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 a_{1}i + a_{2}j + a_{3}k, b_{1}i + b_{2}j + b_{3}k, c_{1}i + c_{2}j + c_{3}k. 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).