Class 12

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).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 10 Vector Algebra Ex 10.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3

Question 1.
Find the angle between two vectors \(\vec{a}\) and \(\vec{b}\) with magnitudes √3 and 2, respectively, having \(\vec{a}\) . \(\vec{b}\) = √6.
Solution.
It is given that, |\(\vec{a}\)| = √3, |\(\vec{b}\)| = 2 and
\(\vec{a}\) . \(\vec{b}\) = √6

∴ √6 = √3 × 2 × cos θ

⇒ cos θ = \(\frac{\sqrt{6}}{\sqrt{3} \times 2}\)

⇒ cos θ = \(\frac{1}{\sqrt{2}}\)

⇒ θ = \(\frac{\pi}{4}\)
Hence, the angle between the given vectors \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{4}\).

Question 2.
Find the angle between the vectors î – 2 ĵ + 3k̂ and 3 î – 2 ĵ + k̂.
Solution.
The given vectors are \(\vec{a}\) = î – 2 ĵ + 3k̂ and \(\vec{b}\) = 3 î – 2 ĵ + k̂

Question 3.
Find the projection of the vector î – ĵ on the vector î + ĵ.
Solution.
Let \(\vec{a}\) = î – ĵ and \(\vec{b}\) = î + ĵ
Now, projection of vector a on S is given by
\(\frac{1}{|\vec{b}|}(\vec{a} \cdot \vec{b})=\frac{1}{\sqrt{1+1}}\{1.1+(-1)(1)\}=\frac{1}{\sqrt{2}}(1-1)=0\)
Hence, the projection of vector \(\vec{a}\) and \(\vec{b}\) is 0.

Question 4.
Find the projection of the vector î + 3ĵ + k̂ on the vector 7î – ĵ + 8k̂.
Solution.
Let \(\vec{a}\) = i + 3j + 7k and \(\vec{b}\) = 7i – j + 8k
Now, projection of vector a and b is given by
\(\frac{1}{|\vec{b}|}(\vec{a} \cdot \vec{b})=\frac{1}{\sqrt{7^{2}+(-1)^{2}+8^{2}}}\{1(7)+3(-1)+7(8)\}\)

= \(\frac{7-3+56}{\sqrt{49+1+64}}=\frac{60}{\sqrt{114}}\)

Question 5.
Show that each of the given three vectors is a unit vector \(\frac{1}{7}\) (2î + 3ĵ + 8k̂), \(\frac{1}{7}\) (3î- 6ĵ + 2k̂), \(\frac{1}{7}\) (6î + 2k̂ – 3k̂)
Also, show that they are mutually perpendicular to each other.
Solution.

Question 6.
Find |\(|\overrightarrow{\boldsymbol{a}}|\)| and |\(|\overrightarrow{\boldsymbol{b}}|\)|, if \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})\) = 8 and \(|\vec{a}|=8|\vec{b}|\).
Solution.

Question 7.
Evaluate the product \((3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})\).
Solution.

Question 8.
Find the magnitude of two vectors \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\) having the same magnitude and such that the angle between them is 60° and their scalar product is \(\frac{1}{2}\).
Solution.
Let θ be the angle between the vectors \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\)

It is given that |\(\overrightarrow{\boldsymbol{b}}\)| = |\(\overrightarrow{\boldsymbol{b}}\)|, \(\overrightarrow{\boldsymbol{a}}\) . \(\overrightarrow{\boldsymbol{b}}\) = \(\frac{1}{2}\) and θ = 60°

We know that \(\overrightarrow{\boldsymbol{a}}\) . \(\overrightarrow{\boldsymbol{b}}\) = |\(\overrightarrow{\boldsymbol{a}}\) . \(\overrightarrow{\boldsymbol{b}}\)| cos θ

Question 9.
Find |\(\vec{x}\)|, if for a unit vector \(\vec{a}\), (\(\vec{x}\) – \(\vec{a}\)) . (\(\vec{x}\) + \(\vec{a}\)) = 12.
Solution.

Question 10.
If \(\vec{a}\) = 2î + 2 ĵ + 3k̂, \(\vec{b}\) = – î + 2 ĵ + k̂ and \(\vec{c}\) = 3î + ĵ are such that \(\vec{a}\) + \(\vec{b}\) is perpendicular to \(\vec{c}\), then find the value of λ.
Solution.

⇒ (2 – λ)3 + (2 + 2λ)1 + (3 + λ)0 = 0
⇒ 6 – 3λ + 2 + 2λ = 0
⇒ – λ + 8 = 0
⇒ λ = 8
Hence, the required value of λ is 8.

Question 11.
Show that |\(\vec{a}\)| \(\vec{b}\) + |\(\vec{b}\)| \(\vec{a}\) is perpendicular to |\(\vec{a}\)| \(\vec{b}\) – |\(\vec{b}\)| \(\vec{a}\), for any two non-zero vectors a and b.
Solution.

Question 12.
If \(\vec{a}\) . \(\vec{a}\) = 0 and \(\vec{a}\) . \(\vec{b}\) = 0, then what can be concluded about the vector 6?
Solution.

Question 13.
If \(\vec{a}\), \(\vec{a}\), \(\vec{c}\) are unit vectors such that \(\vec{a}\) + \(\vec{a}\) + \(\vec{c}\) = 0, find the value of \(\vec{a}\) . \(\vec{b}\) + \(\vec{b}\) . \(\vec{c}\) + \(\vec{c}\) . \(\vec{a}\).
Solution.

Question 14.
If either vector \(\vec{a}=\overrightarrow{0}\) or \(\overrightarrow{\boldsymbol{b}}=\overrightarrow{\mathbf{0}}\), then \(\overrightarrow{\boldsymbol{a}} \cdot \overrightarrow{\boldsymbol{b}}\) = 0. But the converse need not be true. Justify your answer with an example.
Solution.

Question 15.
If the vertices A, B, C of a triangle ABC are (1, 2, 3), (- 1, 0, 0), (0, 1, 2), respectively, then find ∠ABC [∠ABC is the angle
between the vector \(\overrightarrow{\boldsymbol{B A}}\) and \(\overrightarrow{\boldsymbol{B C}}\)].
Solution.

Question 16.
Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, – 1) are collinear.
Solution.

Question 17.
Show that the vectors 2î – ĵ + k̂, î – 3ĵ – 5k̂ and 3î – 4ĵ – 4k̂ form the vertices of a right angled triangle.
Solution.

Question 18.
If \(\overrightarrow{\boldsymbol{a}}\) is a non-zero vector of magnitude ‘a’ and λ, is a non-zero scalar, then λ \(\overrightarrow{\boldsymbol{a}}\) is unit vector if
(A) λ = 1
(B) λ = – 1
(C) a = |λ|
(D) a = \(\frac{1}{|\lambda|}\)
Solution.

Hence, vector λ \(\overrightarrow{\boldsymbol{a}}\) is a unit vector if a = \(\frac{1}{|\lambda|}\).
The correct answer is (D).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 10 Vector Algebra Ex 10.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.2

Question 1.
Compute the magnitude of the following vectors
\(\overrightarrow{\boldsymbol{a}}=\hat{\boldsymbol{i}}+\hat{\boldsymbol{j}}+\hat{\boldsymbol{k}}\); \(\overrightarrow{\boldsymbol{b}}=\mathbf{2} \hat{\boldsymbol{i}}-\mathbf{7} \hat{j}-\mathbf{3} \hat{\boldsymbol{k}}\); \(\vec{c}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}\)
Solution.
The given vectors are

Question 2.
Write two different vectors having same magnitude.
Solution.
Consider the vectors \(\vec{a}=\hat{i}+2 \hat{j}+\hat{k}\) and \(\vec{b}=\hat{i}+\hat{j}+2 \hat{k}\)
\(|\vec{a}|=\sqrt{1^{2}+2^{2}+1^{2}}=\sqrt{6}\),
\([|\vec{b}|=\sqrt{1^{2}+1^{2}+2^{2}}=\sqrt{6}/latex]
Thus, vectors [latex]\vec{a}\) and \(\vec{a}\) have the same magnitude.

Question 3.
Write two different vectors having same direction.
Solution.

The direction cosines of \(\vec{p}\) and \(\vec{q}\) are the same.
Hence, the two vectors have the same direction.

Question 4.
Find the values of x and y so that the vectors \(2 \hat{i}+\mathbf{3} \hat{j}\) and \(x \hat{i}+y \hat{j}\) are equal.
Solution.
The two vectors \(2 \hat{i}+\mathbf{3} \hat{j}\) and \(x \hat{i}+y \hat{j}\) will be equal if their corresponding components are equal.
Hence, the required values of x and y are 2 and 3 respectively.

Question 5.
Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (- 5, 7).
Solution.
The vector with the initial point P(2,1) and terminal point Q(- 5, 7) can be given by,
\(\overline{P Q}=(-5-2) \hat{i}+(7-1) \hat{j}\)

⇒ \(\overline{P Q}=7 \hat{i}+6 \hat{j}\)
Hence, the required scalar components are -7 and 6 while the vector components are – 7î and 6ĵ.

Question 6.
Find the sum of the vectors a = î – 2ĵ + k̂, b = – 2î + 4ĵ + 5k̂ and c = î – 6ĵ – 7k̂.
Solution.
The given vectors are a = î – 2ĵ + k̂, b = – 2î + 4ĵ + 5k̂ and c = î – 6ĵ – 7k̂.
∴ \(\vec{a}+\vec{b}+\vec{c}\) = (1 – 2 + 1) î + (- 2 + 4 – 6) ĵ + (1 + 5 – 7) k̂.
0 . î – 4 ĵ – 1 . k̂ = – 4 ĵ – k̂ .

Question 7.
Find the unit vector in the direction of the vector \(\vec{a}\) = î + ĵ + 2k̂.
Solution.
The unit vector â in the direction of vector \(\vec{a}\) = î + ĵ + 2k̂ is given by

Question 8.
Find the unit vector in the direction of vector \(\overrightarrow{P Q}\), where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.
Solution.
The given points are P (1, 2, 3) and Q (4, 5 6).

Question 9.
For given vectors, \(\vec{a}\) = 2i – j + 2 k and \(\vec{b}\) = – i + j k, find the unit vector in the direction of the vector \(\vec{a}\) + \(\vec{b}\).
Solution.
The given vectors are \(\vec{a}\) = 2 î – ĵ + 2 k̂ and \(\vec{b}\) = – î + ĵ – k̂
\(\vec{a}\) = 2 î – ĵ + 2 k̂
\(\vec{b}\) = – î + ĵ – k̂
.-. \(\vec{a}\) + \(\vec{a}\) = (2 – 1) î +(- 1 + 1) ĵ + (2 – 1) k̂
= 1 î + 0 ĵ + 1 k̂
|\(\vec{a}\) + \(\vec{b}\)| = \(\sqrt{1^{2}+1^{2}}=\sqrt{2}\)
Hence, the unit vector in the direction of (\(\vec{a}\) + \(\vec{b}\)) is \(\frac{(\vec{a}+\vec{b})}{|\vec{a}-\vec{b}|}=\frac{\hat{i}+\hat{k}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}\).

Question 10.
Find a vector in the direction of vector 5î – ĵ + 2k̂ which has magnitude 8 units.
Solution.
Let \(\vec{a}\) = – 5î – ĵ + 2k̂
∴ |\(\vec{a}\)| = \(\sqrt{5^{2}+(-1)^{2}+2^{2}}=\sqrt{25+1+4}=\sqrt{30}\)

⇒ \(\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{5 \hat{i}-\hat{j}+2 \hat{k}}{\sqrt{30}}\)

Hence, the vector in the direction of vector 5î – ĵ + 2k̂ which has magnitude 8 units is given by,

Question 11.
Show that the vectors 2î – 3ĵ + 4k̂ and – 4î + 6ĵ – 8k̂ are collinear.
Solution.
Let \(\vec{a}\) = 2î – 3ĵ + 4k̂ and \(\vec{b}\) = – 4î + 6ĵ – 8k̂.

It is observed that \(\vec{b}\) = – 4î + 6ĵ – 8k̂
= -2 (2î – 3ĵ + 4k̂) = – 2 \(\vec{a}\)
\(\vec{b}\) = λ \(\vec{a}\), where λ = – 2.
Hence, the given vectors are collinear.

Question 12.
Find the direction cosines of the vector î + 2 ĵ + 3 k̂.
Solution.
Let |\(\vec{a}\)| = î + 2 ĵ + 3 k̂
|\(\vec{a}\)| = \(\sqrt{1^{2}+2^{2}+3^{2}}=\sqrt{1+4+9}=\sqrt{14}\)
Hence the direction cosines of \(\vec{a}\) are \(\left(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right)\).

Question 13.
Find the direction cosines of the vector, joining the points A(1, 2,- 3) and B(- 1, – 2, 1) directed from A to B.
Solution.
The given points are A(1, 2, – 3) and B(- 1, – 2, 1).

Question 14.
Show that the vector î + ĵ + k̂ is equally inclined to the axes OX, OY and OZ.
Solution.
Let \(\vec{a}\) = î + ĵ + k̂
Then, |\(\vec{a}\)| = \(\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}\)

Therefore, the direction cosines of \(\vec{a}\) are \(\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\).
Now, let α, β and γ be the angles formed by \(\vec{a}\) with the positive directions of x, y and z axes.

Then, we have cos α = \(\frac{1}{\sqrt{3}}\), cos β = \(\frac{1}{\sqrt{3}}\), cos γ = \(\frac{1}{\sqrt{3}}\).

Hence, the given vector is equally inclined to axes OX, OY and OZ.

Question 15.
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are î + 2ĵ – k̂ and – î + ĵ + k̂ respectively, in the ratio 2 : 1
(i) internally
(ii) externally
Solution.
The position vector of point R dividing the line segment joining two points P and Q in the ratio m : n is given by
Case I:
Internally \(\frac{m \vec{b}+n \vec{a}}{m+n}\)

Case II:
Externally \(\frac{m \vec{b}-n \vec{a}}{m-n}\)
Position vectors of P and Q are given as \(\overrightarrow{O P}\) = î + 2 ĵ – k̂ and \(\overrightarrow{O Q}\) = – î + ĵ + k̂

(i) The position vector of point R which divides the line joining two points P and Q internally in the ratio 2 : 1 is given by
\(\overrightarrow{O R}\) = \(\frac{2(-\hat{i}+\hat{j}+\hat{k})+1(\hat{i}+2 \hat{j}-\hat{k})}{2+1}=\frac{(-2 \hat{i}+2 \hat{j}+2 \hat{k})+(\hat{i}+\hat{j}-\hat{k})}{3}\)

= \(\frac{-\hat{i}+4 \hat{j}+\hat{k}}{3}=-\frac{1}{3} \hat{i}+\frac{4}{3} \hat{j}+\frac{1}{3} \hat{k}\)

The position vector of point R which divides the line joining two points P and Q externally in the ratio 2 : 1 is given by
\(\overrightarrow{O R}\) = \(\frac{2(-\hat{i}+\hat{j}+\hat{k})-1(\hat{i}+2 \hat{j}-\hat{k})}{2-1}\)
= (- 2î + 2ĵ + 2k̂) – (î + 2 ĵ – k̂)
= – 3 î + 3 k̂.

Question 16.
Find the position vector of the mid point of the vector joining the points P( 2, 3, 4) and Q( 4, 1, – 2).
Solution.
The position vector of mid-point R of the vector joining points P(2, 3, 4) and Q(4, 1, – 2) given by
\(\overrightarrow{O R}\) = \(\frac{(2 \hat{i}+3 \hat{j}+4 \hat{k})+(4 \hat{i}+\hat{j}-2 \hat{k})}{2}\)

= \(\frac{(2+4) \hat{i}+(3+1) \hat{j}+(4-2) \hat{k}}{2}\)

= \(\frac{6 \hat{i}+4 \hat{j}+2 \hat{k}}{2}\)

= 3î + 3ĵ + k̂

Question 17.
Show that the points A, B and C with position vectors, a = 3î – 4ĵ – 4k̂, b = 2î – j + k̂ and c = î – 3ĵ – 5k̂, respectively form the vertices of a right angled triangle.
Solution.
Position vectors of points A,B and C are respectively given as

Hence, ABC is a right angled triangle.

Question 18.
In triangle ABC which of the following is not true

(A) \(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=\overrightarrow{0}\)
(B) \(\overrightarrow{\boldsymbol{A B}}+\overrightarrow{\boldsymbol{B C}}-\overrightarrow{\boldsymbol{A C}}=\overrightarrow{\mathbf{0}}\)
(C) \(\overrightarrow{\boldsymbol{A B}}+\overrightarrow{B C}-\overrightarrow{C A}=\overrightarrow{0}\)
(D) \(\overrightarrow{\mathbf{A B}}-\overrightarrow{C B}+\overrightarrow{C A}=\overrightarrow{0}\)
Solution.
On applying the triangle law of addition in the given triangle, we have
\(\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C}\) ………….(i)

⇒ \(\overrightarrow{A B}+\overrightarrow{B C}=-\overrightarrow{C A}\)

⇒ \(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=\overrightarrow{0}\) …………(ii)

∴ The equation given in alternative A is true.
\(\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C}\)

⇒ \(\overrightarrow{A B}+\overrightarrow{B C}-\overrightarrow{A C}=\overrightarrow{0}\)
∴ The equation given in alternative B is true.
From equation (ii), we have
\(\overrightarrow{A B}-\overrightarrow{C B}+\overrightarrow{C A}=\overrightarrow{0}\)
∴ The equation given in alternative D is true.
Now, consider the equation given in alternative C.
\(\overrightarrow{A B}+\overrightarrow{B C}-\overrightarrow{C A}=\overrightarrow{0}\)

⇒ \(\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{C A}\)………………..(iii)
From equations (i) and (iii), we have AC – CA
⇒ \(\overrightarrow{A C}=\overrightarrow{C A}\)

⇒ \(\overrightarrow{A C}=-\overrightarrow{A C}\)

⇒ \(\overrightarrow{A C}+\overrightarrow{A C}=\overrightarrow{0}\)

⇒ \(2 \overrightarrow{A C}=\overrightarrow{0}\)

⇒ \(\overrightarrow{A C}=\overrightarrow{0}\) which is not true.
Hence, the equation given in alternative C is incorrect.
The correct answer is (C).

Question 19.
If a and B are two collinear vectors, then which of the following are incorrect
(A) \(\overrightarrow{\boldsymbol{b}}\) = X \(\overrightarrow{\boldsymbol{a}}\), for some scalar X
(B) \(\overrightarrow{\boldsymbol{a}}\) = ± \(\overrightarrow{\boldsymbol{b}}\)
(C) The respective components of a and B are proportional
(D) Both the vectors \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\) have same direction, but different magnitudes.
Solution.

Thus, the respective components of \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\) are proportional.
However, vector \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\) can have different directions.
Hence, the statement given in D is incorrect.
The correct answer is (D).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter Differential Equations Ex 9.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6

Direction (1 – 12): For each of the given differential equation, find the general solution.

Question 1.
\(\frac{d y}{d x}\) + 2y = sin x
Solution.
The given differential equation is \(\frac{d y}{d x}\) + 2y = sin x
This is in the form of \(\frac{d y}{d x}\) + Py = Q (where P = 2 and Q = sin x).
Now, I.F = e^{∫ P dx} = e^{∫ 2 dx} = e^2x.
The solution of the given differential equation is given by the relation,
y (I.F.) = ∫ (Q × I.F.) dx + C
⇒ ye^2x = ∫ sin x. e^2x dx + C …………….(i)
Let I = ∫ sin x. e^2x dx
⇒ I = sin x . ∫ e^2x dx – ∫ (\(\frac{d}{d x}\) (sin x) . ∫ e^2x dx) dx

Question 2.
\(\frac{d y}{d x}\) + 3y = e^{– 2x}
Solution.
The given differential equation is
\(\frac{d y}{d x}\) + 3y= e^{– 2x}
This is in the form of \(\frac{d y}{d x}\) + Py = Q (where P = 3 and Q = e^{– 2x})
Now, I.F.= e^{∫ P dx} = e^{∫ 3 dx} = e^3x
The solution of the given differential equation is given by the relation,
y (I.F.) = ∫ (Q × I.F.) dx + C
⇒ y e^3x = ∫ (e^{– 2x} × e^3x) + C
⇒ y e^3x = ∫e^x dx + C
⇒ y e^3x = e^x + C
⇒ ye^3x = e^{– 2x} + Ce^{– 3x}
This is the required general solution of the given differential equation.

Question 3.
\(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x²
Solution.
The given differential equation is
\(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x²
This is in the form of \(\frac{d y}{d x}\) + Py = Q (where P = \(\frac{1}{x}\) and Q = x²)
Now, I.F. = e^{∫ P dx}
= e^{∫ \(\frac{1}{x}\) dx}
= e^{log x} = x
The solution of the given differential equation is given by the relation
y (I.F.) = ∫ (Q × I.F.) dx + C
y(x) = ∫ (x² . x) dx + C
xy = ∫ x³ dx + C
⇒ xy = \(\frac{x^{4}}{4}\) + C
This is the required general solution of the given differential equation.

Question 4.
\(\frac{d y}{d x}\) + (sec x) y = tan x 0 ≤ x < 2
Solution.
The given differential equation is
\(\frac{d y}{d x}\) + (sec x) y = tan x
This is in the form of \(\frac{d y}{d x}\) + Py = Q (where P = sec x and Q = tan x)
Now, I.F = e^{∫ P dx}
= e^{∫ sec x}
= e^{log (sec x + tan x)}
= sec x + tan x
The general solution of the given differential equation is given by the relation
y (I.F.) = ∫ (Q × I.F.) dx + C
⇒ y (sec x + tan x) = ∫ tan x (sec x + tan x) dx + C
⇒ y (sec x + tan x) = ∫ sec x tan x dx + ∫ tan² x dx + C
⇒ y (sec x + tan x) = sec x + ∫ (sec² x – 1) dx + C
⇒ y (sec x + tan x) = sec x + tan x – x + C

Question 5.
cos² x \(\frac{d v}{d x}\) + y = tan x (0 ≤ x < \(\frac{\pi}{2}\))
Solution.
Given, cos² x \(\frac{d y}{d x}\) + y = tan x
On dividing by cos² x on both sides, we get
\(\frac{d y}{d x}+\frac{1}{\cos ^{2} x} y=\frac{\tan x}{\cos ^{2} x}\)
The equation is of the form \(\frac{d y}{d x}\) + Py = Q
(where P = \(\frac{1}{\cos ^{2} x}\) = sec² x, Q = \(\frac{\tan x}{\cos ^{2} x}\))
∴ I.F. = e^{∫ P dx}
= e^{∫ sec2 x}
= e^{tan x}
The general solution of the given differential equation is given by
y (I.F.) = ∫ (Q × I.F.) dx + C
i.e., y × e^{tan x} = ∫ \(\frac{\tan x}{\cos ^{2} x}\) e^{tan x} dx + C
= ∫ tan x . e sec² x + c
Put tan x = t, sec² x dx = dt
∴ ye^{tan x} = ∫ t e^t dt + C
Integrating by parts taking t as a first function.
= ye^{tan x} = te^t – ∫ 1 . e^t dt + C
= te^t – e^t + C
= tan x e^{tan x} – e^{tan x} + C (Putting t = tan x)
or y = tan x – 1 + Ce^{– tan x}

Question 6.
x \(\frac{d y}{d x}\) + 2y = x² log x
Solution.
The given differential equation is x \(\frac{d y}{d x}\) + 2y = x² log x
\(\frac{d y}{d x}\) + \(\frac{2}{dx}\) y = x log x
This equation is in the form of a linear differential equation as
\(\frac{d y}{d x}\) + Py = Q (where P = \(\frac{2}{x}\) and Q = x log X)
Now, I.F. = e^{∫ P dx}
= e^{∫ \(\frac{2}{x}\) dx}
= e^{2 log x}
= e^{log x2}
= x²
The general solution of the given differential equation is given by the relation,
y (I.F.) = ∫ (Q × I.F.) dx + C
y . x² = ∫ (x log x . x²) dx + C
x² y = ∫ (x³ log x) dx + C

Question 7.
x log x \(\frac{d y}{d x}\) + y = log x
Solution.
The given differential equation is x log x \(\frac{d y}{d x}\) + y = \(\frac{2}{x}\) log x

⇒ \(\frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x^{2}}\)
This equation is the form of a linear differential equation as

\(\frac{d y}{d x}\) + Py = Q, (where P = \(\frac{1}{x \log x}\) and Q = \(\frac{2}{x^{2}}\))

Now, I.F. = e^{∫ P dx}
= e^{∫ \(\frac{1}{x \log x}\) dx}
= e^{log (log x)} = log x

The general solution of the given differential equation is given by the relation,
y(I.F.) = ∫ (Q × I.F.) dx + C

Substiting the value of ∫ (\(\frac{2}{x^{2}}\) log x) in equation (i), we get
y log x = – \(\frac{2}{x}\) (1 + log x) + C
This is the required general solution of the given differential equation.

Question 8.
(1 + x²) dy + 2xy dx = cot x dx (x ≠ 0)
Solution.
Given, (1 + x²) dy + 2xy dx = cot x dx

⇒ \(\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{\cot x}{1+x^{2}}\)

This equation is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q (where P = \(\frac{2 x}{1+x^{2}}\) and Q = \(\frac{\cot x}{1+x^{2}}\))

Now, I.F. = e^{∫ P dx}
= e^{\(\frac{2 x}{1+x^{2}}\) dx}
= e^{log (1 + x2)}
= 1 + x²
The general solution of the given differential equation is given by the relation,
y × I.F. = ∫ (Q × l.F.) dx
⇒ y (1 + x²) = ∫ [\(\frac{\cot x}{1+x^{2}}\) (1 + x²)] dx + C
⇒ y (1 + x²) = ∫ cot x dx + C
⇒ y (1 + x²) = log |sin x| + C

Question 9.
x \(\frac{d y}{d x}\) + y – x + xy cot x = 0,(x ≠ 0)
Solution.
Given, x \(\frac{d y}{d x}\) + y – x + xy cot x = 0
⇒ x \(\frac{d y}{d x}\) + y (1 + x cot x) = x
⇒ \(\frac{d y}{d x}\) + (\(\frac{1}{x}\) + cot x) y = 1
This is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q
(where P = \(\frac{1}{x}\) + cot x and Q = 1)
Now, I.F. = e^{∫ P dx}
= e^{(\(\frac{1}{x}\) + cot x) dx}
= e^{log x + log (sin x)}
= e^{log (x sin x)}
= x sin x
The general solution of the given differential equation is given by the relation,
y × I.F. = ∫ (Q × I.F.) dx
⇒ y (x sin x) = ∫ (1 × x sin x) dx + C
⇒ y (x sin x) = ∫ (x sin x) dx + C
⇒ y (x sin x) = x ∫ sin x dx – ∫ [\(\frac{d}{d x}\) (x) . ∫ sin x dx] + C
⇒ y (x sin x) = x(- cos x) – ∫ 1 .(- cos x) dx + C
⇒ y (x sin x) = – x cos x + sin x + C

⇒ y = \(\frac{-x \cos x}{x \sin x}+\frac{\sin x}{x \sin x}+\frac{C}{x \sin x}\)

y = – cot x + \(\frac{1}{x}+\frac{C}{x \sin x}\)

Question 10.
(x + y) \(\frac{d y}{d x}\) = 1
Solution.
Given, (x + y) \(\frac{d y}{d x}\) = 1
⇒ \(\frac{d y}{d x}=\frac{1}{x+y}\)

⇒ \(\frac{d y}{d x}\) = x + y

⇒ \(\frac{d y}{d x}\) – x = y
This is a linear differential equation of the form
\(\frac{d x}{d y}\) + P₁x = Q₁ (where P₁ = – 1 and Q₁ = y)
Now, I.F. = e^{∫ P dx}
= e^{∫ – dy}
= e^{– y}
The general solution of the given differentia1 equation is given by the relation,
x (I.F.) = ∫ (Q × I.F.) dy + C
⇒ xe^{– y} = ∫ (y . e^{– y}) dy + C
⇒ xe^{– y} = y ∫ e^{– y} dy – ∫ [\(\frac{d}{d y}\) (y) ∫ e^{– y} dy] dy + C
⇒ xe^{– y} = y(- e^{– y}) – ∫ (- e^{– y}) dy + C
⇒ xe^{– y} = – ye^{– y} + ∫ e^{– y} dy + C
⇒ xe^{– y} = – ye^{– y} – e^{– y} + C
⇒ x = – y – 1 + C e^y
⇒ x + y + 1 = Ce^y

Question 11.
y dx + (x – y²) dy = 0
Solution.
Given y dx + (x – y²) dy = 0
⇒ y dx = (y² – x) dy
⇒ \(\frac{d x}{d y}\) = \(\frac{y^{2}-x}{y}=y-\frac{x}{y}\)

⇒ \(\frac{d x}{d y}+\frac{x}{y}\) = y
This is a linear differential equation of the form
\(\frac{d x}{d y}\) + P₁x = Q₁ (where P₁ = \(\frac{1}{y}\) and Q₁ = y)
Now, I.F. = e^{∫ P₁ x dx}
= e^{∫ \(\frac{1}{y}\) dy}
= e^{log y} = y
The general solution of the given differential equation is given by the relation,
⇒ x × I.F.= ∫ (Q₁ × I.F.) dy + C
⇒ xy = ∫ (y.y) dy + C
⇒ xy = ∫ y² dx + C
⇒ xy = \(\frac{y^{3}}{3}\) + C
⇒ x = \(\frac{y^{2}}{3}+\frac{C}{y}\).

Question 12.
(x + 3y²) \(\frac{d y}{d x}\) = y (y > 0)
Solution.
Given, (x + 3y²) \(\frac{d y}{d x}\) = y

⇒ \(\frac{d y}{d x}=\frac{y}{x+3 y^{2}}\)

⇒ \(\frac{d x}{d y}=\frac{x+3 y^{2}}{y}=\frac{x}{y}+3 y\)

⇒ \(\frac{d x}{d y}-\frac{x}{y}\) = 3y
This is a linear differential equation of the form
\(\frac{d x}{d y}\) + P₁ x = Q₁
(where P₁ = – \(\frac{1}{y}\) and Q₁ = 3y)
Now, I.F. = e^{∫ P₁ dx}
= e^{– ∫ \(\frac{d y}{y}\)}
= e^{– log y}
= e^{log (\(\frac{1}{y}\))}
= \(\frac{1}{y}\)
The general solution of the given differential equation is given by the relation.
x × I.F.= ∫ (Q₁ × I.F.) dy + C
⇒ x × \(\frac{1}{y}\) = ∫ (3y × \(\frac{1}{y}\)) dy + C
⇒ \(\frac{x}{y}\) = 3y + C
⇒ x = 3y² + Cy.

Direction (13 – 15): For each of the differential equation, find a particular solution satisfying the given condition.

Question 13.
\(\frac{d y}{d x}\) + 2y tan x = sin x; y = 0 when x = \(\frac{\pi}{3}\)
Solution.
The given differential equation is \(\frac{d y}{d x}\) + 2y tan x = sin x
This is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q (where P = 2 tan x and Q = sin x)
Now, I.F. = e^{∫ P dx}
= e^{∫ 2 tan x dx}
= e^{2 log |sec x|}
= e^{log (sec2 x)}
= sec² x
The general solution of the given differential equation is given by the relation,
x (I.F.) = ∫ (Q × I.F.) dx + C
⇒ y (sec² x) = ∫ (sin x . sec² x) dx + C
⇒ y sec² x = ∫ (sec x . tan x) dx + C
⇒ y sec² x = sec x + C ……………(i)
Now, y = 0 at x = \(\frac{\pi}{3}\)
Therefore,
0 × sec² \(\frac{\pi}{3}\) = sec\(\frac{\pi}{3}\) + C
⇒ 0 = 2 + C
⇒ C = – 2
Substituting C = – 2 in equation (i), we get
y sec² x = sec x – 2
y = cos x – 2 cos² x
Hence, the required solution of the given differential equation is y = cos x – 2 cos² x.

Question 14.
(1 + x²) \(\frac{d y}{d x}\) + 2xy = \(\frac{1}{1+x^{2}}\); y = 0 when x = 1
Solution.
Given, (1 + x²) \(\frac{d y}{d x}\) + 2xy = \(\frac{1}{1+x^{2}}\)

⇒ \(\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}\)
This is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q (where P = \(\frac{2 x}{1+x^{2}}\) and Q = \(\frac{1}{\left(1+x^{2}\right)^{2}}\))
Now, I.F. = e^{∫ P dx}
= e^{\(\int \frac{2 x d x}{1+x^{2}}\)}
= e^{log (1 + x2)} = 1 + x²
The general solution of the given differential equation is given by the relation,
x (I.F.) = ∫ (Q × I.F.)dx + C
⇒ y(1 + x²) = \(\left[\frac{1}{\left(1+x^{2}\right)^{2}} \cdot\left(1+x^{2}\right)\right]\) dx + C
⇒ y (1 + x²) = ∫ \(\frac{1}{1+x^{2}}\) dx + C
y (1 + x²) = tan^-1 x + C ……………..(i)
Now, y = 0 at x = 1
Therefore, 0 = tan^-1 1 + C
⇒ C = – \(\frac{\pi}{4}\)
Substituting C = – \(\frac{\pi}{4}\) in equation (i), we get
y (1 + x²) = tan^-1 x – \(\frac{\pi}{4}\)
This is the required general solution of the given differential equation.

Question 15.
\(\frac{d y}{d x}\) – 3y cot x = sin 2x; y = 2 when x = \(\frac{\pi}{2}\).
Solution.
The given differential equation is \(\frac{d y}{d x}\) – 3y cot x = sin 2x
This is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q (where P = – 3 cot x and Q = sin 2x)
Now, I.F. = e^{∫ P dx}
= e^{– 3 ∫ cot x dx}
= e^{– 3 log |sin x|}
= e^{log \(\left|\frac{1}{\sin ^{3} x}\right|\)}

= \(\frac{1}{\sin ^{3} x}\)

The general solution of the given differential equation is given by the relation,
y(I.F.) = ∫ (Q × I.F.) dx + C
⇒ \(y \cdot \frac{1}{\sin ^{3} x}=\int\left[\sin 2 x \cdot \frac{1}{\sin ^{3} x}\right] d x+C\)

⇒ y cosec³ x = 2 ∫ (cot x cosec x) dx + C
⇒ y cosec³ x = 2 cosec x + c
⇒ y = \(-\frac{2}{\operatorname{cosec}^{2} x}+\frac{3}{\operatorname{cosec}^{3} x}\)
y = – 2 sin² x + C sin³ x ………..(i)
Now, y = 2 at x = \(\frac{\pi}{2}\)
Therefore, we get
2 = – 2 + C
⇒ C = 4
Substituting C = 4 in equation (i), we get
y = – 2 sin² x + 4 sin³ x
y = 4 sin³ x – 2 sin² x
This is the required particular solution of the given differential equation.

Question 16.
Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.
Solution.
Let F(x, y) be the curve passing through the origin.
At point (x, y), the slope of the curve will be \(\frac{d y}{d x}\).
According to the given information,
\(\frac{d y}{d x}\) = x + y
⇒ \(\frac{d y}{d x}\) – y = x
This is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q (where P = – 1 and Q = x)
Now, I.F. = e^{∫ P dx}
= e^{∫ (- 1)} dx
= e^{– x}
The general solution of the given differential equation is given by the relation,
y(I.F.) = ∫ (Q × I.F.) dx + C
ye^{– x} = ∫ xe^{– x} dx + C ……………(i)
Now, ∫ xe^{– x} dx = x ∫ e^{– x} dx – ∫ \(\left[\frac{d}{d x}(x) \cdot \int e^{-x} d x\right]\) dx
= – xe^{– x} – ∫ – e^{– x} dx
= – xe^{– x} + (- e^{– x})
= – e^{– x} (x + 1)
Substituting in equation (i), we get
ye^{– x} = – e^{– x} (x + 1) + C
⇒ y = – (x + 1) + Ce^x
⇒ x + y + 1 = Ce^x ………….(ii)
The curve passes through the origin.
Therefore, equation (ii) becomes
1 = C
Substituting C = 1 in equation (ii), we get
⇒ x + y + 1 = e^x
Hence, the required equation of curve passing through the origin is x + y+ 1 = e^x.

Question 17.
Find the equation of a curve passing through the point (0, 2) given that the sum of the coordiDntes of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.
Solution.
According to question, we have x + y = \(\frac{d y}{d x}\) + 5
or \(\frac{d y}{d x}\) + (- 1) y = x – 5
It is a linear differential equation of the form \(\frac{d y}{d x}\) + Py = Q
∴ P = – 1 andQ = x – 5 and
I.F. = e^{∫ P dx}
l.F. = e^{– x}
The general equation of the curve is given by
y. I.F = ∫ Q × I.F dx + C
⇒ y . e^{– x} = ∫ (x – 5) e^{– x} dx + C
⇒ y . e^{– x} = (x – 5) ∫ e^{– x} dx – ∫ [\(\frac{d}{d x}\) (x – 5) . ∫ e^{– x} dx] dx + C
⇒ y . e^{– x} = (x – 5) (- e^{– x}) – ∫ (- e^{– x}) dx + C
⇒ y . e^{– x} = (5 – x) e^{– x} – e^{– x} + C ……………(ii)
The curve passes through the point (0, 2), therefore
2 e^{– 0} = (5 – 0) e^{– 0} – e⁰ + C
⇒ 2 = 5 – 1 + C
⇒ C = 2 – 4 = – 2
On putting the value of C in equation (ii), we get
e^{– x} y = (5 – x) e^{– x} – e^{– x} – 2
⇒ y = 4 – x – 2e^{– x}
which is the required equation of the curve in reference.

Question 18.
The integrating factor of the differential equation x \(\frac{d y}{d x}\) – y = 2x²
(A) e^{– x}
(B) e^{– y}
(C) \(\frac{1}{x}\)
(D) x
Solution.
The given differential equation is x \(\frac{d y}{d x}\) – y = 2x²
⇒ \(\frac{d y}{d x}\) – \(\frac{y}{x}\) = 2x
This is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q
(where P = – \(\frac{1}{x}\) and Q = 2x)
The integrating factor (1.F.) is given by the relation e^{∫ P dx}
∴ I.F. = e^{∫ \(\frac{1}{x}\) dx}
= e^{∫ – log x}
= e^{log (x-1)}
= x^-1
= \(\frac{1}{x}\)
Hence, the correct answer is (C).

Question 19.
The integrating factor of the differential equation (1 – y²) \(\frac{d x}{d y}\) + yx = ay (- 1 < < 1) is
(A) \(\frac{1}{y^{2}-1}\)

(B) \(\frac{1}{\sqrt{y^{2}-1}}\)

(C) \(\frac{1}{1-y^{2}}\)

(D) \(\frac{1}{\sqrt{1-y^{2}}}\)
Solution.
The given differential equation is (1 – y²) \(\frac{d x}{d y}\) + yx = ay

⇒ \(\frac{d x}{d y}+\frac{y x}{1-y^{2}}=\frac{a y}{1-y^{2}}\)

This is a linear differential equation of the form
\(\frac{d x}{d y}\) + P₁ x = Q₁
(where P₁ = \(\frac{y}{1-y^{2}}\) and Q₁ = \(\frac{a y}{1-y^{2}}\))
The integrating factor (I.Fj is given by the relation e^{∫ P₁ dx}
∴ I.F.= e^{∫ P₁ dy}

= \(e^{\int \frac{y}{1-y^{2}} d y}=e^{-\frac{1}{2} \log \left(1-y^{2}\right)}=e^{\log \left[\frac{1}{\sqrt{1-y^{2}}}\right]}=\frac{1}{\sqrt{1-y^{2}}}\)
Hence, the correct answer is (D).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 10 Vector Algebra Ex 10.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.1

Question 1.
Represent graphically a displacement of 40 km, 30° east of north.
Solution.
The displacement is 30° east of north so, we have to draw a straight line making 30° with north.

Here, vector \(\overrightarrow{O P}\) represents the displacement of 40 km, 30° east of north.

Question 2.
Classify the following measures as scalars and vectors.
(i) 10 kg
(ii) 2 metres north-west
(iii) 40°
(iv) 40 watt
(v) 10^{– 19} coulomb
(vi) 20 m/sec²
Solution.
(i) 10 kg is a scalar quantity because it involves only magnitude.
(ii) 2 metres north-west is a vector quantity as it involves both magnitude and direction.
(iii) 40° is a scalar quantity as it involves only magnitude.
(iv) 40 watts is a scalar quantity as it involves only magnitude.
(v) 10^{– 19} coulomb is a scalar quantity as it involves only magnitude.
(vi) 20 m/sec² is a vector quantity as it involves magnitude as well as direction.

Question 3.
Classify the following as scalar and vector quantities.
(i) time period
(ii) distance
(iii) force
(iv) velocity
(v) work done
Solution.
(i) Time period is a scalar quantity as it involves only magnitude.
(ii) Distance is a scalar quantity as it involves only magnitude.
(iii) Force is a vector quantity as it involves both magnitude and direction.
(iv) Velocity is a vector quantity as it involves both magnitude as well as direction.
(v) Work done is a scalar quantity as it involves only magnitude.

Question 4.
In Figure, (a square) identify the following vectors.
(i) Coinitial
(ii) Equal
(iii) Collinear but not equal
Solution.

(i) Vectors \(\vec{a}\) and \(\vec{d}\) are coinitial because they have the same initial point.
(ii) Vectors \(\vec{b}\) and \(\vec{d}\) are equal because they have the same magnitude and direction.
(iii) Vectors \(\vec{a}\) and \(\vec{c}\) are collinear but not equal. This is because although they are parallel, their directions are not the same.

Question 5.
Answer the following as true or false.
(i) \(\vec{a}\) and – \(\vec{a}\) are collinear.
Solution.
True

(ii) Two collinear vectors are always equal in magnitude.
Solution.
False

(iii) Two vectors having same magnitude are collinear.
Solution.
False

(iv) Two collinear vectors having the same magnitude are equal.
Solution.
False.

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 9 Differential Equations Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 9 Differential Equations Miscellaneous Exercise

Question 1.
For each of the differential equations given below, indicate its order and degree (if defined).
(i) \(\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}\) – 6y = log x

(ii) \(\left(\frac{d y}{d x}\right)^{3}-4\left(\frac{d y}{d x}\right)^{2}\) + 7y = sin x

(iii) \(\frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)\) = 0
Solution.
(i) The given differential equation is \(\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}\) – 6y = log x

\(\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}\) – 6y – log x = 0
The highest order derivative present in the differential equation is \(\frac{d^{2} y}{d x^{2}}\).
Thus, its order is 2.
The highest power raised to \(\frac{d^{2} y}{d x^{2}}\) is one.
Hence, its degree is 1.

(ii) The given differential equation is \(\) + 7y = sin x

\(\left(\frac{d y}{d x}\right)^{3}-4\left(\frac{d y}{d x}\right)^{2}\) + 7y – sin x = 0
The highest order derivative present in the differential equation is \(\frac{d y}{d x}\).
Thus, its order is 1.
The highest power raised to \(\frac{d y}{d x}\) is three.
Hence, its degree is 3.

(iii) The given differential equation is \(\frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)\) = o
The highest order denvative present in the differential equation is \(\frac{d^{4} y}{d x^{4}}\).
Thus, its order is 4.
However, the given differential equation is not a polynomial equation.
Hence, its degree is not defined.

Question 2.
For each of the questions given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
(i) y = ae^x + e^-x + x² : \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}\) – xy + x² – 2 = 0

(ii) y = e^x (a cos x + sin x) : \(\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0\)

(iii) y = x sin 3x : \(x \frac{d^{2} y}{d x^{2}}\) + 9y – 6 cos 3x = 0

(iv) x² = 2y² log y : (x² + y²) \(\frac{d y}{d x}\) – xy = 0
Solution.
(i) Given, y = ae^x + be^{– x} + x²
Differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}=a \frac{d}{d x}\left(e^{x}\right)+b \frac{d}{d x}\left(e^{-x}\right)+\frac{d}{d x}\left(x^{2}\right)\)

\(\frac{d y}{d x}\) = ae^x – be^{– x} + x²
Again, differentiating both sides w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = ae^x + be^{– x} + x²
Now, on substituting the values of \(\frac{d y}{d x}\) and \(\frac{d^{2} y}{d x^{2}}\) in the differential equation, we get
L.H.S. = x\(\frac{d^{2} y}{d x^{2}}\) + 2 \(\frac{d y}{d x}\) – xy + x² – 2
= x (ae^x + be^{– x} + 2) + 2 (ae^x – be^{– x} + 2x) – x (ae^x + be^{– x} + x²) + x² – 2

= (axe^x + bxe^{– x} + 2x) + (2ae^x – 2be^{– x} + 4x) – (axe^x + bxe^-x + x³) + x² – 2

= 2ae^x – 2be^-x + x² + 6x – 2 ≠ 0
= L.H.S. ≠ R.H.S.
Hence, the given function is not a solution of the corresponding differential equation.

(ii) Given, y = e^x (a cos x + b sin x)
= a e^x cos x + b e^x sin x
Differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = a . \(\frac{d}{d x}\) (e^x cos x) + b . \(\frac{d}{d x}\) (e^x sin x)

⇒ \(\frac{d y}{d x}\) = a (e^x cos x – e^x sin x) + b . (e^x sin x + e^x cos x)

⇒ \(\frac{d y}{d x}\) = (a + b) e^x cos x + (b – a) e^x sin x
Again, differentiating both sides w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = (a + b) . \(\frac{d}{d x}\) . (e^x cos x) + (b – a) \(\frac{d}{d x}\) (e^x sin x)

\(\frac{d^{2} y}{d x^{2}}\) = (a + b) . [e^x cos x – e^x sin x + (b – a) e^x sin x + e^x cos x]

\(\frac{d^{2} y}{d x^{2}}\) = e^x [(a + b) (cos x – sin x) + (b – a) (sin x + cos x)]

\(\frac{d^{2} y}{d x^{2}}\) = e^x [a cos x – a sin x + b cos x – b sin x + b sin x + b cos x – a sin x – a cos x]

\(\frac{d^{2} y}{d x^{2}}\) = [2e^x (b cos x – a sin x)] d2 dy
Now, on substituting the values of \(\frac{d^{2} y}{d x^{2}}\) and \(\frac{d y}{d x}\) in the L.H.S. of the given differential equation, we get

\(\frac{d^{2} y}{d x^{2}}\) + 2 \(\frac{d y}{d x}\) + 2y = 2 e^x (b cos x – a sin x) – 2 e^x [(a + b) cos x + (b – a) sin x] + 2 e^x (a cos x + b sin x)
= e^x [(2b c0s x – 2a sin x) – (2a cos x + 2b cos x) – 2b sin x – 2a sin x) + (2a cos x + 2b sin x)]
= e^x [(2b – 2a – 2h + 2a) cos x] + e^x [(- 2a – 2b + 2a + 2b) sin x]
=0
Hence, the given function is a solution of the corresponding differential equation.

(iii) Given, y = x sin 3x
Differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (x sin 3x)
= sin 3x + x . cos 3x . 3
= \(\frac{d y}{d x}\) = sin 3x + 3x cos 3x
Again differentiating both sides w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) (sin 3x) + 3 \(\frac{d}{d x}\) (x cos 3x)

\(\frac{d^{2} y}{d x^{2}}\) = 3 cos 3x + 3 [cos 3x + x (- sin3x). 3]

\(\frac{d^{2} y}{d x^{2}}\) = 6 cos 3x – 9 x sin 3x

Substituting the value of \(\frac{d^{2} y}{d x^{2}}\) in the L.H.S. of the given differential equation, we get

\(\frac{d^{2} y}{d x^{2}}\) + 9y – 6 cos 3x = (6 cos 3x – 9 x sin 3x) + 9x sin 3x – 6 cos 3x = 0
Hence, the given function is a solution of the corresponding differential equation.

(iv) Given, x² = 2y² log y
Differentiating both sides w.r.t. x, we get
2x = 2 . \(\frac{d}{d x}\)
= [y² log y]

⇒ x = [2y . log y . \(\frac{d y}{d x}\) + y² . \(\frac{1}{y}\) . \(\frac{d y}{d x}\)]

x = \(\frac{d y}{d x}\) (2y log y + y)

⇒ \(\frac{d y}{d x}\) = \(\frac{x}{y(1+2 \log y)}\)
Substituting the value of \(\frac{d y}{d x}\) in the L.H.S. of the given differential equation, we get

(x² + y²) \(\frac{d y}{d x}\) – xy = (2y² log y + y²) . \(\frac{x}{y(1+2 \log y)}\) – xy

= y² (1 + 2 log y) . \(\frac{x}{y(1+2 \log y)}\) – xy

= xy – xy = 0

Hence, the given function is a solution of the corresponding differential equation.

Question 3.
Form the differential equation representing the family of curves given by (x – a)² + 2y² = a², where a is an arbitrary constant.
Solution.
Given family of curves (x – a)² + 2y² = a²
⇒ x² + a² – 2ax + 2y² = a²
⇒ 2y² = 2ax – x²
Differentiating with respect w.r.t. x, we get
2y \(\frac{d y}{d x}\) = \(\frac{2 a-2 x}{2}\)

Question 4.
Prove that x² – y² = c (x² + y²)² is the general solution of differential equation (x³ – 3xy²)dx = (y² – 3x²y) dy, where c is a parameter.
Solution.
The differential equation is (x³ – 3xy²)dx = (y³ – 3x²y) dy
∴ \(\frac{d y}{d x}=\frac{x^{3}-3 x y^{2}}{y^{3}-3 x^{2} y}\) ……………(i)
This is a homogeneous equation.
Let y = vx
Differentiating w.r.t. x, we get

x² – y² = c (x² + y²)²
Hence, x² – y² = c (x² + y²)² is the solution of the differential equation.
(x³ – 3xy²)dx = (y³ – 3x²y) dy where c is a parameter.

Question 5.
Form the differential equation of the family of circles in the first quadrant which touch the coordinates axes.
Solution.
The equation of a circle in the first quadrant with centre (a, a) and radius
(a) which touches the coordinate axis is
(x – a)² + (y – a)² = a²

Differentiating eq (i) w.r.t x, we get
2 (x – a) + 2 (y – a) \(\frac{d y}{d x}\) = 0
⇒ (x – a) + (y – a) y’ = 0
⇒ x – a + yy’ – ay’ = 0
⇒ x + yy’ – a (1 – y’) = 0
⇒ a = \(\frac{x+y y^{\prime}}{1+y^{\prime}}\)
Substituting the value of a in equation (i), we get
\(\left[x-\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right)\right]^{2}+\left[y-\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right)\right]^{2}+\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right)^{2}\)

⇒ \(\left[\frac{(x-y) y^{\prime}}{\left(1+y^{\prime}\right)}\right]^{2}+\left[\frac{y-x}{1+y^{\prime}}\right]^{2}+\left[\frac{x+y y^{\prime}}{1+y^{\prime}}\right]^{2}\)

⇒ (x – y)² . y’² + (x – y) = (x +yy’)

⇒ (x – y)² [1 + (y’)²] = (x + yy’)²
Hence, the required differential equation of the family of circles is (x – y)² [1 + (y’)²] = (x + yy’)^2.

Question 6.
Find the general solution of the differential equation \(\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}\) = 0
Solution.
Given, differential equation is \(\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}\) = 0
⇒ \(\frac{d y}{d x}=-\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}\)

⇒ \(\frac{d y}{\sqrt{1-y^{2}}}=\frac{d x}{\sqrt{1-x^{2}}}\)
Integrating both sides, we get
sin^-1 y = – sin^-1 x + C
⇒ sin^-1 x + sin^-1 y = C.

Question 7.
Show that the general solution of the differential equation \(\frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}\) is given by (x + y + 1) = A (1 – x – y – 2xy) where A is parameter.
Solution.

Hence, the given result is proved.

Question 8.
Find the equation of the curve passing through the point (o, \(\frac{\pi}{4}\)) whose differential equation is sin x cos y dx + cos x sin y dy = 0.
Solution.
The differential equation of the given curve is sin x cos y dx + cos x sin y dy = 0
⇒ \(\frac{\sin x \cos y d x+\cos x \sin y d y}{\cos x \cos y}=0\)

⇒ tan x dx + tan y dy = 0
Integrating both sides, we get
log (sec x) + log (sec y) = log C
log (sec x sec y) = log C
⇒ sec x . sec y = C
The curve passes through point (0, \(\frac{\pi}{4}\))
∴ 1 × √2 = C
⇒ C = √2
On substituting the value of C in equation (i), we get
sec x . sec y = √2
sec x . \(\frac{1}{cos y}\) = √2

⇒ cos y = \(\frac{\sec x}{\sqrt{2}}\)

Hence, the required equation of the curve is cos y = \(\frac{\sec x}{\sqrt{2}}\).

Question 9.
Find the particular solution of the differential equation (1 + e^2x) dy + (1 + y²) e^x dx = 0, given that y = 1 when x = 0.
Solution.
The differential equation is
(1 + e^2x) dy + (1 + y²) e^x dx = 0
On separating the variables, we get

\(\frac{d y}{1+y^{2}}+\frac{e^{x} d x}{1+e^{2 x}}\) = 0

Integrating both sides, we get
\(\int \frac{d y}{1+y^{2}}+\int \frac{e^{x}}{1+e^{2 x}} d x\) = C

Put e^x = t,
e^x dx = dt
∴ tan^-1 y + \(\int \frac{d t}{1+t^{2}}\) = C
tan^-1 y + tan^-1 t = C
i.e., tan^-1 y + tan^-1 e^x = C
Put y = 1, x = 0
∴ tan^-1 1 + tan^-1 1 = C or
2 tan^-1 1 = C
2 × \(\frac{\pi}{4}\) = C
∴ C = \(\frac{\pi}{2}\)
The particular solution of the given differential equation is
tan^-1 y + tan^-1 e^x = \(\frac{\pi}{2}\)

Question 10.
Solve the differential equation \(y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y\) (y ≠ 0)
Solution.
Given differential equation is \(y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y\)

From equation (î) and equation (ii), we get
\(\frac{d z}{d y}\) = 1
⇒ dz = dy
Integrating both sides, we get
z = y + C
⇒ e^{\(\frac{x}{y}\)} = y + C.

Question 11.
Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = – 1, when x = 0.
(Hint: put x – y = t).
Solution.
Given differential equation is (x – y) (dx + dy) = dx – dy

Integrating both sides, we get
t + |log t| = 2x + C
⇒ (x – y) + log(x – y) = 2x + C
log |x – y| = x + y + C …………….(iii)
Now, y = – 1 at x = 0.
Therefore, equation (iii) becomes
log 1 = 0 – 1 + C
⇒ C = 1
Substituting C = 1 in equation (iii) we get
log |x – y| = x + y + 1
This is the required particular solution of the given differential equation.

Question 12.
Solve the differential equation \(\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{d y}\) = 1 (x ≠ 0)
Solution.

Question 13.
Find a particular solution of the differential equation \(\frac{d y}{d x}\) + y cot x = 4x cosec x (x ≠ 0),
given that y = 0 when x = \(\frac{\pi}{2}\).
Solution.
The given differential equation is \(\frac{d y}{d x}\) + y cot x = 4x cosec x
This equation is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q, (where P = cot x and Q = 4x cosec x)
Now, I.F. = e^{∫ P dx}
= e^{∫ cot x dx}
= e^{log |sin x|}
= sin x
The general solution of the given differential equation is given by
y (I.F.) = ∫ (Q × I.F.) dx + C
⇒ y sin x = ∫ (4x cosec x . sin x)dx + C
⇒ y sin x = 4 ∫ x dx + C
⇒ y sin x = 4\(\frac{x^{1+1}}{1+1}\) + C
⇒ y sin x = 4 . \(\frac{x^{2}}{2}\) + C
⇒ y sin x = 2x² + C …………..(i)
Now, y = 0 at x = \(\frac{\pi}{2}\)
Therefore, equation (i) becomes
0 = 2 × \(\frac{\pi^{2}}{4}\) + C

= C = \(\frac{\pi^{2}}{2}\)
Substituting C = – \(\frac{\pi^{2}}{2}\) in equation (i), we get
y sin x = 2x – \(\frac{\pi^{2}}{2}\).
This is the required particular solution of the given differential equation.

Question 14.
Find a particular solution of the differential equation (x + 1) \(\frac{d y}{d x}\) = 2e^{– y} – 1, given that y = 0 when x = 0.
Solution.
Given differential equation is (x + 1) \(\frac{d y}{d x}\) = 2e^{– y} – 1

Question 15.
The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?
Solution.
Let the population at any instant (t) be y.
It is given that the rate of increase of population is proportional to the number of inhabitants at any instant.
∴ \(\frac{d y}{d t}\) ∝ y
⇒ \(\frac{d y}{d t}\) = ky (k is a constant)
⇒ \(\frac{d y}{y}\) = k dt
Integrating both sides, we get
log y = kt + C …………..(i)
In the year 1999, t = 0 and y = 20000.
Therefore, we get
log 20000 C …………..(ii)
In the year 2004, t = 5 and y = 25000.
Therefore, we get
log 25000 = k.5 + C
⇒ log 25000 = 5k + log 20000
⇒ 5k = \(\log \left(\frac{25000}{20000}\right)=\log \left(\frac{5}{4}\right)\)
⇒ k = \(\frac{1}{5} \log \left(\frac{5}{4}\right)\) …………(i)
In the year 2009, t = 10 years.
Now, on substituting the values oft, k, and C in equation (i), we get
log y = 10 × \(\frac{1}{5} \log \left(\frac{5}{4}\right)\) + log (20000)
⇒ log y = log \(\left[20000 \times\left(\frac{5}{4}\right)^{2}\right]\)
⇒ y = 20000 × \(\frac{5}{4}\) × \(\frac{5}{4}\)
⇒ y = 31250
Hence, the population of the village in 2009 will be 31250.

Question 16.
The general solution of the differential equation \(\frac{y d x-x d y}{y}\) = 0
(A) xy = C
(B) x = Cy²
(C) y = Cx
(D) y = Cx²
Sol.
The given differential equation is \(\frac{y d x-x d y}{y}\) = 0
⇒ \(\frac{y d x-x d y}{y}\) = 0

⇒ \(\frac{1}{x} d x-\frac{1}{y} d x\) = 0
Integrating both sides, we get
log |x| – log |y| = log k

log |\(\frac{x}{y}\)| = log k
⇒ \(\frac{x}{y}\) = k
⇒ y = \(\frac{1}{k}\) x
⇒ y = Cx, where C = \(\frac{1}{k}\)
Hence, the correct answer is (C).

Question 17.
The general solution of a differential equation of the type \(\frac{d y}{d x}\) + P₁x = Q₁ is
(A) \(y e^{\int P_{1} d y}=\int\left(Q_{1} e^{\int P_{1} d y}\right) d y\) + C

(B) \(y \cdot e^{j P_{1} d x}=\int\left(Q_{1} e^{\int P_{1} d x}\right) d x\) + C

(C) \(x e^{\int P_{1} d y}=\int\left(Q_{1} e^{\int P_{1} d y}\right) d y\) + C

(D) \(x e^{\int P_{1} d x}=\int\left(Q_{1} e^{\int P_{1} d x}\right) d y\) + C
Solution.
The integrating factor of the given differential equation \(\frac{d y}{d x}\) + P₁x = Q₁ is e^{∫ P dx}
The general solution of the differential equation is given by
x (I.F.) = ∫ (Q × I.F.) dy + C
x . e^{∫ P dx} = ∫ (Q₁ e^{∫ P dx} ) dy + C
Hence, the correct answer is (C).

Question 18.
The general solution of the differential equation e^x dy + (ye^x +2x) dx = 0 is
(A) xe^y + x² = C
(B) xe^y + y² = C
(C) ye^x + x² = C
(D) ye^y + x² = C
Solution.
The given differential equation is e^x dy + (ye^x +2x) dx = 0
⇒ e^x \(\frac{d y}{d x}\) + y e^x + 2x = 0
⇒ \(\frac{d y}{d x}\) + y = – 2x e^{– x}
This is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q, (where P = 1 and Q = – 2xe^{– x})

Now, I.F.= e^{∫ P dx}
= e^{∫ 1 dx}
= e^x
The general solution of the given differential equation is given by
y(I.F.) = ∫ (Q × I.F.)dx + C
⇒ ye^x = ∫ (- 2xe^{– x} . e^x) dx + C
⇒ ye^x = – ∫ 2x dx + C
⇒ ye^x = – x² + C
⇒ ye^x + x² = C
Hence, the correct answer is (C).