direction (6 – 9): Determine P(E/F):

Question 6.
A coin is tossed three times, where
(i) E : head on third toss, F : heads on first two tosses.
(ii) E : at least two heads, F : at most two heads
(iii) E : at most two tails, F : at least on tail
Solution.
When a coin is tossed three times, then the sample space S is
S = {HHH, HHT, HTH, HIT, THH, THT, TTH, TTT}
It, can be seen that the sample space has 8 elements.

(i) E = {HHH, HTH, THH, TTH}
F = {HHH, HHT}
∴ E ∩ F = {HHH}
P(F) = \(\frac{2}{8}=\frac{1}{4}\) and P(E ∩ F) = \(\frac{1}{8}\)
P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{8}}{\frac{1}{4}}=\frac{4}{8}=\frac{1}{2}\)

(ii) E = {HHH, HHT, HTH, THH}
F = {HHT, HTH, HTT, THH, THT, TTH, TTT}
∴ E ∩ F – {HHT, HTH, THH}
Clearly, P(E ∩ F) = \(\frac{3}{8}\) and P(F) = \(\frac{7}{8}\)
P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{3}{8}}{\frac{7}{8}}=\frac{3}{7}\)

(iii) E = {HHH, HHT, HTT, HTH, THH, THT, TTH}
F = {HHT, HTT, HTH, THH, THT TTH, TTT}
∴ E ∩ F = {HHT, HTT, HTH, THH, THT, TTH} v
P(F) = \(\frac{7}{8}\) and P(E ∩ F) = \(\frac{6}{8}\)

Therefore, P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{6}{8}}{\frac{7}{8}}=\frac{6}{7}\)

Question 7.
Two coins are tossed once, where
(i) E : tail appears on one coin, F : one coin shows head
(ii) E : no tail appears, F : no head appears
Solution.
If two coins are tossed once, then the sample space S is,
S = {HH, HT, TH, TT}
E : tail appears on one coin – {TH, HT}
F : one coin shows head = {HT, TH}
E ∩ F : {TH, HT}
∴ P(E ∩ F) = \(\frac{2}{4}=\frac{1}{2}\),

P(F) = \(\frac{2}{4}=\frac{1}{2}\)

P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{1 / 2}{1 / 2}\) = 1

(ii) E = set of events having no tail = {HH}
F = set of events having no head = {TT} and E ∩ F = Φ
P(F) = 1 and P(E ∩ F) = 0
∴ P(E / F) = \(\frac{P(E \cap F)}{P(F)}=\frac{0}{1}\) = 0

Question 8.
A die is thrown three times,
E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses.
Solution.
If a die is thrown three times, then the number of elements in the sample space will be 6 × 6 × 6 = 216

and F = {(6, 5, 1) (6, 5, 2), (6, 5, 3) (6, 5, 4) (6, 5, 5), (6, 5, 6)}

∴ E ∩ F = {(6, 5, 4)}

P(F) = \(\frac{6}{216}\) and P(E ∩ F) = \(\frac{1}{216}\)
[∵ From three dice, number of exhaustive cases = 6 × 6 × 6 = 216]
∴ P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{216}}{\frac{6}{216}}=\frac{1}{6}\)

Question 9.
Mother, father and son line up at random for a family picture
E : son on one end, F : father in middle.
Solution.
If mother (M), father (P), and son (S) line up for family picture, then the sample space will be
S = {MFS, MSF, FMS, FSM, SMF, SFM}
⇒ E = {MFS, FMS, SMF, SFM}
F = {MFS, SFM}
E ∩ F = {MFS, SFM}
P(E ∩ F) = \(\frac{2}{6}=\frac{1}{3}\)
P(F) = \(\frac{2}{6}=\frac{1}{3}\)
P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{1 / 3}{1 / 3}\) = 1

Question 10.
A black and a red dice are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Solution.
Let the first observation be from the black die and second from the red die.
When two dice (one black and another red) are rolled the sample space S has 6 × 6 = 36 number of elements
(a) Let
A : Obtaining a sum greater than 9 = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)
B : Black die results in a 5 = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
∴ A ∩ B = {(5, 5), (5, 6)}
The conditional probability of obtaining a sum is greater than 9, given that the black die resulted in a 5 is given by P(A/B).
∴ P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{\frac{2}{36}}{\frac{6}{36}}=\frac{2}{6}=\frac{1}{3}\)

(b) E : Sum of the observations is 8 = {2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
F : Red die resulted in a number less than 4
=
⇒ E ∩ F – {(5, 3), (6, 2)}
Similarly, P(F) = \(\frac{18}{36}\) and P(E ∩ F) = \(\frac{2}{36}=\frac{1}{18}\)
The conditional probability of obtaining the sum equal to 8 given that the red die resulted in a number less then 4 is given by P(E/F).
Therefore, P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{2}{36}}{\frac{18}{36}}=\frac{2}{18}=\frac{1}{9}\).

Question 11.
A fair die is rolled. Consider events E = {1, 3, 5}, F (2, 3) and G = {2, 3, 4, 5}. Find
(i) P(E/F) and P(F/E)
(ii) P(E/G) and P(G/E)
(iii) P((E ∪ F)/G) and P((E ∩ F)/G)
Solution.
When a fair die is rolled, the sample space S will be S = {1, 2, 3, 4, 5, 6}

(i)

(ii) E ∩ G = {3, 5}

(iii) E ∪ F = {1, 2, 3, 5}
(E ∪ F) ∩ G = {12, 3, 5} ∩ {2, 3, 4, 5} = {2, 3, 5} £ ∩ F = {3}
(E ∩ F) ∩ G = {3} ∩ {2, 3, 4, 5} = {3}

Question 12.
Assume that each horn child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that
(i) the youngest is a girl,
(ii) at least one is a girl?
Solution.
Let b and g represent the boy and the girl child respectively. If a family has two children, the sample space will be
S = {(b, b), (b, g), (g, b), (g, g)}
Let A be the event that both children are girls.
∴ A = {(g, g)}

(i) Let B be the event that the youngest child is a girl.
B = [(b, g), (g, g)] =$
⇒ A ∩ B = {(g, g)}
⇒ P(B) = \(\frac{2}{4}=\frac{1}{2}\) and P(A ∩ B) = \(\frac{1}{4}\)

The conditional probability that both are girls, given that the youngest child is a girl, is given by P(A/B).

P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}\)
Therefore, the required probability is \(\frac{1}{2}\).

(ii) Let C be the event that at least one child is a girl.
∴ C = {(b, g), (g, b)(g, g)}
⇒ A ∩ C = {g, g}
⇒ P(C) = \(\frac{3}{4}\) and P(A ∩ C) = \(\frac{1}{4}\).
The conditional probability that both are girls, given that at least one child is a girl, is given by P(A/C).

Therefore, P(A/C) = \(\frac{P(A \cap C)}{P(C)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}\).

Question 13.
An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question?
Solution.

Let us denote E = easy questions, M = multiple choice questions, D =
difficult questions, and T = True/False questions
Total number of questions = 1400
Total number of multiple choice questions = 900
Therefore, probability of selecting an easy multiple choice questions is
P(E ∩ M) = \(\frac{500}{1400}=\frac{5}{14}\)
Probability of selecting a multiple choice questions, P(M) is
\(\frac{900}{1400}=\frac{9}{14}\)
Therefore, the required probability is \(\frac{5}{9}\).

Question 14.
Given that the two numbers appearing on throwing the two dice are different. Find the probability of the event, “the sum of numbers on the dice is 4”.
Solution.
When dice is thrown, number of observations in the sample space = 6 × 6 = 36
Let A be the event that the sum of the numbers on the dice is 4 and B be the event that the two numbers appearing on throwing the two dice are different.

P(B) = \(\frac{30}{36}=\frac{5}{6}\) and

P(A ∩ B) = \(\frac{2}{36}=\frac{1}{18}\)

Let P(A/B) represent the probability that the sum of the numbers on the dice is 4, given that the two numbers appearing on throwing the two dice are different.

∴ P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{18}}{\frac{5}{6}}=\frac{1}{15}\)
Therefore, the required probability is \(\frac{1}{15}\).

 

Question 15.
Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.
Solution.
The outcomes of the given experiment can be represented by the following set.
The sample space of the experiment is,

Direction (16 – 17): Choose the correct answer.

Question 16.
If P(A) = \(\frac{1}{2}\), P(B) = 0, then P(A/B) is
(A) 0
(B) \(\frac{1}{2}\)
(C) not defined
(D) 1
Solution.
It is given that P(A) = \(\frac{1}{2}\) and P(B) = 0
P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{P(A \cap B)}{0}=\infty\)
Therefore, P(A/B) is not defined.
Thus, the correct answer is (C).

Question 17.
If A and B are events such that P(A/B) = P(B/A), then
(A) A ⊂ B but A±B
(B) A = B
(C) A ∩ B = Φ
(D) P(A) = P(B)
Solution.
It is given that, P(A/B) = P(B/A)
⇒ \(\frac{P(A \cap B)}{P(B)}=\frac{P(A \cap B)}{P(A)}\)
⇒ P(B) = P(A).
Thus, the correct answer is (D).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 12 Linear Programming Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 12 Miscellaneous Exercise

Question 1.
A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol.

How many packets of each food should be used to maximise the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?
Solution.
Let the diet contain x and y packets of foods P and Q respectively.
Therefore, x ≥ 0 and y ≥ 0.
The mathematical formulation of the given problem is as follows.
Maximise Z = 6x + 3y ……………..(i)
subject to the constraints, 4x + y ≥ 80 ………… (ii)
x + 5y ≥ 115 …………..(iii)
3x + 2y ≤ 150 ……………(iv)
x, y ≥ 0 …………….(v)
The feasible region determined by the system of constraints is as follows.

The comer points of the feasible region are A (15, 20), B (40, 15) and C (2, 72).
The values of Z at these corner points are as follows.

Thus, the maximum value of Z is 285 at (40, 15).

Therefore to maximise the amount of vitamin A in the diet, 40 packets of food P and 15 packets of food Q should be used. The maximum amount of vitamin A in the diet is 285 units.

Question 2.
A farmer mixes two brands P and Q of cattle feed. Brand P costing ₹ 250 per bag contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing ₹ 200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B and 3 units of element C.

The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?
Solution.
Let the farmer mix x bags of brand P and y bags of brand Q.
The given information can be compiled in a table as follows

The given problem can be formulated as follows.
Minimise Z = 250x + 200y …………(i)
subject to the constraints, 3x + 1.5y ≥ 18 …………(ii)
2.5x + 11.25y ≥ 45 …………(iii)
2x + 3y ≥ 24 …………..(iv)
x, y ≥ 0 ………….(v)
The feasible region determined by the system of constraints is as follows.

The comer points of the feasible region are A (18, 0) B (9, 2) C (3, 6) and D (0, 12).
The values of Z at these corner points are as follows.

As the feasible region is unbounded, therefore, 1950 may or may not be minimum value of Z.
For this, we draw a graph of the inequality, 250x + 200y < 1950 or 5x + 4y < 39 and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 5x + 4y < 39.
Therefore, the minimum value of Z is 2000 at (3, 6).
Thus, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimise the cost to ₹ 1950.

Question 3.
A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:

One kg of food X costs ₹ 16 and one kg of food Y costs ₹ 20. Find the least cost of the mixture which will produce the required diet?
Solution.
Let the mixture contain x kg of food X and y kg of food Y.
The mathematical formulation of the given problem is as follows,
Minimise Z = 16x + 20y ………….(i)
subject to the constraints, x + 2y ≥ 10 ……… (ii)
x + y ≥ 6 …………….(iii)
3x + y ≥ 8 ………….(iv)
x, y ≥ 0 ………….(v)
The feasible region determined by the system of constraints is as follows.

The corner points of the feasible region are A (10, 0) B (2, 4), C (1, 5) and D (0, 8).
The values of Z at these corner points are as follows.

As the feasible region is unbounded, therefore, 112 may or may not be the minimum value of Z.
For this, we draw a graph of the.. inequality, 16x + 20y <112 or 4x + 5y < 28 and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 4x + 5y < 28.
Therefore, the minimum value of Z is 112 at (2, 4).
Thus, the mixture should contain 2 kg of food X and 4 kg of food Y.
The mixture is ₹ 112.

Question 4.
A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below.

Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is ₹ 7.50 and that on each toy of type B is ₹ 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.
Solution.
Let x and y toys of type A and type B respectively be manufactured in a day.
The given problem can be formulated as follows.
Maximise Z = 7.5x + 5y ……………….(i)
subject to the constraints, 2x + y ≤ 60 …………….(ii)
x ≤ 20 ………………..(iii)
2x + 3y ≤ 120 ……………..(iv)
x, y ≥ 0 …………………(v)
The feasible region determined by the constraints is as follows.
The corner points of the feasible region are A (20, 0), B (20, 20) C (15, 30) and D (0, 40).

The values of Z at these comer points are as follows.

The values of Z is 262.5 at (15, 30).
Thus, the manufacturer should manufacture 15 toys of type A and 30 toys of type B to maximise the profit.

Question 5.
An aeroplane can carry a maximum of 200 passengers. A profit of ₹ 1000 is made on each executive class ticket and a profit of ₹ 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?
Solution.
Let the airline sell x tickets of executive class and y tickets of economy class.
The mathematical formulation of the given problem is as follows.
Maximise Z = 1000x + 600y ……………..(i)
subject to the constraints, x + y ≤ 200 …………(ii)
x ≥ 20 …………(iii)
y – 4x ≥ 0 ……….(iv)
x, y ≥ 0 ……………..(v)
The feasible region determined by the constraints is as follows.

The corner points of the feasible region are A (20, 80), B (40, 160) and C (20, 180).
The values of Z at these corner points are as follows.

The maximum value of Z is 136000 at (40, 160).
Thus, 40 tickets of executive class and 160 tickets of economy class should be sold to maximise the profit and the maximum profit is ₹ 136000.

Question 6.
Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table.

How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost? Solution.
Let godown A supply x and y quintals of grain to the shops D and E respectively. Then, (100 – x – y) will be supplied to shop F.

The requirement at shop D is 60 quintals since x quintals are transported from godown A.
Therefore, the remaining (60 – x) quintals will be transported from godown. B.
Similarly, (50 – y) quintals and 40 – (100 – x – y) = (x + y – 60) quintals will be transported from godown B to shop E and F respectively.
The given problems can be represented diagrammatically in above figure x ≥ 0, y ≥ 0 and 100 – x – y ≥ 0
=» x ≥ 0, y ≥ 0 and x + y ≤ 100 60 – x ≥ 0, 50 – y ≥ 0 and x + y – 60 ≥ 0
⇒ x ≤ 60, y ≤ 50 and x + y ≥ 60
Total transportation cost Z is given by,
Z = 6x + 3y + 2.5 (100 – x – y) + 4 (60 – x) + 2 (50 – y) + 3 (x + y – 60)
= 6x + 3y + 250 – 2.5x – 2.5y + 240 – 4x +100 – 2y + 3x + 3y -180
= 2.5x + 1.5y + 410
The given problem can be formulated as
Minimise Z = 2.5x + 1.5y + 410 …………..(i)
subject to the constraints, x + y ≤ 100 ………..(ii)
x ≤ 60 …………..(iii)
y ≤ 50 …………(iv)
x + y ≥ 60 ………….(v)
x, y ≥ 0 ……………(vi)
The feasible region determined by the constraints is as follows.

The values of Z at these corner points are as follows.

The minimum value of Z is 510 at (10, 50).
Thus, the amount of grain transported from A to D, E and F is 10 quintals, 50 quintals, and 40 quintals respectively and from B to D, E and F is 50 quintals, 0 quintals, and 0 quintals respectively.
The minimum cost is ₹ 510.

Question 7.
An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500 L, 3000 L and 3500 L respectively. The distance (in km) between the depots and the petrol pumps is given in the following table:

Assuming that the transportation cost of 10 litres of oil is ₹ 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?
Solution.
Let x and y litres of oil be supplied from A to the petrol pumps, D and E. Then, (7000 – x – y) will be supplied from A to petrol pump F.
The requirement at petrol pump D is 4500 L.
Since xL are transported from depot A, the remaining (4500 – x) L will be transported from petrol pump B.
Similarly, (3000 – y) L and 3500 – (7000 – x – y) = (x + y -3 500) L will be transported from depot B to petrol E and F respectively.
The given problems can be represented diagrammatically as follows.
x ≥ 0, y ≥ 0 and (7000 – x – y) ≥ 0
⇒ x ≥ 0, y ≥ 0 and x + y ≤ 7000
4500 – x ≥ 0, 3000 – y ≥ 0 and x + y – 3500 ≥ 0
⇒ x ≤ 4500, y ≤ 3000 and x + y ≥ 3500
Cost of transporting 10 L of petrol = ₹ 1
Cost of transporting 1 L of petrol = ₹ \(\frac{1}{10}\)

Therefore total transportation cost is given by,
Z = \(\frac{7}{10}\) x + \(\frac{6}{10}\) y + \(\frac{3}{10}\) (7000 – x – y) + \(\frac{3}{10}\) (4500 – x) + \(\frac{4}{10}\) (3000 – y) + \(\frac{2}{10}\) (x + y – 3500)
= 0.3x + 0.1 y + 3950
The problem can be formulated as follows
Minimise Z = 0.3x + 0.1y + 3950 …………….(i)
subjectto the constraints, x + y ≤ 7000 ………….(ii)
x ≤ 4500 ……………(iii)
y ≤ 3000 ………….(iv)
x + y ≥ 350 …………….(v)
x, y ≥ 0 ………….(vi)
The feasible region determined by the constraints is as follows.

The corner points of the feasible region are A (3500, 0), B (4500, 0) C (4500, 2500), D (1000, 3000) and E (500, 3000).
The values of Z at these corner points are as follows.

The minimum value of Z is 4400 at (500, 3000).
Thus, the oil supplied from depot A is 500 L, 3000 L and 3500 L, and from depot B is 4000 L, 0L and 0 L to petrol pumps D, E and F respectively. The minimum transportation cost is ₹ 4400.

Question 8.
A fruit grower can use two types of fertiliser in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid atleast 270 kg of potash and at most 310 kg of chlorine.

If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?

Solution.
Let the fruit grower use x bags of brand P and y bags of brand Q.
The problem can be formulated as follows.
Minimise Z = 3x + 3.5y
subject to the constraints, x + 2y ≥ 240 ………….(ii)
x + 0.5y ≥ 90 ……………….(iii)
1.5x + 2y ≤ 310 …………….(iv)
x, y ≥ 0
The feasible region determined by the system of constraints is as follows.

The corner points are A(140, 50), B(20, 140) and C(40, 100).

The values of Z at these corner points are as follows.
The minimum value of Z is 470 at (40, 100).
Thus, 40 bags of brand P and 100 bags of brand Q should be added to the garden to minimise the amount of nitrogen.
The minimum amount of nitrogen added to the garden is 470 kg.

Question 9.
Refer to question 8. If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?
Solution.
Let the fruit grower use x bags of brand P and y bags of brand Q.
The problem can be formulated as follows.
Maximise Z = 3x + 3.5y ……………(i)
subject to the constraints, x + 2y ≥ 240 …………..(ii)
x + 0.5y ≥ 90 …………(iii)
1.5x + 2y ≤ 310 …………..(iv)
x, y ≥ 0 ……….(v)
The feasile region determined y the system of constraints is as follows.

The corner points are A (1. 40, 50), B(20, 140) and C(40, 100).
The values of Z at these corner points are as follows.

The maximum value of Z is 595 at (140, 50).
Thus, 140 bags of brand P and 50 bags of brand Q should be used to maximise the amount of nitrogen.
The maximum amount of nitrogen added to the garden is 595 kg.

Question 10.
A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A.

Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of ₹ 12 and 116 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?
Solution.
Let x and y be the number of dolls of type A and B respectively that are produced per week.
The given problem can be formulated as follows.
Maximise Z = 12x + 16y ……………(i)
subject to the constraints, x + y ≤ 1200……….(ii)
y ≤ \(\frac{x}{2}\)
⇒x ≥ 2y ………..(iii)
x – 3y ≤ 600 ……………(iv)
x, y ≥ 0 …………….(v)
The feasible region determined by the system of constraints is as follows.

The corner points are A(600, 0), B(1050, 150) and C(800, 400).
The Y’ values of Z at these corner points are as follows.

The maximum value of Z is 16000 at (800, 400).
Thus, 800 and 400 dolls of type A and type B should be produced respectively to get the maximum profit of ₹ 16000.

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 12 Linear Programming Ex 12.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.2

Question 1.
Reshma wishes to mix two types of food P and Q in such a way that vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs ₹ 60/kg and food Q costs ₹ 80/kg. Food P contains 3 units/kg of vitamin A and 5 units/kg of vitamin B while food Q contains 4 units/kg of vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture?
Solution.
Let the mixture contains x kg of food P and y kg of food Q.
Therefore, x ≥ 0 and y ≥ 0
The given information can be compiled in a table as follows.

The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B.
Therefore, the constraints are 3x + 4y ≥ 8; 5x + 2y ≥ 11.
Total cost, Z of purchasing food is Z = 60x + 80 y
The mathematical formulation of the given prolem is

Minimise Z = 60x + 80y …………(i)
subject to the constraints,
3x + 4y ≥ 8 ……………(ii)
5x + 2y ≥ 1 …………….(iii)
x,y ≥ 0 …………..(iv)
The feasible region determined by the system of constraints is as follows.
It can be seen that the feasible region is unbounded.
The corner points of the feasible region are A (\(\frac{8}{3}\), 0), (2, \(\frac{1}{2}\)) and C (0, \(\frac{11}{2}\)).
The value of Z at these corner points are as follows.

As the feasible region is unbounded, therefore, 160 may or may not be the minimum value of Z.
For this, we graph the inequality, 60x + 80 y < 160 or 3x + 4y < 8 and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 3x + 4y< 8.
Therefore, the minimum cost of the mixture will be ₹ 160 at the line segment joining the points (\(\frac{8}{3}\), 0) and (2, \(\frac{1}{2}\)).

Question 2.
One kind of cake requires 200g of flour and 25g of fat and another kind of cake requireds 100g of flour and 50g of fat. Find the maximum number of cakes which can he made from 5kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.
Solution.
Let there be x cakes of first kind and y cakes of second kind.
Therefore, x ≥ 0 and y ≥ 0
The given information can be compiled in a table as follows

∴ 200x +100y ≤ 5000
⇒ 2x + y ≤ 50
25x + 50 y ≤ 1000
⇒ x + 2y ≤ 40 Total number of cakes, Z, that can be made are, Z = x + y
The mathematical formulation of the given problem is
Maximise Z = x + y ……………(i)
subject to the constraints,
2x + y ≤ 50 …………..(ii)
x + 2y < 40 ……………(iii)
x , y ≥ 0 ………………(iv)

The feasible region determined by the system of constraints is as follows

The corner points are A(25, 0), B(20, 10), C(0, 20) and 0(0, 0). The values of Z at these comer points are as follows

Thus, the maximum number of cakes that can be made are 30 (20 of one kind and 10 of the other kind).

Question 3.
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftsman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.
(i) What number of rackets and bats must be made if the factory is to work at full capacity?
(ii) If the profit on a racket and on a bat is ₹ 20 and ₹ 10 respectively, find the maximum profit of the factory when it works at full capacity.
Solution.
(i) Let the number of rackets and the number of bats to be made by x and y respectively.
The machine time is not available for more than 42 hours.
∴ 1.5x + 3y ≤ 42 …………(i)
The craftsman’s time is not available for more than 42 hours.
∴ 3x + y ≤ 24 ……………(ii)
The factory is to work at full capacity.
Therefore, 1.5x + 3y = 42;
3x + y = 24
On solving these equations, we get x = 4 and y = 12
Thus, 4 rackets and 12 bats must be made.

(ii) The given information can be compiled in a tables as follows

∴ 1.5x + 3y ≤ 42;
3x + y ≤ 24;
x, y ≥ 0
The profit on a racket is ₹ 20 and on a bat is ₹ 10.
∴ Z = 20x +10y
The mathematical formulation of the given problem is
Maximise Z = 20x+10y …………….(i)
subject to the constraints,
1.5x + 3y ≤ 42 …………(ii)
3x + y ≤ 24 …………..(iii)
x, y ≥ 0 ……………(iv)
The feasible region determined by the system of constraints is as follows

The corner points are A (8, 0) B (4, 12), C (0, 14) and O (0, 0).
The values of Z at these comer points are as follows

Thus, the maximum profit of the factory when it works to its full capacity is ₹ 200.

Question 4.
A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of holts. He earns a profit, of ₹ 17.50 per package on nuts and ₹ 7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at the most 12 hours a day?
Solution.
Let the manufacturer produce x packages of nuts and y packages of bolts.
Therefore, x ≥ 0 and y ≥ 0
The given information can be compiled in a table as follows.

The profit on a package of nuts if ₹ 17.50 and on a package of bolts is ₹ 7.
Therefore, the constraints are x + 3y ≤ 12; 3x + y ≤ 12.
Total profit Z = 17.5x + 7y
The mathematical formulation of the given problem is
Maximise Z = 17.5x + 7y ……………(i)
subject to the constraints, x + 3y ≤12 ……….(ii)
3x + y ≤ 12………….(iii)
x, y ≥ 0 …………(iv)
The feasible region determined by the system of constraints is as follows.

The corner points are A(4, 0), B(3, 3) and C(0, 4).
The values of Z at these corner points are as follows.

The maximum value of Z is ₹ 73.50 at (3, 3).
Thus, 3 packages of nuts and 3 packages of bolts should be produced each day to get the maximum profit ₹ 73.50.

Question 5.
A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B.

Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of ₹ 7 and screws B at a profit of ₹ 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit.
Solution.
Let the factory manufacture x screws of type A and y screws of type B on each day.
Therefore, x ≥ 0 and y ≥ 0.
The given information can be compiled in a table as follows.

The profit on a package of screws A is ₹ 7 and on the package of screws B is ₹ 10.
Therefore the constraints are 4x + 6y ≤ 240; 6x + 3y ≤ 240
Total profit, Z = 7x + 10y
The mathematical formulation of the given problem is
Maximise Z = 7x+10y …………..(i)
subjectto the constraints, 4x + 6y ≤ 240 …………..(ii)
6x + 3y ≤ 240 …………..(iii)
x, y ≥ 0 …………….(iv)
The feasible region determined by the system of constraints is shown in previous page graph.

The comer points are A (40, 0), B (30, 20) and C (0, 40).
The values of Z at these comer points are as follows.

The maximum value of Z is 410 at (30, 20).
Thus, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit ₹ 410.

Question 6.
A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade.

On any day, the sprayer is available for the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is ₹ 5 and that from a shade is ₹ 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit?
Solution.
Let the cottage industry manufacture x pedestal lamps and y wooden shades.
Therefore, x ≥ 0 and y ≥ 0.
The given information can be compiled in a table as follows.

The profit on a lamp is ₹ 5 and on the shades is ₹ 3.
Therefore, the constraints are 2x + y ≤ 12; 3x + 2y ≤ 20
Total profit, Z = 5x + 3y ,
The mathematical formulation of the given problem is
Maximise Z = 5x + 3y …………(i)
subject to the constraints,
2x + y ≤ 12 ……………(ii)
3x + 2y ≤ 20 ………(iii)
x, y ≥ 0 …………..(iv)
The feasible region determined by the system of constraints is as follows.

The maximum value of Z at these comer points are as follows.

The maximum value of Z is 32 at (4, 4).
Thus, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maximise his profits.

Question 7.
A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is ₹ 5 each for type A and ₹ 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximise the profit.
Solution.
Let the company manufacture x souvenirs of type A and y souvenirs of type B.
Therefore, x ≥ 0 and y ≥ 0.
The given information can be compiled in the table as follows.

The profit on type A souvenirs is ₹ 5 and on type B souvenirs is ₹ 6.
Therefore, the constraints are
5x + 8y ≤ 200;
10x +8y ≤ 200 i.e., 5x +4y ≤ 120

The mathematical formulation of the given problem is
Maximise Z = 5x + 6y ………….(i)
subject to the constraints,
5x + 8y ≤ 200 ……….(ii)
5x + 4y ≤ 120 ……….(iii)
x,y ≥ 0 …………..(iv)
The feasible region determined by the system of constraints is as follows.
The corner points are A(24, 0), B(8, 20) and C(0, 25).
The values of Z and these comer points are as follows.

The maximum value of Z is 200 at (8, 20).
Thus, 8 souvenirs of type A and 20 souvenirs of type B should be produced each day to get the maximum profit of ₹ 160.

Question 8.
A merchant plans to sell two types of personal computers — a desktop model and a portable model that will cost ₹ 25000 and ₹ 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than ₹ 70 lakhs and if his profit on the desktop model is₹ 4500 and on portable model is ₹ 5000.
Solution.
Let the merchant stock x desktop models and y portable models.
Therefore, x ≥ 0 and y ≥ 0
The cost of a desktop model is ₹ 25000 and of a portable model ₹ 40000.
However, the merchant can invest a maximum of ₹ 70 lakhs.
∴ 25000x + 40000y ≤ 7000000
⇒ 5x + 8y ≤ 1400
The monthly demand of computers will not exceed 250 units.
∴ x + y ≤ 250
The profit on a desktop model is ₹ 4500 and the profit on a portable model is ₹ 5000.
Total profit, Z = 4500x + 5000y
Thus, the mathematical formulation of the given problem is
Maximise Z = 4500x + 5000y ………………..(i)
subject to the constraints, 5x + 8y < 1400 ……………(ii)
x + y ≤ 250 ……………..(iii)
x, y ≥ 0 ……………(iv)
The feasible region determined by the system of constraints is as follows.

The corner points are A (250, 0), B (200, 50) and C (0, 175).
The values of Z at these corner points are as follows.

The maximum value of Z is 1150000 at (200, 50).
Thus, the merchant should stock 200 desktop models and 50 portable models to get the maximum profit of ₹ 1150000.

Question 9.
A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F₁ and F₂ are available. Food F₁ costs ₹ 4 per unit food and F₂ costs ₹ 6 per unit. One unit of food F₁ contains 3 units of vitamin A and 4 units of minerals. One unit of food F₂ contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements?
Solution.
Let the diet contain x units of food F₁ and y units of food F₂.
Therefore, x ≥ 0 and y ≥ 0.
The given information can be compiled in a table as follows.

The values of Z at these comer points are as follows.
The cost of food F₁is 4 per unit and of food F₂ is 6 per unit.
Therefore, the constraints are 3x + 6y ≥80; 4x + 3y ≥100; x, y ≥ 0
Total cost of the diet, Z = 4x + 6y
The mathematical formulation of the given problem is
Minimise Z = 4x + 6y ………… …(i)
subject to the constraints, 3x + 6y ≥80 …………… (ii)
4x + 3y ≥ 100 …………..(iii)
x, y ≥ 0 ……………(iv)
The feasible region determined by the constraints is as follows.

It can be seen that the feasible region is unbounded.
The corner points of the feasible region are A (\(\frac{8}{3}\), o) B (2, \(\frac{1}{2}\)) and C(o, \(\frac{11}{2}\))
The comer points are A (\(\frac{80}{3}\), B (24, \(\frac{4}{3}\)) and C (o, \(\frac{100}{3}\))

The value of Z at these corner points are as follows.

As the feasible region is unbounded, therefore, 104 may or may not be the minimum value of Z.
For this, we draw a graph of the inequality, 4x + 6y < 104 or 2x + 3y < 52, and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 2x +3y < 52.
Therefore, the minimum cost of the mixture will be ₹ 104.

Question 10.
There are two types of fertilisers F₁ and F₂ F₁ consists of 10% nitrogen and 6% phosphoric acid and F₂ consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F₁ costs ₹ 6per kg and F₂ costs ₹ 5 kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
Solution.
Let the farmer buy x kg of fertiliser F₁ and y kg of fertiliser F₂.
Therefore, x ≥ 0 and y ≥ 0.
The given information can be compiled in a table as follows.

F₁ consists of 10% nitrogen and F₂ consists of 5% nitrogen.
However, the farmer requires adeast 14 kg of nitrogen.
∴ 10% of x + 5% of y ≥ 14
⇒ \(\frac{x}{10}+\frac{y}{20}\) ≥ 14
⇒ 2x + y ≥ 280
F₁ consists of 6% phosphoric acid and F₂ consists of 10% phosphoric acid. However, the farmer requires at least 14 kg of phosphoric acid.
∴ 6 % of x +10 % of y ≥ 14
⇒ \(\frac{6 x}{100}+\frac{10 y}{100}\) ≥ 14
⇒ 3x + 56y ≥ 700
Total cost of fertilisers, Z = 6x + 5y
The mathematical formulation of the given problem is
Minimise Z = 6x + 5y ………….(i)
subject to the constraints,
2x + y ≥ 280 ………..(ii)
3x + 5y > 700 ……….(iii)
x,y ≥ 0 ………….(iv)
The feasible region determined by the system of constraints is as follows.

It can be seen that the feasible region is unbounded.
The comer points are A (\(\frac{700}{3}\), o) B (100, 80) and C (0, 280).
The values of Z of these points are as follows.

As the feasible region is unbounded, therefore, 1000 may or may not be the minimum value of Z.
For this, we draw a graph of the inequality, 6x + 5y < 1000 and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 6x + 5y < 1000
Therefore, 100 kg of fertiliser F₁ and 80 kg of fertlizer F₂ should be used to minimise the cost.
The minimum cost is ₹ 1000.

Question 11.
The comer points of the feasible region determined by the following system of liner inequalities.
2x + y ≤ 10, x + 3y ≤ 15, x, y > 0 are (0, 0), (5, 0), (3, 4) and (0, 5).
Let Z = px + qy, where p, q> 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is
(A) p = q
(B) p = 2q
(C) p = 3q
(D) q = 3p
Solution.
The maximum value of Z is unique.
It is given that the maximum value of Z occurs at two points, (3, 4) and (0, 5).
∴ Value of Z at (3, 4) = Value of Z at (0, 5)
⇒ p(3) + q(4) = p(0) + q(5)
⇒ 3p + 4q = 5q
⇒ q = 3P
Hence, the correct answer is (D).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 12 Linear Programming Ex 12.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1

Direction (1 – 6): Solve the following linear programming problems graphically.

Question 1.
Maximise Z = 3x + 4y
subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥ 0.
Solution.
The feasible region determined by the constraints, x + y ≤ 4, x ≥ 0, y ≥ 0, is as follows.

The corner points of the feasible region are 0 (0, 0), A (4, 0) and B (0, 4).

Therefore, the maximum value of Z is 16 at the point B(0, 4).

Question 2.
Minimise Z = – 3x + 4y
Subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0
Solution.
The feasible region determined by the system of constraints, x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0 and y ≥ 0 is as follows.

The comer points of the feasible region are O (0, 0), A (4, 0), B (2, 3) and C (0, 4).
The values of Z at these comer points are as follows

Therefore, the minimum value of Z is – 12 at the point (4, 0).

Question 3.
Maximise Z = 5x + 3y
subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.
Solution.
The feasible region determined by the system of constraints, 3 + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0 and y ≥ 0 are as follows.
The corner points of the feasible region are O (0, 0), A (2, 0), B (0, 3) and C (\(\frac{20}{19}\), \(\frac{45}{19}\))
The values of Z at these comer points are as follows.

Therefore, the maximum value of Z is \(\frac{235}{19}\) at the point (\(\frac{20}{19}\), \(\frac{45}{19}\)).

Question 4.
Minimise Z = 3x + 5y
such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.
Solution.
The feasible region determined by the system of constraints, x + 3y ≥ 3, x + y ≥ 2 and x, y ≥ 0 is as follows

It can be seen that the feasible region is unbounded.
The corner points of the feasile region are A (3, 0), (\(\frac{3}{2}\), \(\frac{1}{2}\)) and C (0, 2)
The values of Z at these comer points are as follows

As the feasible region is unbounded, therefore, 7 may or may not be the minimum value of Z.
For this, we draw the graph of the inequality, 3x + 5y < 7 and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 3x + 5y < 7.
Therefore, the minimum value of Z is 7 at (\(\frac{3}{2}\), \(\frac{1}{2}\)).

Question 5.
Maximise Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.
Solution. T
he feasible region determined by the constraints, x + 2y ≤ 10, 3x + y ≤ 15, x ≥ 0 and y ≥ 0 is as follows.
The corner points of the feasible region are A(5, 0) B(4, 3) and C(0, 5).

The values of Z at these comer points are as follows.

Therefore, the maximum value of Z is 18 at the point B(4, 3).

Question 6.
Minimise Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.
Solution.
The feasible region determined by the constraints, 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0 and y ≥ 0 is as follows.

The corner points of the feasible region are A(6, 0) and B(0, 3).
The values of Z at these comer points are as follows

Comer point Z = x + 2y.
It can be seen that the value of Z at points A and B is same.
If we take any other point such as (2, 2) on line x + 2y = 6, then Z = 6.
Thus, the minimum value of Z occurs for more than 2 points.
Therefore, the value of Z is minimum at every point on the line, x + 2y = 6.

Direction (7 – 10): Show that the minimum of Z occurs at more than two points.

Question 7.
Minimise and Maximise Z = 5x + 110y subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0.
Solution.
The feasible region determined by the constraints, x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x ≥ 0, and y ≥ 0 is as follows

The corner points of the feasible region are A (60, 0), B (120, 0), C (60, 30) and D (40, 20).
The values of Z at the corner points are as follows

The minimum value of Z is 300 at (60, 0) and the maximum value of Z is 600 at all the points on the line segment joining (120, 0) and (60, 30).

Question 8.
Minimise and maximise Z = x + 2y subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0. Solution. The feasible region determined by the constraints, x + 2y ≥ 100, 2x – y ≤ 0,2x + y ≤ 200, x > 0 and y ≥ 0 is as follows

The comer points of the feasible region are A (0, 50), B (20, 40), C (50, 100) and D (0, 200).
The values of Z at the corner points are as follows

The minimum value of Z is 400 at (0, 200) and the maximum value of Z is 100 at all the points on the line segment joining (0, 50) and (20, 40).

Question 9.
Maximise Z = – x + 2y,
subject to the constraints : x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.
Solution.
The feasible region determined by the constraints, x ≥ 3, x + y ≥ 5, x + 2y ≥ 6 and y ≥ 0 is as given below

It can be seen that the feasible region is unbounded
The values of Z at corner points A(6,0), B(4,1) and C(3,2) are as follows.

As the feasible region is unbounded, therefore, Z = 1 may or may not be the maximum value.
For this, we graph the inequality, – x + 2y > 1 and check whether the resulting half plane has points in common with the feasible region or not. The resulting feasible reason has points in common with the feasible region.
Therefore, Z = 1 is not the maximum value. Hence, Z has no maximum value.

Question 10.
Maximise Z = x + y,
subject to x – y ≤ – 1, – x + y ≤ 0, x, y ≥ 0.
Solution.
The region determined by the constraints, x – y ≤ – 1, – x + y ≤ 0, x, y ≥ 0 is as follows.

There is no feasible region and thus, Z has no maximum value.

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 11 Three Dimensional Geometry Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise

Question 1.
Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5,-1), (4, 3,-1).
Solution.
Let OA be the line joining the origin, 0(0, 0, 0), and the point, A(2, 1, 1).
Also, let BC be the line joining the points, B (3, 5, – 1) and C(4, 3, – 1).
The direction ratios of OA are 2, 1 and 1 and of BC are (4 – 3) = 1, (3 – 5) = – 2 and (- 1 + 1) = 0
OA is perpendicular to BC, if a<sub1a₂ + b₁b₂ + c₁c₂ = 0
∴ a₁a₂ + b₁b₂ + c₁c₂ = 2 × 1 + 1 (- 2) + 1 × 0
= 2 – 2 = 0
Thus, OA is perpendicular to BC.

Question 2.
If l₁, m₁ and n₁ and l₂, m₂, n₂ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m₁n₂ – m₂n₁, n₁l₂ – n₂l₁, l₁m₂ – l₂m₁.
Solution.
Given, lines are respectively parallel to unit vector

Question 3.
Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.
Solution.
The angle θ between the lines with direction cosines, a, b, c and b – c, c – a, a – b is given by
cos θ = \(\frac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^{2}+b^{2}+c^{2}}+\sqrt{(b-c)^{2}+(c-a)^{2}+(a-b)^{2}}}\)
⇒ cos θ = 0
⇒ θ = cos^-1 0
⇒ θ = 90°
Thus, the angle between the lines is 90°.

Question 4.
Find the equation of a line parallel to x-axis and passing through the origin.
Solution.
The line parallel to x-axis and passing through the origin is x-axis itself.
Let A be a point on x-axis.
Therefore, the coordinates of A are given by (a, 0,0), where a ∈ R.
Direction ratios of OA are (a – 0), 0, 0 = a, 0,0
The equation of OA is given by,
\(\frac{x-0}{a}=\frac{y-0}{0}=\frac{z-0}{0}\)

⇒ \(\frac{x}{1}=\frac{y}{0}=\frac{z}{0}\) = a
Thus, the equation of line parallel to x-axis and passing through origin is \(\frac{x}{1}=\frac{y}{0}=\frac{z}{0}\) = a

Question 5.
If the coordinates of the points A, B, C,D be (1, 2, 3), (4, 5, 7), (- 4, 3, – 6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.
Solution.
The coordinates of A, B, C and D are (1, 2, 3),(4, 5, 7), (- 4, 3, – 6)and (2, 9, 2) respectively.
The direction ratios of AB are (4 – 1) = 3, (5 – 2) = 3 and (7 – 3) = 4
The direction ratios of CD are (2 – (- 4)) = 6, (9 – 3) = 6 and (2 – (- 6)) = 8
It can be seen that,

\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{2}\)

Therefore, AB is parallel to CD.
Thus, the angle between AB and CD is either 0° or 180°.

Question 6.
If the lines \(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and \(\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}\) are perpendicular, find the value of k.
Solution.
The direction of ratios of the lines, \(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and \(\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}\) are – 3, 2k, 2 and 3k, 1, – 5 respectively.
Itis known that two lines with direction ratios, a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular, if
a₁a₂ + b₁b₂ + c₁c₂ = 0
∴ – 3 (3k) + 2k × 1 + 2 (- 5) = 0
⇒ – 9k + 2k – 10 = 0
⇒ 7k = – \(\frac{10}{7}\)
⇒ k = – \(\frac{10}{7}\)
Therefore, for k = – \(\frac{10}{7}\), the given lines are perpendicular to each other.

Question 7.
Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \(\vec{r}\) . (î + 2ĵ – 5k̂) + 9 = 0.
Solution.
Direction ratios of the normal of the plane \(\vec{r}\) . (î + 2ĵ – 5k̂) + 9 = 0 are 1, 2, – 5.
The equation of line passing through \(\overrightarrow{r_{1}}\) and with direction ratios b₁, b₂, b₃ is
\(\vec{r}=\vec{r}_{1}+\lambda\left(b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\right)\)
Hence, the line passing through (1, 2, 3) and having the direction ratios 1, 2, – 5 is \(\vec{r}\) = î + 2ĵ + 3k̂ + λ (î + 2ĵ – 5k̂)

Question 8.
Find the equation of the plane passing through (a, b, c) and parallel to the plane \(\vec{r}\) . (î + ĵ + k̂) = 2.
Solution.
Any plane parallel to the plane, \(\overrightarrow{r_{1}}\) . [(î + ĵ + k̂) = 2, is of the form
\(\vec{r}\) . (î + ĵ + k̂) = λ …………….(i)
The plane passes through the point (a, b, c).
Therefore, the position vector \(\vec{r}\) of this point is \(\vec{r}\) = aî + bĵ + ck̂
Therefore, equation (i) becomes
(aî + bĵ + ck̂) . (î + ĵ + k̂) = λ
⇒ a + b + c = λ
Substituting λ = a + b + c in equation (i), we get
\(\vec{r}\) . (î + ĵ + k̂) = a + b + c …(ii)
This is the vector equation of the required plane.
Substituting \(\vec{r}\) = xî + yĵ + zk̂ in equation (ii), we get
(xî + yĵ + zk̂) . (î + ĵ + k̂) = a + b + c
⇒ x + y + z = a + b + c.

Question 9.
Find the Shortest distance between lines \(\vec{r}\) = 6î + 2ĵ + 2k̂ + λ (î – 2ĵ + 2k̂) and r = – 4î – k̂ + μ (3î – 2ĵ – 2k̂).
Solution.

Question 10.
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane.
Solution.
It is known that the equation of the line passing through the points, (x₁, y₁, z₁) and (x₂, y₂, z₂) is
The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by \(\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}\)

⇒ \(\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}\)= k (say)

⇒ x = 5 – 2k, y = 3k + 1, z = 6 – 5k
Any point on the line is of the form (5 – 2k, 3k + 1, 6 – 5k).
The equation of YZ-plane is x = 0
Since the line passes through YZ-plane.
5 – 2k = 0
⇒ k = \(\frac{5}{2}\)
⇒ 3k + 1 = 3 × \(\frac{5}{2}\) + 1 = \(\frac{17}{2}\);

6 – 5k = 6 – 5 × \(\frac{5}{2}\) = – \(\frac{13}{2}\)

Therefore, the required point is (0, \(\frac{17}{2}\), \(-\frac{13}{2}\)).

Question 11.
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.
Solution.
It is known that the equation of the ime passing through the points, (x₁, y₁, z₁) and (x₂, y₂, z₂) is \(\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}\)

The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by \(\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}\)

⇒ \(\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}\) = k(say)
⇒ x = 5 – 2k, y = 3k + 1, z = 6 – 5k
Any point on the line is of the form (5 – 2k, 3k + 1, 6 – 5k).
Since the line passes through ZX-plane.
3k + 1 = 0
⇒ k = – \(\frac{1}{3}\)
⇒ 5 – 2k = 5 – 2(- \(\frac{1}{3}\)) = \(\frac{17}{3}\);

6 – 5k = 6 – 5(- \(\frac{1}{3}\)) = \(\frac{23}{3}\)

Therefore, the required point is (\(\frac{17}{3}\), 0, \(\frac{23}{3}\)).

Question 12.
Find the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses the plane 2x + y + z = 7.
Solution.
It is known that the equation of the line through the points, (x₁, y₁, z₁) and (x₂, y₂, z₂)
Since the line passing through the points, (3, – 4,- 5) and (2, – 3, 1), its equation is given by \(\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}\)

⇒ \(\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}\) = k (say)
⇒ x = 3 – k, y = k – 4, z = 6k – 5
Therefore, any point on the line is of the form (3 – k, k – 4, 6 k – 5).
This point lies on the plane, 2x + y + z = 7
∴ 2 (3 – k) + (k – 4) + (6k – 5) = 7
⇒ 5k – 3 = 7
⇒ k = 2
Hence, the coordinates of the required point are (3 – 2, 2 – 4, 6 × 2 – 5), i.e., (1, – 2, 7).

Question 13.
Find the equation of the plane passing through the point (- 1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
Solution.
The equation of the plane passing through the point (- 1, 3, 2) is
a (x + 1) + b (y – 3) + c (z- 2) = 0 ……………(i)
where, a, b, c are the direction ratios of normal to the plane.
It is known that two planes, a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0, are perpendicular, if
a₁a₂ + b₁b₂ + c₁c₂ = 0
Plane (i) is perpendicular to the plane, x + 2y + 3z = 5
∴ a . 1 + b . 2 + c . 3 = 0
⇒ a + 2b + 3c = 0 …………..(ii)
Also, plane (i) is perpendicular to the plane, 3x + 3y +z = 0
∴ a . 3 + b . 3 + c . 1 = 0
⇒ 3a + 3b + c = 0 …………(iii)
From equations (ii) and (iii), we get
\(\frac{a}{2 \times 1-3 \times 3}=\frac{b}{3 \times 3-1 \times 1}=\frac{c}{1 \times 3-2 \times 3}\)
⇒ \(\frac{a}{-7}=\frac{b}{8}=\frac{c}{-3}\) = k (say)
⇒ a = – 7k, b = 8k, c = – 3k -7
Substituting the value of a, b and c in equation (i), we get
– 7k (x + 1) + 8k (y – 3) – 3k(z – 2) = 0
⇒ (- 7x – 7) + (8y – 24) – 3z + 6 = 0
⇒ – 7x + 8y – 3z – 25 = 0
⇒ 7x – 8y + 3z + 25 = 0
This is the required equation of the plane.

Question 14.
If the points (1, 1, p) and (- 3, 0, 1) be equidistant from the plane \(\vec{r}\) = (3î + 4ĵ – 12k̂) + 13 = 0, then find the value of p.
Solution.
The position vector through the point (1, 1, p) is \(\overrightarrow{a_{1}}\) = î + ĵ + pk̂
Similarly, the position vector through the point (- 3, 0, 1) is \(\overrightarrow{a_{2}}\) = – 4î + k̂
The equation of the given plane is \(\vec{r}\) . (3î + 4ĵ – 12k̂) + 13 = 0

Question 15.
Find the equation of the plane passing through the line of intersection of the planes \(\vec{r}\) . (î + ĵ + k̂) = 1 and \(\vec{r}\) . (2î + 3ĵ – k̂) + 4 = 0 and parallel to x-axis.
Solution.
The given planes are \(\vec{r}\) . (î + ĵ + k̂) = 1
⇒ \(\vec{r}\) . (î + ĵ + k̂) – 1 = 0
\(\vec{r}\) . (2î + 3ĵ – k̂) + 4 = 0
The equation of any plane passing through the line of intersection of these planes is
[\(\vec{r}\) . (î + ĵ + k̂) – 1] + λ [\(\vec{r}\) . (2î + 3ĵ – k̂) + 4] = 0
\(\vec{r}\) . [(2λ + 1)î + (3λ + 1)ĵ + (1 – λ) k̂] + (4λ +1) = 0 ………….(i)
Its direction ratios are (2λ + 1), (3λ + 1) and (1 – λ)
The required plane is parallel to x-axis.
Therefore, its normal is perpendicular to x-axis.
The direction ratios of x-axis are 1, 0 and 0.
∴ 1 (2λ + 1) + 0 (3λ + 1) + 0 (1 – λ) = 0
⇒ 2λ + 1 = 0
⇒ λ = – \(\frac{1}{2}\)
Substituting λ = – \(\frac{1}{2}\) in equation (i), we get
⇒ \(\vec{r} \cdot\left[-\frac{1}{2} \hat{j}+\frac{3}{2} \hat{k}\right]\) + (- 3) = 0
⇒ \(\vec{r}\) (ĵ – 3k̂) + 6 = 0
Therefore, its Cartesian equation is y – 3z + 6 = 0.
This is the equation of the required plane.

Question 16.
If O be the origin and the coordinates of P be (1, 2,- 3), then find the equation of the plane passing through P and perpendicular to OP.
Solution.
The coordinates of the points, O and P are (0, 0, 0) and (1, 2, – 3) respectively.
Therefore, the direction ratios of OP are (1 – 0) = 1, (2 – 0) = 2 and (- 3 – 0) = – 3
Here, the direction ratios of normal are 1, 2 and – 3 and the point P is (1, 2, – 3).
Thus, the equation of the required plane is 1 (x – 1) + 2 (y – 2) – 3 (z + 3) = 0
⇒ x – 1 + 2y – 4 – 3z – 9 = 0
⇒ x + 2y – 3z – 14 = 0.

Question 17.
Find the equation of the plane which contains the line of intersec¬tion of the planes \(\vec{r}\) . (î + 2ĵ + 3k̂) – 4 = 0, \(\vec{r}\) . (2î + ĵ – k̂) + 5 = 0 and which is perpendicular to the plane \(\vec{r}\) . (5î + 3ĵ – 6k̂) + 8 = 0.
Solution.
The equations of the given planes are
\(\vec{r}\) . (î + 2ĵ + 3k̂) – 4 = 0 …………..(i)
\(\vec{r}\) . (2î + ĵ – k̂) + 5 = 0 ………….(ii)
The equation of the plane passing through the line intersection of the plane given in equation (i) and equation (ii) is
[\(\vec{r}\) . (î + 2ĵ + 3k̂) – 4 ] + λ [\(\vec{r}\) . (2î + ĵ – k̂) + 5] = 0
\(\vec{r}\) . [(2λ + 1)î + (λ + 2)ĵ + (3 – λ)k̂] + (5λ – 4) = 0 ……………(iii)
The plane in equation (iii) is perpendicular to the plane,
\(\vec{r}\) . (5î + 3ĵ – 6k̂) + 8 = 0
∴ 5(2λ + 1) + 3 (λ + 2) – 6 (3 – λ) = 0
⇒ 19λ – 7 = 0
⇒ λ = \(\frac{7}{19}\)
Sustituting λ = \(\frac{7}{19}\) in equation (iii), we get
⇒ \(\vec{r} \cdot\left[\frac{13}{19} \hat{i}+\frac{45}{19} \hat{j}+\frac{50}{19} \hat{k}\right] \frac{-41}{19}\) = 0
⇒ \(\vec{r}\) . (33î + 45ĵ + 50k̂) – 41 = 0
This is the vector equation of the required plane.
The Cartesian equation of this plane can be geted by substituting \(\vec{r}\) = xî + yĵ + zk̂ in equation (iii).
(xî + yĵ + zk̂) . (33î + 45ĵ + 50k̂) – 41 = 0
⇒ 33x + 45y + 50z – 41 = 0.

Question 18.
Find the distance of the point (- 1, – 5, 10) from the point of intersection of the line \(\vec{r}\) = 2î – ĵ + 2k̂ + λ (3î + 4ĵ + 2k̂) and the plane \(\vec{r}\) . (î – ĵ + k̂) = 5.
Solution.
The equation of the given line is
\(\vec{r}\) = 2î – ĵ + 2k̂ + λ (3î + 4ĵ + 2k̂) ………….(i)
The equation of the given plane is
\(\vec{r}\) . (î – ĵ + k̂) = 5 ………….(ii)
Substituting the value of r from equation (i) in equation (ii), we get
[2î – ĵ + 2k̂ + λ (3î + 4ĵ + 2k̂)] . (î – ĵ + k̂) = 5
⇒ (3λ + 2)î + (4λ – 1)ĵ + (2λ + 2)k̂] . (î – ĵ + k̂) = 5
⇒ (3λ + 2) – (4λ – 1) + (2λ + 2) = 5
⇒ λ = 0
Substituting this value in equation (i), we get the equation of the line as
\(\vec{r}\) = 2î – ĵ + 2k̂
And r is position vector of the point (2, – 1, 2)
The distance d between the points (2, – 1, 2) and (- 1, – 5, – 10) is
d = \(\sqrt{(-1-2)^{2}+(-5+1)^{2}+(-10-2)^{2}}\)
= \(\sqrt{9+16+144}=\sqrt{169}\)
= 13 units.

Question 19.
Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes \(\vec{r}\) . (î – ĵ + 2k̂) = 5 and \(\vec{r}\) . (3î + ĵ + k̂) = 6.
Solution.
Let the required line be parallel to vector \(\vec{b}\) given by
\(\vec{b}\) = b₁ î + b₂ ĵ + b₃ k̂
The position vector of the point (1, 2, 3) is \(\vec{a}\) = î + 2 ĵ + 3 k̂
The equation of line passing through (1, 2, 3) and parallel to \(\vec{b}\) is given by
\(\vec{r}=\vec{a}+\lambda \vec{b}\)
⇒ \(\vec{r}\) (î + 2 ĵ + 3 k̂) + λ (b₁ î + b₂ ĵ + b₃ k̂) ……………(i)
The equations of the given planes are
\(\vec{r}\) . (î – ĵ + 2k̂) = 5 …………….(ii)
\(\vec{r}\) . (3î + ĵ + k̂) = 6 …………..(iii)
The line in equation (i) and plane in equation (ii) are parallel. Therefore, the normal to the plane of equation (ii) and the given line are perpendicular.
⇒ (î – ĵ + 2k̂) . λ (b₁ î + b₂ ĵ + b₃ k̂) = 0

Question 20.
Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines \(\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}\) and \(\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}\).
Solution.
Let the required line be parallel to the vector \(\vec{b}\) given by,

Question 21.
Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then \(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{p^{2}}\).
Solution.
The equation of a plane having Intercepts a, b, c with x, y and z axes respetively is given by \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1 …………….(i)

The distance (p) of the plane from the origin is given by

Direction (22 – 23): Choose the correct answer.
Question 22.
Distance between the two planes 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is
(A) 2 units
(B) 4 units
(C) 8 units
(D) units
Solution.
The equations of the planes are
2x + 3y + 4z = 4 ……………(i)
4x + 6y + 8z = 112
⇒ 2x + 3y + 4z = 6 ……………(ii)
It can be seen that the given planes are parallel.
k is known that the distance between two parallel planes, ax + by + cz = d₁ and ax + by + cz = d₂ given by
D = \(\frac{d_{2}-d_{1}}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

D = \(\left|\frac{6-4}{\sqrt{(2)^{2}+(3)^{2}+(4)^{2}}}\right|\)

D = \(\frac{2}{\sqrt{29}}\)
Thus, the distance between the lines is \(\frac{2}{\sqrt{29}}\) units.
Hence, the correct answer is (D).

Question 23.
The planes 2x – y + 4z = 5 and 5x – 25y + 10z = 6are
(A) perpendicular
(B) parallel
(C) intersect y-axis
(D) passes through (o, o, \(\frac{5}{4}\))
Solution.
The equations of the planes are
2x – y + 4z = 5 ………….(i)
5x – 25y + 10z = 6 …………(ii)
It can be seen that
\(\frac{a_{1}}{a_{2}}=\frac{2}{5}\);

\(\frac{b_{1}}{b_{2}}=\frac{-1}{-2.5}=\frac{2}{5}\);

\(\frac{c_{1}}{c_{2}}=\frac{4}{10}=\frac{2}{5}\)

⇒ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\).

Therefore, the given planes are parallel.
Hence, the correct answer is (B).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 11 Three Dimensional Geometry Ex 11.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3

Question 1.
In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
(a) z = 2
(b) x + y + z = 1
(c) 2x + 3y – z = 5
(d) 5y + 8 = 0
Solution.
(a) The equation of the plane is z = 2 or 0x + 0y + z = 2 ……………..(i)
The direction ratios of normal are 0, 0 and 1.
∴ \(\sqrt{0^{2}+0^{2}+1^{2}}\) = 1
Dividing both sides of equation (i) by 1, we get
0 . x + 0 . y + 1 . z = 2
This is of the form lx + my + nz = d, where 1, m, n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.
Therefore, the direction cosines are 0, 0 and 1 and the distance of the plane from the origin is 2 units.

(b) x + y + z = 1 ……………..(i)
The direction ratios of normal are 1, 1 and 1.
∴ \(\sqrt{(1)^{2}+(1)^{2}+(1)^{2}}\) = √3
Dividing both sides of equation (1) by we get
\(\frac{1}{\sqrt{3}} x+\frac{1}{\sqrt{3}} y+\frac{1}{\sqrt{3}} z=\frac{1}{\sqrt{3}}\) ………………(ii)
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
Therefore, the direction cosines of the normal are and and the \(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\) distance of normal from the origin is \(\frac{1}{\sqrt{3}}\) units.

(c) 2x + 3y – z = 5 ………………(i)
The direction ratios of normal are 2, 3 and – 1.
∴ \(\sqrt{(2)^{2}+(3)^{2}+(-1)^{2}}=\sqrt{14}\)
Dividing both sides of equation (i) by √14, we get

\(\frac{2}{\sqrt{14}} x+\frac{3}{\sqrt{14}} y-\frac{1}{\sqrt{14}} z=\frac{5}{\sqrt{14}}\)

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal are \(\frac{2}{\sqrt{14}}\), \(\frac{3}{\sqrt{14}}\) and \(\frac{- 1}{\sqrt{14}}\) and the distance of normal from the origin is \(\frac{5}{\sqrt{14}}\) units.

(d) 5y + 8 = 0
⇒ 0x – 5y + 0z = – 8 ………………..(i)
The direction ratios of normal are 0, – 5 and 0.
∴ \(\sqrt{0+(-5)^{2}+0}\) = 5
Dividing both sides of equation (i) by 5, we get – y = \(\frac{8}{5}\)
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
Therefore, the direction cosines of the normal to the plane are 0,- 1 and 0 and the distance of normal from the origin is \(\frac{8}{5}\) units.

Question 2.
Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector 3î + 5ĵ – 6k̂.
Solution.
The normal vector is, \(\vec{n}\) = 3î + 5ĵ – 6k̂

This is the vector equation of the required plane.

Question 3.
Find the Cartesian equation of the following planes
(a) \(\vec{r}\) (î + ĵ – k̂) = 2
(b) \(\vec{r}\) (2î + 3ĵ – 4k̂) = 1
(c) \(\vec{r}\) [(s – 2t) î + (3 – t) ĵ + (2s + t) k̂] = 15
Solution.
(a) It is given that equation of the plane is \(\vec{r}\) (î + ĵ – k̂) = 2 …………. (i)
For any arbitrary point P(x, y, z) on the plane, position vector \(\vec{r}\) is given by,
\(\vec{r}\) = xî + yĵ – zk̂
Substituting the value, of \(\vec{r}\) in equation (i), we get
(xî + yĵ – zk̂) (î + ĵ- k̂) = 2
⇒ x + y – z = 2
This is the Cartesian equation of the plane.

(b) Given vector equation is \(\vec{r}\) . (2î +3ĵ – 4k̂) = 1
For any arbitrary point P(x, y, z) on the plane, position vector \(\vec{r}\) is given by
\(\vec{r}\) = xî + yĵ – zk̂
Substituting the value of r in equation (i), we get
(xî + yĵ – zk̂ ) . (2î +3ĵ – 4k̂) = 1
⇒ 2x + 3y – 4z = 1
This is the Cartesian equation of the plane.

(c) Given vector equation is \(\vec{r}\) . [(s – 2t) î + (3 – t ) ĵ + (2s + t) k̂] = 15 ………….. (i)
For any arbitrary point P (x, y, z) on the plane, position vector is given by
\(\vec{r}\) = xî + yĵ – zk̂
Substituting the value, of in equation (i), we get
(xî + yĵ – zk̂) . [(s – 2t) î + (3 – t) k̂ + (2s + t) k̂] = 15
⇒ (s – 2t) x + (3 – t) y + (2s + t) z = 15
This is the Cartesian equation of the plane.

Question 4.
In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
(a) 2x + 3y + 4z – 12 = 0
(b) 3y + 4z – 6 = 0
(c) x + y + z = 1
(d) 5y + 8 = 0
Solution.
(a) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x, y, z).
2x + 3y + 4z – 12 = 0
⇒ 2x + 3y + 4z = 12 ………..(i)
The direction ratios of normal are 2, 3 and 4.
∴ \(\sqrt{(2)^{2}+(3)^{2}+(4)^{2}}=\sqrt{29}\)

Dividing both sides of equation (i) by we get
∴ \(\frac{2}{\sqrt{29}} x+\frac{3}{\sqrt{29}} y+\frac{4}{\sqrt{29}} z=\frac{12}{\sqrt{29}}\)

This equation is of the form lx + my + nz = d, where 1, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by (ld, md, nd)
Therefore, the coordinates of the foot of perpendicular are
\(\left(\frac{2}{\sqrt{29}}, \frac{12}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{12}{\sqrt{29}}, \frac{4}{\sqrt{29}}, \frac{12}{\sqrt{29}}\right)\) i.e., \(\left(\frac{24}{29}, \frac{36}{49}, \frac{48}{29}\right)\).

(b) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x₁, y₁, z₁).
3y + z – 6 = 0
⇒ 0x + 3y + 4z = 6 ………………(i)
The direction ratios of normal are 0, 3 and 4.
∴ \(\sqrt{0+3^{2}+4^{2}}\) = 5
Dividing the both sides of equation (i) by 5, we get
0 . x + \(\frac{3}{5}\) y + \(\frac{4}{5}\) z = \(\frac{6}{5}\)
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ld, md, nd).
Therefore, the coordinates of the foot of the perpendicular are \(\left(0, \frac{3}{5} \cdot \frac{6}{5}, \frac{4}{5} \cdot \frac{6}{5}\right)\) i.e., \(\left(0, \frac{18}{25}, \frac{24}{25}\right)\)

(c) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x₁, y₁, z₁).
x + y + z = 1 ………………(i)
The direction ratios of normal are 1, 1 and 1.
∴ \(\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}\)

Dividing the both sides of equation (i) by √3, we get
\(\frac{1}{\sqrt{3}} x+\frac{1}{\sqrt{3}} y+\frac{1}{\sqrt{3}} z=\frac{1}{\sqrt{3}}\)

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by (ld, md, nd).
Therefore, the coordinates of the foot of the perpendicular are \(\left(\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}}\right)\) i.e., \(\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)\).

(d) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x₁, y₁, z₁).
5y + 8 = 0
⇒ 0 . x – 5y + 0 . z = 8 ……………(i)
The direction ratios of normal are 0, – 5 and 0.
∴ \(\sqrt{0+(-5)^{2}+0}\) = 5
Dividing both sides of equation (i) by 5, we get
– y = \(\frac{8}{5}\)
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ld, md, nd) Therefore, the coordinates of the foot of the perpendicular are (0, – 1 (\(\frac{8}{5}\), 0) i.e., (0, – \(\frac{8}{5}\), 0).

Question 5.
Find the vector and Cartesian equations of the planes
(a) that passes through the point (1, 0, – 2) and the normal to the plane is î + ĵ – k̂.
(b) that passes through the point (1, 4, 6) and the normal vector to the plane is î – 2 ĵ + k̂.
Solution.
(a) The position vector of point (1, 0, – 2) is \(\vec{a}\) = î – 2 k̂
The normal vector N perpendicular to the plane is \(\vec{N}\) = î + ĵ – k̂
The vector equation of the plane is given by, \((\vec{r}-\vec{a}) \cdot \vec{N}\) = 0
⇒ [\(\vec{r}\) – (î – 2k̂)] . (î + ĵ – k̂) = 0 ……………(i)
\(\vec{r}\) is the position vector of any point P(x, y, z) in the plane.
∴ \(\vec{r}\) = xî + yĵ + zk̂
Therefore, equation (i) becomes
[(xî + yĵ + zk̂) – (î – 2k̂) . {î + ĵ – k̂) = 0
⇒ [(x – 1)î + yĵ + (z + 2)k̂] . (î + ĵ – k̂) = 0
⇒ (x – 1) + y – (z + 2) = 0
⇒ x + y – z – 3 = 0
⇒ x + y – z = 3
This is the Cartesian equation of the required plane.

(b) The position vector of the point (1, 4, 6) is \(\vec{a}\) = î + 4ĵ + 6k̂
The normal vector N perpendicular to the plane is \(\vec{N}\) = î – 2ĵ + k̂
The vector equation of the plane is given by, (\(\vec{r}\) – \(\vec{a}\)) . \(\vec{N}\) = 0
⇒ [\(\vec{r}\) – ( î + 4ĵ + 6k̂)] + 4; +6fc)] . (î – 2ĵ + k̂ ) = 0 ……………(i)
\(\vec{r}\) is the position vector of any point P(x, y, z) in the plane.
∴ \(\vec{r}\) = xî + yĵ + zk̂
Therefore, equation (i) becomes
[(xî + yĵ + zk̂) – (î + 4ĵ + 6k̂)] (î – 2ĵ + k̂) = 0
⇒ [(x – 1)î +(y – 4)ĵ + (z – 6)k̂] . (î – 2ĵ + k̂) = 0
⇒ (x – 1) – 2 (y – 4) + (z – 6) = 0
⇒ x – 2y + z + 1 = 0
This is the Cartesian equation of the required plane.

Question 6.
Find the equations of the planes that passes through three points.
(a) (1, 1, – 1), (6, 4, – 5), (- 4, – 2, 3)
(b) (1, 1, 0),(1, 2, 1), (- 2, 2, – 1)
Solution.
Solution.
(a) The given points are A(1, 1, – 1), B(6, 4, – 5) and C(- 4, – 2, 3).
\(\left|\begin{array}{ccc}
1 & 1 & -1 \\
6 & 4 & -5 \\
-4 & -2 & 3
\end{array}\right|\) =(12 – 10) – (18 – 20) – (- 12 + 16)
= 2 + 2 – 4 = 0
Since, A, B, C are collinear points, there will be infinite number of planes passing through the given points.

(b) The given points are A(1, 1, 0), B(1, 2, 1) and C(- 2, 2, – 1)
\(\left|\begin{array}{ccc}
1 & 1 & 0 \\
1 & 2 & 1 \\
-2 & 2 & -1
\end{array}\right|\) = (- 2 – 2) – (2 + 2) = – 8 ≠ 0
Therefore, a plane will pass through the points A, B and C.
It is known that the equation of the plane through the points, (x₁, y₁, z₁),
(x₂, y₂, z₂)and (x₃, y₃, z₃) is

\(\left|\begin{array}{ccc}
x-x_{1} & y-y_{1} & z-z_{1} \\
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1}
\end{array}\right|\) = 0

\(\left|\begin{array}{ccc}
x-1 & y-1 & z \\
0 & 1 & 1 \\
-3 & 1 & -1
\end{array}\right|\) = 0
⇒ (- 2) (x – 1) – 3 (y – 1) + 3z = 0
⇒ – 2x – 3y + 3z + 2 + 3 = 0
⇒ – 2x – 3y + 3z = – 5
⇒ 2x + 3y – 3z = 5
This is the Cartesian equation of the required plane.

Question 7.
Find the intercepts cut off by the plane 2x + y – z = 5.
Sol.
Given plane, 2x + y – z = 5 ……………(i)
Dividing both sides of equation (i) by 5, we get
\(\frac{2}{5} x+\frac{y}{5}-\frac{z}{5}\) = 1

\(\frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-5}\) = 1 ………………….(ii)

It is known that the equation of a plane in intercept form is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1,
where a, b, c are the intercepts cut off by the plane at x, y and z axes respectively.
Therefore, for the given equation, a = \(\frac{5}{2}\), b = 5 and c = – 5.

Question 8.
Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.
Solution.
The equation of the plane ZOX is y = 0
Any plane parallel to it is of the form, y = a
Since, the y-intercept of the plane is 3,
∴ a = 3
Thus, the equation of the required plane y = 3.

Question 9.
Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).
Solution.
The equation of any plane through the intersection of the planes, 3x – y + 2z – 4 = 0 and x + y + z- 2 = 0 is
(3x- y + 2z – 4) + α (x + y + z -2) = 0 where α ∈ R ……………..(i)
The plane passes through the point (2, 2, 1).
Therefore, this point will satisfy equation (i).
∴ (3 × 2 – 2 + 2 × 1 – 4) + α (2 + 2 + 1 – 2) = 0
⇒ 2 + 3α = 0
⇒ α = – \(\frac{2}{3}\)
Substituting α = – \(\frac{2}{3}\) in equation (i), we get
(3x – y + 2z – 4) – \(\frac{2}{3}\) (x + y + z – 2) = 0
⇒ 3 (3x -y + 2z – 4) – 2 (x + y + z – 2) = 0
⇒ (9x – 3y + 6z – 12) – 2(x + y + z – 2) = 0
⇒ 7x – 5y + 4z – 8 = 0
This is the required equation of the plane.

Question 10.
Find the vector equation of the plane passing through the intersection of the planes \(\overrightarrow{\boldsymbol{r}}\) . (2î + 7ĵ – 3k̂) = 7, \(\overrightarrow{\boldsymbol{r}}\) . (2î + 5ĵ + 3k̂) = 9 and through the point (2,1, 3).
Solution.
The equations of the planes are \(\overrightarrow{\boldsymbol{r}}\) . (2î + 7ĵ – 3k̂) = 7 and \(\overrightarrow{\boldsymbol{r}}\) . (2î + 5ĵ + 3k̂) = 9

⇒ \(\overrightarrow{\boldsymbol{r}}\) . (2î + 2ĵ – 3k̂) – 7 = 0 …………..(i)

⇒ \(\overrightarrow{\boldsymbol{r}}\) . (2î + 5ĵ + 3k̂) – 9 = 0 …………..(ii)

The equation of any plane through the intersection of the planes given in equations (i) and (ii) is given by
⇒ [\(\vec{r}\) . (2î + 2ĵ – 3k̂)] – 7] + λ [\(\vec{r}\) . (2î + 5ĵ + 3k̂) – 9] = 0, where λ ∈ R
\(\vec{r}\) . [(2î + 2ĵ – 3k̂) + λ (2î + 5ĵ + 3k̂)] = 9λ. + 7
\(\vec{r}\) . [(2 + 2λ) î + (2 + 5λ) ĵ + (3λ – 3) k̂] = 9λ + 7 ………………(iii)
The plane passes through the point (2, 1, 3).
Therefore, its position vector is given by, \(\vec{r}\) = 2î + 2ĵ – 3k̂
Substituting in equation (iii), we get
(2î + ĵ – 3k̂) . [(2 + 2λ)î + (2 + 5λ)ĵ + (3λ – 3)k̂] = 9λ + 7
(2 + 2λ) + (2 + 5λ) + (3λ – 3) = 9λ + 7
⇒ 18λ – 3 = 9λ + 7
⇒ 9λ = 10
⇒ λ = \(\frac{10}{9}\)
Substituting λ = \(\frac{10}{9}\) in equation (iii), we get
\(\vec{r} \cdot\left(\frac{38}{9} \hat{i}+\frac{68}{9} \hat{j}+\frac{3}{9} \hat{k}\right)\) = 17
⇒ \(\vec{r}\) = (38î +68ĵ + 3k̂) = 153
This is the vector equation of the required plane.

Question 11.
Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.
Solution.
The equation of the plane through the intersection of the planes, x + y + z = 1 and 2x + 3y + 4z = 5 is
(x + y + z – 1) + λ (2x + 3y + 4z – 5) = 0
⇒ (2λ + 1) x + (3λ + 1) y + (4λ + 1) z – (5λ + 1) = 0 ……………..(i)
The direction ratios, a₁, b₁, c₁ of this plane are (2λ + 1), (3λ + 1) and (4λ + 1).
The plane in equation (i) is perpendicular to x – y + z = 0
Its direction ratios, a₂, b₂, c₂ are 1, – 1 and 1.
Since the planes are perpendicular, a₁a₂ + b₁b₂ + c₁c₂ = 0
⇒ (2λ + 1) – (3λ + 1) + (4λ + 1) = 0
⇒ 3λ + 1 = 0
⇒ λ = – \(\frac{1}{3}\)
Substituting λ = – \(\frac{1}{3}\) in equation (i), we get
\(\frac{1}{3} x-\frac{1}{3} z+\frac{2}{3}\) = 0
⇒ x – z + 2 = 0
This is the required equation of the plane.

Question 12.
Find the angle between the planes whose vector equations are \(\overrightarrow{\boldsymbol{r}}\) . (2î + 3ĵ – 3k̂) = 5 and \(\overrightarrow{\boldsymbol{r}}\) . (3î – 3ĵ + 5k̂) = 3.
Solution.
We know that the angle 0 between the planes

Question 13.
In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.
(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4=0
(b) 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0
(c) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 62 -1 = 0
(d) 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0
(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0
Solution.
(a) The equations of the planes are 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
Here, a₁ = 7, b₁ = 5, c₁ = 6 and a₂ = 3, b₂ = – 1, c₂ = – 10
∴ a₁a₂ + b₁b₁ + c₁c₂ = 7 × 3 + 5 × (- 1) + 6 × (- 10) = – 44 ≠ 0
Therefore, the given planes are not perpendicular each other

(b) The equations of the planes are 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0
Here, a₁ = 2, b₁ = 1, c₁ = 3 and a₂ = 1, b₂ = – 2, c₂ = 0
∴ a₁a₂ + b₁b₂ + c₁c₂ = 2 × 1 + 1 × (- 2) + 3 × 0 = 0
Therefore, the given planes are perpendicular to each other.

(c) The equations of the planes are 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0
Here, a₁ = 2, b₁ = – 2, c₁ = 4 and a₂ = 3, b₂ = – 3, c₂ = 6
∴ a₁a₂ + b₁b₂ + c₁c₂ = 2 × 3 + (- 2) (- 3) + 4 × 6
= 6 + 6 + 24 = 36 ≠ 0
Thus, the given planes are not perpendicular to each other.

(d) The equation of the planes are 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0
Here, a₁ = 2, b₁ = – 1, C₁ = 3 and a₂ = 2, b₂ = – 1, c₂
∴ a₁a₂ + b₁b₂ + c₁c₂ = 2 × 2 + (- 1) × (- 1) + 3 × 3 = 14 ≠ 0
Thus, the given planes are not perpendicular to each other.
\(\frac{a_{1}}{a_{2}}=\frac{2}{2}\) = 1,

\(\frac{b_{1}}{b_{2}}=\frac{-1}{-1}\) = 1 and

\(\frac{c_{1}}{c_{2}}=\frac{3}{3}\) = 1

∴ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
Thus, the given lines are parallel to each other.

(e) The equations of the planes are 4x + 8y + z – 8 = 0 and y + z – 4 = 0
Here a₁ = 4, b₁ = 8, c₁ = 1 and a₂ = 0, b₂ = 1, c₂ = 1
a₁a₂ + b₁b₂ + c₁c₂ = 0
Therefore, the given lines are not perpendicular to each other.
\(\frac{a_{1}}{a_{2}}=\frac{4}{0}\),

\(\frac{b_{1}}{b_{2}}=\frac{8}{1}\) = 8,

\(\frac{c_{1}}{c_{2}}=\frac{1}{1}\) = 1,

∴ \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Therefore, the given lines are not parallel to each other.
The angle between the planes is given by,
θ = cos^-1 \(\left|\frac{4 \times 0+8 \times 1+1 \times 1}{\sqrt{4^{2}+8^{2}+1^{2}} \times \sqrt{0^{2}+1^{2}+1^{2}}}\right|\)

θ = cos^-1 \(\left|\frac{9}{9 \times \sqrt{2}}\right|\)

θ = cos^-1 \(\left(\frac{1}{\sqrt{2}}\right)\)

θ = 45.

Question 14.
In the following cases, find the distance of each of the given points from the corresponding given plane.

Solution.
We know that the distance between a point P(x₁, y₁, z₁) and a plane, Ax + By + Cz = D, is given by
d = \(\left|\frac{A x_{1}+B y_{1}+C z_{1}-D}{\sqrt{A^{2}+B^{2}+C^{2}}}\right|\) …………………(i)

(a) The given point is (0, 0, 0) and the plane is 3x – 4y + 12z = 3

∴ d = \(\left|\frac{3 \times 0-4 \times 0+12 \times 0-3}{\sqrt{(3)^{2}+(-4)^{2}+(12)^{2}}}\right|=\frac{3}{\sqrt{169}}=\frac{3}{13}\)

(b) The given point is(3,- 2, 1) and the plane is 2x – y + 2z – 3 = 0

∴ d = \(\left|\frac{2 \times 3-(-2)+2 \times 1+3}{\sqrt{(2)^{2}+(-1)^{2}+(2)^{2}}}\right|=\left|\frac{13}{3}\right|=\frac{13}{3}\)

(c) The given point is (2, 3, – 5) and the plane is x + 2y – 2z = 9

∴ d = \(\left|\frac{2+2 \times 3-2(-5)-9}{\sqrt{(1)^{2}+(2)^{2}+(-2)^{2}}}\right|=\frac{9}{3}\) = 3

(d) The given point is (- 6, 0, 0) and the plane is lx – 3y + 6z – 2 = 0.

∴ d = \(\left|\frac{2(-6)-3 \times 0+6 \times 0-2}{\sqrt{(2)^{2}+(-3)^{2}+(6)^{2}}}\right|=\left|\frac{-14}{\sqrt{49}}\right|=\frac{14}{7}\) = 2

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 11 Three Dimensional Geometry Ex 11.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2

Question 1.
Show that the three lines with direction cosines \(\frac{12}{13}\), \(\frac{-3}{13}\), \(\frac{-4}{13}\); \(\frac{4}{13}\), \(\frac{12}{13}\), \(\frac{3}{13}\); \(\frac{3}{13}\), \(\frac{-4}{13}\), \(\frac{12}{13}\) are mutually perpendicular.
Solution.
Two lines with direction cosines, l₁, m₁, n₁ and l₂ m₂, n₂ are perpendicular to each other, if l₁l₂ + m₁m₂ + n₁n₂ = 0

(i)

Therefore, the lines are perpendicular.

(ii)

Therefore, the lines are perpendicular.

(iii)
Therefore, the lines are perpendicular.
Thus, all the lines are mutually perpendicular.

Question 2.
Show that the line through the points (1, – 1, 2) (3, 4, – 2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Solution.
Let AB be the linejoining the points, (1, – 1, 2)and(3, 4, – 2) and CD be the line joining the points, (0, 3, 2) and (3, 5, 6).
The direction ratios, a₁, b₁, c₁ of AB are (3 – 1), (4 – (- 1)) and (- 2 – 2) i.e., 2, 5 and – 4.
The direction ratios, a₂, b₂, c₂ of CD are (3 – 0), (5 – 3) and (6 – 2 ) i.e., 3, 2 and 4.
AB and CD will be perpendicular to each other, if a₁a₂ + b₁b₂ + c₁c₂ = 0
a₁a₂ + b₁b₂ + c₁c₂ = 2 × 3 + 5 × 2 + (- 4) × 4
= 6 + 10 – 16 = 0
Therefore, AB and CD are perpendicular to each other.

Question 3.
Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (- 1, – 2, 1), (1, 2, 5).
Solution.
Let AB be the line through the points, (4, 7, 8) and (2, 3, 4) and CD be the line through the points (- 1, – 2, 1) and (1, 2, 5).
The direction ratios a₁, b₁, c₁ of AB are (2 – 4), (3 – 7) and (4 – 8) i.e., – 2, – 4 and – 4.
The direction ratios a₂, b₂, c₂ of CD are (1 – (- 1)), (2 – (- 2)) and (5 – 1) i.e., 2, 4 and 4. .
AB will be parallel to CD, if \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
\(\frac{a_{1}}{a_{2}}=\frac{-2}{2}\) = – 1;

\(\frac{b_{1}}{b_{2}}=\frac{-4}{4}\) = – 1;

\(\frac{c_{1}}{c_{2}}=\frac{-4}{4}\) = – 1

∴ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
Thus, AB is parallel to CD.

Question 4.
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3î + 2ĵ – 2k̂.
Solution.
It is given that the line passes through the point A(1, 2, 3).
Therefore, the position vector through A is \(\vec{a}\) = î + 2ĵ + 3k̂, \(\vec{b}\) = 3î + 2ĵ – 2k̂
It is known that the line which passes through point A and parallel to \(\vec{b}\) is given by
\(\vec{r}\) = \(\vec{a}\) + λ \(\vec{b}\), where λ is a constant.
⇒ \(\vec{r}\) = î + 2ĵ + 3k̂ + λ (3î + 2ĵ – 2k̂)
This is the required equation of the line.

Question 5.
Find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2î – ĵ + 4k̂ and is in the direction î + 2ĵ – k̂.
Solution.
It is given that the line passes through the point with position vector
\(\vec{a}\) = 2î – ĵ + 4k̂ ………………….(i)
\(\vec{b}\) = î + 2ĵ – k̂ ………………..(ii)
It is known that a line through a point with position vector a and parallel to b is given by the equation,
\(\vec{r}\) = \(\vec{a}\) + λ \(\vec{b}\)
⇒ \(\vec{r}\) = 2î – ĵ + 4k̂ + λ (î + 2ĵ – k̂)
This is the required equation of the line is vector form.
\(\vec{r}\) = xî – yĵ + zk̂
\(\vec{r}\) = xî – yĵ + zk̂ = (λ + 2)î + (2λ – 1)ĵ + (- λ + 4) k̂
Eliminating λ, we obtain the Cartesian form equation as
\(\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}\)
This is the required equation of the given line in Cartesian form.

Question 6.
Find the Cartesian equation of the line which passes through the point (- 2, 4, – 5) and parallel to the line given by
\(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\).
Solution.
It is given that the line passes through the point (- 2, 4, – 5) and is parallel to \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\)

The direction ratios of the line, \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\), are 3, 5 and 6.

The required line is parallel to \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\)

Therefore, its direction ratios are 3k, 5k and 6k where k ≠ 0.

It is known that the equation of the line through the point (x₁, y₁, z₁) and with direction ratios, a, b, c is given by \(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\).

Therefore, the equation of the required line is \(\frac{x+2}{3 k}=\frac{y-4}{5 k}=\frac{z+5}{6 k}\)

\(\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}\) = k

Question 7.
The Cartesian equation of a line is \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\). Write its vector form.
Solution.
The Cartesian equation of a line is \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\) …………..(i)
The given line passes through the point (5, – 4, 6).
The position vector of this point is \(\vec{a}\) = 5î – 4ĵ + 6k̂
Also, the direction ratios of the given line are 3, 7 and 2.
This means that the line is in the direction of vector, \(\vec{b}\) = 3î + 7ĵ + 2k̂
It is known that the line through position vector \(\vec{a}\) and in the direction of the vector \(\vec{b}\) is given by the equation,
\(\vec{r}\) = \(\vec{a}\) + λ \(\vec{b}\), λ ∈ R
=*> r = (5î – 4ĵ + 6k̂) + λ (3î + 7ĵ + 2k̂)
This is the required equation of the given line in vector form.

Question 8.
Find the vector and the Cartesian equations of the lines that passes through the origin and (5, – 2, 3).
Solution.
The required line passes through the origin.
Therefore, its position vector is given by,
\(\vec{a}=\overrightarrow{0}\) …………..(i)
The direction ratios of the line through origin and (5, – 2, 3) are
(5 – 0) = 5 (- 2 – 0) = – 2 (3 – 0) = 3
The line is parallel to the vector given by the equation, \(\vec{b}\) = 5î – 2ĵ+ 3k̂
The equation of the line in vector form through a point with position vector \(\vec{a}\) and parallel to \(\vec{b}\) is,
\(\vec{r}\) = \(\vec{a}\) + λ \(\vec{b}\), λ ∈ R
⇒ \(\vec{r}\) = \(\vec{0}\) + λ (5î – 2ĵ+ 3k̂)
⇒ \(\vec{r}\) = λ (5î – 2ĵ+ 3k̂)

The equation of the line through the point (x₁, y₁, z₁) and direction ratios a, b, c is given by, \(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)

Therefore, the equation of the required line in the Cartesian form is
\(\frac{x-0}{5}=\frac{y-0}{-2}=\frac{z-0}{3}\)

⇒ \(\frac{x}{5}=\frac{y}{-2}=\frac{z}{3}\)

Question 9.
Find the vector and the Cartesian equations of the line that passes through the points (3, – 2, – 5), (3, – 2, 6).
Solution.
Let the line passing through the points, P(3, – 2, – 5) and Q(3, – 2, 6), be PQ.
Since PQ passes through P(3, – 2, – 5), its position vector is given by
\(\vec{a}\) = 3î – 2ĵ – 5k̂
The direction ratios of PQ are given by,
(3 – 3) = 0, (- 2 + 2) = 0, (6 + 5) = 11
The equation of the vector in the direction of PQ is
\(\vec{b}\) = 0î – 0ĵ + k̂ = 11k̂
The equation of PQ in vector form is given by,
\(\vec{r}\) = \(\vec{a}\) + λ \(\vec{b}\), λ ∈ R
⇒ \(\vec{r}\) = (3î – 2ĵ – 5k̂) + 11λk̂
The equation of PQ in Cartesian form is
\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\) i.e.,

\(\frac{x-3}{0}=\frac{y+2}{0}=\frac{z+5}{11}\)

Question 10.
Find the angle between the following pairs of lines
(i) \(\vec{r}\) = 2î – 5ĵ + k̂ + λ (3î + 2ĵ + 6k̂) and
\(\vec{r}\) = 7î – 6k̂ + µ (î + 2ĵ + 2k̂)

(ii) \(\vec{r}\) = 3î + ĵ – 2k̂ + λ (î – ĵ – 2k̂) and
r = 2î – ĵ – 56k̂ + µ (3î – 5ĵ – 4k̂)
Solution.
(i) Let Q be the angle between the given lines.
The angle between the given pairs of lines is given by, cos θ
The given lines are parallel to the vectors, \(\vec{b}_{1}\) = 3î + 2ĵ+ 6k̂ and \(\vec{b}_{2}\) = î + 2ĵ + 2k̂ respectively.
∴ \(\left|\vec{b}_{1}\right|=\sqrt{3^{2}+2^{2}+6^{2}}\) = 7;

\(\left|\vec{b}_{2}\right|=\sqrt{(1)^{2}+(2)^{2}+(2)^{2}}\) = 3.

\(\vec{b}_{1}\) . \(\vec{b}_{2}\) = (3î + 2ĵ+ 6k̂) . (î + 2ĵ + 2k̂)
= 3 × 1 + 2 × 2 + 6 × 2
= 3 + 4 + 12 = 19
⇒ cos θ = \(\\left|\frac{\overrightarrow{b_{1}} \cdot \overrightarrow{b_{2}}}{\mid \overrightarrow{b_{1}} \| \vec{b}_{2}}\right|\)

θ = cos^-1 \(\left(\frac{19}{21}\right)\)

(ii) The given lines are parallel to the vectors \(\vec{b}_{1}\) = î – ĵ – 2k̂ and \(\vec{b}_{2}\) = 3î – 5ĵ – 4k̂, respectively.

Question 11.
Find the angle between the following pair of lines
(i) \(\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}\) and \(\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}\)

(ii) \(\frac{x}{2}=\frac{y}{2}=\frac{z}{1}\) and \(\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}\)
Solution.

⇒ cos θ = \(\frac{26}{9 \sqrt{38}}\)

θ = cos^-1 \(\left(\frac{26}{9 \sqrt{38}}\right)\)

(ii)

Question 12.
Find the values of p so that the lines \(\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}\) and \(\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}\) are at right angles.
Solution.
The given equations are not in the standard form.
The equations of the given lines can be written as
\(\) …………….(i)

and \(\) ……………..(ii)

The direction ratios of the given lines are – 3, \(\frac{2 p}{7}\), 2 and \(-\frac{3 p}{7}\), 1, – 5.
Two lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular to each other, if
a₁ a₂ + b₁b₂ + c₁c₂ = 0
∴ (- 3) (\(-\frac{3 p}{7}\)) + (\(\frac{2 p}{7}\)) (1) + (2) – (5) = 0
⇒ 9p + 2p – 70 = 0
⇒ 11p = 70
⇒ p = \(-\frac{70}{11}\)

Question 13.
Show that the lines \(\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\) and \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) are perpendicular to each other.
Solution.
The equations of the given lines are \(\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\) and \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\).

The direction ratios of the given lines are 7, – 5, 1 and 1, 2, 3 respectively.

Two lines with direction ratios, a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular to each other, if
a₁a₂ + b₁b₂ + c₁c₂ = 0
∴ 7 × 1 + (- 5) × 2 + 1 × 3
= 7 – 10 + 3 = 0
Therefore, the given lines are perpendicular to each other.

Question 14.
Find the shortest distance between the lines \(\overrightarrow{\boldsymbol{r}}\) = (î + 2ĵ + k̂) + λ (î – ĵ + k̂) and \(\overrightarrow{\boldsymbol{r}}\) = 2î – ĵ – k̂ + μ (2î + ĵ + 2k̂).
Solution.
The equations of the given lines are \(\overrightarrow{\boldsymbol{r}}\) = (î + 2ĵ + k̂) + λ (î – ĵ + k̂) and \(\overrightarrow{\boldsymbol{r}}\) = 2î – ĵ – k̂ + μ (2î + ĵ + 2k̂).
It is known that the shortest distance between the lines, \(\vec{r}\) = \(\vec{a}_{1}\) + λ \(\vec{b}_{1}\) and \(\vec{r}\) = \(\vec{a}_{2}\) + λ \(\vec{b}_{2}\) is given by
d = \(\left|\frac{\left(\overrightarrow{b_{1}} \times \vec{b}_{2}\right) \cdot\left(\overrightarrow{a_{2}}-\overrightarrow{a_{2}}\right)}{\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|}\right|\) ……………….(i)

Compare the given equations, we get

Question 15.
Find the shortest distance between the lines \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\) and \(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\).
Solution.

Question 16.
Find the shortest distance between the lines whose vector equations are \(\vec{r}\) = (î + 2ĵ + 3k̂) + λ (î – 3ĵ + 2k̂) and \(\vec{r}\) = 4î + 5ĵ + 6k̂ + µ (2î+ 3ĵ + k̂)
Solution.

Question 17.
Find the shortest distance between the lines whose vector equations are \(\overrightarrow{\boldsymbol{r}}\) = (1 – t) î + (t- 2) ĵ + (3 – 21) k̂ and \(\overrightarrow{\boldsymbol{r}}\) = (s + 1) î + (2s – 1) ĵ – (2s + 1) k̂.
Solution.
The given lines are \(\overrightarrow{\boldsymbol{r}}\) = (1 – t) î + (t – 2) ĵ + (3 – 2t) k̂
\(\vec{r}\) = (î – 2ĵ + 3 k̂) +1(- î + ĵ – 2 k̂) …………(i)
\(\vec{r}\) = (s + 1) î + (2s – 1) ĵ – (2s + 1) k̂
\(\vec{r}\) = (î – ĵ + k̂) + s (î + 2ĵ – 2k̂) …………(ii)

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 11 Three Dimensional Geometry Ex 11.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.1

Question 1.
If a line makes angles 90°, 135°, 45° with x, y and z-axes respectively, find its direction cosines.
Solution.
Let direction cosines of the line be l, m and n.
I = cos 90° = 0, m = cos 135° = – \(\frac{1}{\sqrt{2}}\), n = cos 45° = \(\frac{1}{\sqrt{2}}\)

Therefore, the direction cosines of the line are 0, \(\frac{1}{\sqrt{2}}\) and \(\frac{1}{\sqrt{2}}\).

Question 2.
Find the direction cosines of a line which makes equal angles with the coordinate axes.
Solution.
Let the direction cosines of the line make an angle a with each of the coordinate axes.
∴ l = cos α, m = cos α, n = cos α
We know that l² + m² + n² = 1
⇒ cos² α + cos² α + cos² α = 1
⇒ 3 cos² α = 1
⇒ cos² α = \(\frac{1}{3}\)
⇒ cos α = ± \(\frac{1}{\sqrt{3}}\)
Thus, the direction cosines of the line, which is equally inclined to the coordinate axes, are ± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\) and ± \(\frac{1}{\sqrt{3}}\).

Question 3.
If a line has the direction ratios -18, 12, – 4, then what are its direction cosines?
Solution.
If a line has direction ratios of – 18,12 and – 4, then its direction cosines are

Question 4.
Show that the points (2, 3, 4), (- 1, – 2, 1), (5, 8, 7) are collinear.
Solution.
Let the given points are A (2, 3, 4), B (- 1, – 2, 1), C(5, 8, 7)
Direction ratios of AB are x₂ – x₁; y₂ – y₁, z₂ – z₁
i.e., (- 1 – 2), (- 2 – 3), (1 – 4) or – 3, – 5, – 3
Direction ratios of BC are
5 – (-1), 8 – (- 2), (7 – 1) or 6, 10, 6
which are – 2 times the direction ratios of AB.
∴ AB and BC have the same direction ratios.
∴ AB || BC, but B is a common point of AB and BC.
Hence A,B,C are collinear.

Question 5.
Find the direction cosines of the sides of the triangle whose vertices are (3, 5, – 4), (- 1, 1, 2) and (- 5, – 5,- 2).
Solution.

The vertices of ∆ABC are A(3, 5, – 4), B(- 2, 1, 2) and C(- 5, – 5, -2).
The direction ratios of side AB are (- 1, – 3), (1, – 5) and (2 – (- 4)) i.e„ – 4, – 4 and 6.
Then, \(\sqrt{(-4)^{2}+(-4)^{2}+(6)^{2}}\)
= \(\sqrt{16+16+36}\)
= \(\sqrt{68}=2 \sqrt{17}\)

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 10 Vector Algebra Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 10 Vector Algebra Miscellaneous Exercise

Question 1.
Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of tf-axis.
Solution.
Let OP make 30 ° with x-axis, 60 ° with y-axis and 90 ° with z-axis and it lies in XY-plane.
∴ Direction cosines of \(\overrightarrow{O P}\) are cos 30°, cos 60° and cos 90°.

which is required unit vector in XY-plane.

Question 2.
Find the scalar components and magnitude of the vector joining the points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂).
Solution.
The vector joining the points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) can be obtained by
\(\overrightarrow{P Q}\) = Position vector of Q – Position vector of P
= (x₂ – x₁) î + (y₂ – y₁) ĵ + (z₂ – z₂) k̂

\(|\overrightarrow{P Q}|=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}\)

Hence, the scalar components and the magnitude of the vector joining the given points are respectively {(x₂ – x₁), (y₂ – y₁), (z₂ – z₁)} and \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}\).

Question 3.
A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl’s displacement from her initial point of departure.
Solution.
Let O and B be the initial and final position of the girl respectively.
Then, the girl’s position can be shown as

Question 4.
If \(\overrightarrow{\boldsymbol{a}}=\overrightarrow{\boldsymbol{b}}+\overrightarrow{\boldsymbol{c}}\), then is it true that \(|\overrightarrow{\boldsymbol{a}}|=|\overrightarrow{\boldsymbol{b}}|+|\overrightarrow{\boldsymbol{c}}|\)? Justify your answer.
Solution.

Question 5.
Find the value of x for which x (î + ĵ + k̂) is a unit vector.
Solution.

Question 6.
Find a vector of magnitude 5 units and parallel to the resultant of the vectors \(\vec{a}\) = 2 î + 3 ĵ – k̂ and \(\vec{b}\) = î – 2 ĵ + k̂.
Solution.
Given, \(\vec{a}\) = 2 î + 3 ĵ – k̂ and \(\vec{b}\) = î – 2 ĵ + k̂.
Let \(\vec{c}\) be the resultant of \(\vec{a}\) and \(\vec{b}\)
Then, \(\vec{c}\) = \(\vec{a}\) + \(\vec{b}\)
= (2 + 1) î + (3 – 2) ĵ + (- 1 + 1) k̂
= 3 î + ĵ

Question 7.
If \(\vec{a}\) = î + ĵ + k̂, \(\vec{b}\) = 2î – ĵ + 3k̂ and c = î – 2 ĵ + k̂, find a unit vector parallel to the vector 2\(\vec{b}\) – \(\vec{b}\) + 3\(\vec{c}\).
Solution.

Question 8.
Show that the points A(1, – 2, – 8), B (5, 0, – 2) and C(1, 3, 7) are collinear, and find the ratio in which B divides AC.
Solution.

⇒ 5 (λ + 1)î – 2 (λ + 1)k̂ = (11λ + 1) î + (3λ -2) ĵ + (7λ – 8) k̂
On equating the corresponding components, we get
5(λ + 1) = 11λ + 1
⇒ 5λ + 5 = 11λ +1
⇒ 6λ = 4
⇒ λ = \(\frac{4}{6}=\frac{2}{3}\)
Hence, point B divides AC in the ratio 2 : 3.

Question 9.
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are (\(2 \vec{a}+\vec{b}\)) and (\(\overrightarrow{\boldsymbol{a}}-\mathbf{3} \vec{b}\)) externally in the ratio 1 : 2. Also, show that P is the mid point of the line segment RQ.
Solution.

Question 10.
The two adjacent sides of a parallelogram are 2î – 4ĵ + 5k̂ and î – 2ĵ – 3k̂. Find the unit vector parallel to its diagonal. Also, find its area.
Solution.

= 22î + 11ĵ
∴ Area of parallelogram = \(\sqrt{(22)^{2}+(11)^{2}}\)
= \(11 \sqrt{4+1}=11 \sqrt{5}\) sq. unit

Thus, the unit vector parallel to its diagonal is \(\frac{1}{7}\) (3î – 6ĵ + 21k̂) and area of parallellogram = 11√5 sq. units.

Question 11.
Show that the direction cosines of a vector equally inclined to the axes, OX, OY and OZ are \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\).
V3 V3 V3
Let a vector be equally inclined to OX, OY and OZ and let it makes an angle α with each of these three, then, the direction cosines of the vector are cos α, cos α and cos α
Now, cos² α + cos² α + cos² α = 1
⇒ 3 cos² α = 1
⇒ cos α = ± \(\frac{1}{\sqrt{3}}\)
Hence, the direction cosines of the vector which are equally inclined to the axes are either \(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\) or – \(\frac{1}{\sqrt{3}}\), – \(\frac{1}{\sqrt{3}}\), – \(\frac{1}{\sqrt{3}}\).

Question 12.
Let \(\vec{a}\) = î + 4ĵ + 2k̂, \(\vec{b}\) = 3î – 2ĵ + 7k̂ and \(\vec{c}\) = 2î – ĵ + 4k̂. Find a vector \(\vec{d}\) which is perpendicular to both \(\vec{a}\) and \(\vec{b}\), and \(\overrightarrow{\boldsymbol{c}} \cdot \overrightarrow{\boldsymbol{d}}\) = 15.
Solution.

Question 13.
The scalar product of the vector î + ĵ + k̂ with a unit vector along the sum of vectors 2î + 4ĵ – 5k̂ and λî + 2ĵ + 3k̂ is equal to one. Find the value of λ.
Solution.
(2î + 4ĵ – 5k̂) + (λî + 2ĵ + 3k̂) = (2 + λ)î + 6ĵ – 2k̂

Question 14.
If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are mutually perpendicular vectors of equal magnitudes, show that the vector \(\overrightarrow{\boldsymbol{a}}+\overrightarrow{\boldsymbol{b}}+\overrightarrow{\boldsymbol{c}}\) is equally inclined to \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\).
Solution.

Question 15.
Prove that \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}\), if and only if \(\overrightarrow{\boldsymbol{a}}\), \(\overrightarrow{\boldsymbol{b}}\) are perpendicular, given \(\vec{a} \neq \overrightarrow{\mathbf{0}}\), \(\vec{b} \neq \overrightarrow{\mathbf{0}}\).
Solution.

Direction (16 – 19): Choose the correct answer.

Question 16.
If θ is the angle between two vectors \(\vec{a}\) and \(\vec{b}\), then \(\vec{a}\) . \(\vec{b}\) ≥ 0 only when
(A) 0 < θ < \(\frac{\pi}{2}\)
(B) 0 ≤ θ ≤ \(\frac{\pi}{2}\)
(C) 0 < θ < π
(D) 0 ≤ θ ≤ π
Solution.
Let θ be the angle between two vectors \(\vec{a}\) and \(\vec{b}\).
Then, without loss of generality, \(\vec{a}\) and \(\vec{b}\) are non-zero vector so that |\(\vec{a}\)| and |\(\vec{b}\)| are positive.

 

The correct answer is (B).

Question 17.
Let \(\vec{a}\) and \(\vec{b}\) be two unit vectors and 0 is the angle between them. Then \(\vec{a}\) + \(\vec{b}\) is a unit vector if
(A) θ = \(\frac{\pi}{4}\)

(B) θ = \(\frac{\pi}{3}\)

(C) θ = \(\frac{\pi}{2}\)

(D) θ = \(\frac{2 \pi}{3}\)
Solution.
Let \(\vec{a}\) and \(\vec{b}\) be two unit vectors and θ be the angle between them.
Then, |\(\vec{a}\)| = |\(\vec{b}\)| = 1
Now, \(\vec{a}\) + \(\vec{b}\) is a unit vector if |\(\vec{a}\) + \(\vec{b}\)| = 1.

The correct answer is (D).

Question 18.
The value of î (ĵ × k̂) + ĵ . (î × k̂) + k̂. (î × ĵ) is
(A) 0
(B) – 1
(C) 1
(D) 3
Solution.
We have, î (ĵ × k̂) + ĵ . (î × k̂) + k̂. (î × ĵ) = î . î + ĵ . (- ĵ) + k̂ . k̂
= 1 – ĵ . ĵ + 1
= 1 – 1 + 1 = 1
The correct answer is (C).

Question 19.
If θ is the angle between any two vectors \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\), then |\(\vec{a} \cdot \vec{b}\)| = |\(\overrightarrow{\boldsymbol{a}} \times \overrightarrow{\boldsymbol{b}}\)|, when θ is equal to
(A) 0
(B) \(\frac{\pi}{4}\)
(C) \(\frac{\pi}{2}\)
(D) π
Solution.
Let θ be the angle between two vectors \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\).
Then, without loss of generality, \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\) are non-zero vectors, so that |\(\overrightarrow{\boldsymbol{a}}\)| and |\(\overrightarrow{\boldsymbol{b}}\)| are positive.

The correct answer is (B).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 10 Vector Algebra Ex 10.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.4

Question 1.
Find \(|\overrightarrow{\boldsymbol{a}} \times \overrightarrow{\boldsymbol{b}}|\), if \(\overrightarrow{\boldsymbol{a}}\) = î – 7ĵ + 7k̂ and \(\overrightarrow{\boldsymbol{b}}\) = 3î – 2ĵ + 2k̂.
Solution.

Question 2.
Find a unit vector perpendicular to each of the vector \(\overrightarrow{\boldsymbol{a}}+\overrightarrow{\boldsymbol{b}}\) and \(\overrightarrow{\boldsymbol{a}}-\overrightarrow{\boldsymbol{b}}\), where \(\overrightarrow{\boldsymbol{a}}\) = 3î + 2ĵ + 2k̂ and \(\overrightarrow{\boldsymbol{a}}\) = î + 2ĵ – 2k̂.
Solution.
Given, \(\overrightarrow{\boldsymbol{a}}\) = 3î + 2ĵ + 2k̂ and \(\overrightarrow{\boldsymbol{a}}\) = î + 2ĵ – 2k̂.

Question 3.
If a unit vector \(\overrightarrow{\boldsymbol{a}}\) makes angles \(\frac{\pi}{3}\) with î, \(\frac{\pi}{4}\) with ĵ and an acute angle θ with k, then find θ and hence, the components of \(\overrightarrow{\boldsymbol{a}}\).
Solution.

Question 4.
Show that \((\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})\)
Solution.

Question 5.
Find λ and µ if (2î + 6ĵ + 21k̂) x (î + λĵ + µk̂) = 0.
Solution.

On comparing the corresponding components, we have
6µ – 27λ = 0,
2µ – 27 = 0,
2λ – 6 = 0
Now, 2λ – 6 = 0
⇒ λ = 3
2µ – 27 = 0
⇒ µ = \(\frac{27}{2}\)
Hence, λ = 3 and µ = \(\frac{27}{2}\).

Question 6.
Given that \(\overrightarrow{\boldsymbol{a}} \cdot \vec{b}\) = 0 and \(\overrightarrow{\boldsymbol{a}} \times \overrightarrow{\boldsymbol{b}}\) = 0. What can you conclude about the vectors \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\)?
Solution.

Question 7.
Let the vectors \(\overrightarrow{\boldsymbol{a}},\), \(\overrightarrow{\boldsymbol{b}},\), \(\overrightarrow{\boldsymbol{c}},\) be given as a₁i + a₂j + a₃k, b₁i + b₂j + b₃k, c₁i + c₂j + c₃k. Then show that \(\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}\).
Solution.

Question 8.
If either \(\vec{a}=\overrightarrow{0}\) or \(\overrightarrow{\boldsymbol{b}}=\overrightarrow{\mathbf{0}}\), then \(\vec{a} \times \vec{b}=\overrightarrow{0}\). Is the converse true? Justify your answer with an example.
Solution.

Question 9.
Find the area of the triangle with vertices A ( 1, 1, 2), B (2, 3, 5) and C(1, 5, 5).
Solution.
The vertices of triangle are given asA(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).
The adjacent sides \(\overrightarrow{A B}\) and \(\overrightarrow{B C}\) of ∆ABC are given as

Question 10.
Find the area of the parallelogram whose adjacent sides are determined by the vectors \(\overrightarrow{\boldsymbol{a}}\) = î – ĵ + 3k̂ and \(\overrightarrow{\boldsymbol{b}}\) = 2î – 7ĵ + k̂.
Solution.
The area of the parallelogram whose adjacent sides are \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\) is \(|\vec{a} \times \vec{b}|\).
Adjacent sides are gives as

Question 11.
Let the vectors \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\) be such that \(|\overrightarrow{\boldsymbol{a}}|\) = 3 and \(|\vec{b}|=\frac{\sqrt{2}}{3}\) then \(\overrightarrow{\boldsymbol{a}} \times \overrightarrow{\boldsymbol{b}}\)
is a unit vector, if the angle between \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\) is
(A) \(\frac{\pi}{6}\)

(B) \(\frac{\pi}{4}\)

(C) \(\frac{\pi}{3}\)

(D) \(\frac{\pi}{2}\)
Solution.

Question 12.
Area of a rectangle having vertices A, B, C and D with position vectors – î + \(\frac{1}{2}\) ĵ + 4k̂, î + \(\frac{1}{2}\) ĵ + 4k̂, î – \(\frac{1}{2}\)ĵ + 4k̂ and – î – ĵ + 4k̂ respectively is
(A) \(\frac{1}{2}\)
(B) 1
(C) 2
(D) 4
Solution.
The position vectors of vertices A, B, C and D of rectangle ABCD are given as

Now, it is known that the area of a parallelogram whose adjacent sides are \(\vec{a}\) and \(\vec{b}\) is |\(\vec{a} \times \vec{b}\)|.

Hence, the area of the given rectangle is |\(\overrightarrow{A B} \times \overrightarrow{B C}\)| = 2 sq unit.
The correct answer is (C).