Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 10 Vector Algebra Ex 10.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 10 Vector Algebra Ex 10.4

Question 1.

Find |\(\vec{a}\) Ã— \(\vec{b}\)|, if \(\vec{a}\) = \(\hat {i} \) – 7\(\hat {j} \) + 7\(\hat {k} \)

and \(\vec{b}\) = 3\(\hat {i} \) – 2\(\hat {j} \) + 2\(\hat {k} \).

= (- 14 + 14)\(\hat {i} \) – (2 – 21)\(\hat {j} \) + (- 2 + 21)\(\hat {k} \)

= – (- 19)\(\hat {j} \) + 19\(\hat {k} \) = 19\(\hat {j} \) + 19\(\hat {k} \)

|\(\vec{a}\) Ã— \(\vec{b}\)| = \(\sqrt{(19)^{2}+(19)^{2}}\) = 19\(\sqrt{1+1}\) = 19\(\sqrt{2}\).

Question 2.

Find a unit vector perpendicular to each of the vectors \(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – \(\vec{b}\), where

\(\vec{a}\) = 3\(\hat {i} \) + 2\(\hat {j} \) + 2\(\hat {k} \) and \(\vec{b}\) = \(\hat {i} \) + 2\(\hat {j} \) – 2\(\hat {k} \).

Solution:

We have:

\(\vec{a}\) = 3\(\hat {i} \) + 2\(\hat {j} \) + 2\(\hat {k} \) and \(\vec{b}\) = \(\hat {i} \) + 2\(\hat {j} \) – 2\(\hat {k} \).

âˆ´ \(\vec{a}\) + \(\vec{b}\) = (3 + 1)\(\hat {i} \) – (2 + 2)\(\hat {j} \) + (2 – 2)\(\hat {k} \)

= 4\(\hat {i} \) + 4\(\hat {j} \)

âˆ´ \(\vec{a}\) – \(\vec{b}\) = (3 – 1)\(\hat {i} \) + (2 – 2)\(\hat {j} \) + (2 + 2)\(\hat {k} \)

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

Now, (\(\vec{a}\) + \(\vec{b}\)) Ã— (\(\vec{a}\) – \(\vec{b}\))

= (16 – 0)\(\hat {i} \) – (16 – 0)\(\hat {j} \) + (0 – 8)\(\hat {k} \)

= 16\(\hat {i} \) – 16\(\hat {j} \) – 8\(\hat {k} \).

âˆ´ A unit vector perpendicular to both the vectors (\(\vec{a}\) + \(\vec{b}\)) and (\(\vec{a}\) – \(\vec{b}\)) is given by

Question 3.

If a unit vector \(\vec{a}\) makes angle \(\frac{Ï€}{3}\) with \(\hat {i} \), \(\frac{Ï€}{4}\) with \(\hat {j} \)

and an acute angle Î¸ with \(\hat {k} \), then find Î¸ and hence the components of \(\vec{a}\).

Solution:

Let \(\vec{a}\) = a_{1}\(\hat {i} \) + a_{2}\(\hat {j} \) + a_{3}\(\hat {k} \)

such that |a_{1}\(\hat {i} \) + a_{2}\(\hat {j} \) + a_{3}\(\hat {k} \)| = 1.

As per question:

(a_{1}\(\hat {i} \) + a_{2}\(\hat {j} \) + a_{3}\(\hat {j} \)).\(\hat {i} \) = |\(\vec{a}\)| |\(\hat {i} \)|cos \(\frac{Ï€}{3}\)

â‡’ (a_{1}) (1) = (1)(1) Ã— \(\frac{1}{2}\)

â‡’ a_{1} = \(\frac{1}{2}\)

Also (a_{1}\(\hat {i} \) + a_{2}\(\hat {j} \) + a_{3}\(\hat {k} \)). \(\hat {j} \)

= |\(\vec{a}\)| |\(\hat {i} \)| cos \(\frac{Ï€}{3}\)

â‡’ (a_{1})(1) = (1)(1) Ã— \(\frac{1}{\sqrt{2}}\) â‡’ a_{2} = \(\frac{1}{\sqrt{2}}\).

As |\(\vec{a}\)| = 1, therefore \(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\) = 1

Question 4.

Show that (\(\vec{a}\) – \(\vec{b}\)) Ã— (\(\vec{a}\) + \(\vec{b}\)) = 2(\(\vec{a}\) Ã— \(\vec{b}\)).

Solution:

(\(\vec{a}\) – \(\vec{b}\)) Ã— (\(\vec{a}\) + \(\vec{b}\))

= \(\vec{a}\) Ã— \(\vec{a}\) + \(\vec{a}\) Ã— \(\vec{b}\) – \(\vec{b}\) Ã— \(\vec{a}\) – \(\vec{b}\) Ã— \(\vec{b}\)

= \(\vec{a}\) Ã— \(\vec{b}\) + \(\vec{a}\) Ã— \(\vec{b}\) = 2(\(\vec{a}\) Ã— \(\vec{b}\)).

[âˆµ \(\vec{a}\) Ã— \(\vec{a}\) = \(\vec{b}\) Ã— \(\vec{b}\) = \(\vec{0}\)

and – \(\vec{b}\) Ã— \(\vec{a}\) = \(\vec{b}\) Ã— \(\vec{b}\)]

Question 5.

Find Î» and Âµ, if (2\(\hat {i} \) + 6\(\hat {j} \) + 27\(\hat {k} \)) Ã— (\(\hat {i} \) + Î»\(\hat {j} \) + Âµ\(\hat {k} \)) = \(\vec{0}\)

Solution:

(2\(\hat {i} \) + 6\(\hat {j} \) + 27\(\hat {k} \)) Ã— (\(\hat {i} \) + Î»\(\hat {j} \) + Âµ\(\hat {k} \)) = \(\vec{0}\)

â‡’ (6Âµ – 27Î»)\(\hat {i} \) – (2Âµ – 27)\(\hat {j} \) + (2Î» – 6)\(\hat {k} \) = \(\vec{0}\)

Comparing, we have:

6Âµ – 27Î» = 0 ………………….. (1)

2Âµ – 17 = 0 â‡’ Âµ = \(\frac{27}{2}\) …………………. (2)

and 2Î» – 6 = 0 â‡’ Î» = \(\frac{6}{2}\) = 3 …………………. (3)

Putting for Î», Âµ in (1), We see that it is satisfied as

6(\(\frac{27}{2}\)) – 27(3) = 81 – 81 = 0

Hence, Î» = 3 and Âµ = \(\frac{27}{2}\).

Question 6.

Given that \(\vec{a}\).\(\vec{b}\) = 0 and \(\vec{a}\) Ã— \(\vec{b}\). What can you conclude about the vectors a and b?

Solution:

Given \(\vec{a}\).\(\vec{b}\) = 0 and \(\vec{a}\) Ã— \(\vec{b}\) = \(\vec{0}\).

â‡’ (\(\vec{a}\) = 0 or \(\vec{b}\) = \(\vec{0}\) or \(\vec{a}\) âŠ¥\(\vec{b}\)) and (\(\vec{a}\) = \(\vec{0}\) or \(\vec{b}\) = \(\vec{0}\) or \(\vec{a}\) || \(\vec{b}\))

â‡’ Either \(\vec{a}\) = 0 or \(\vec{b}\) = \(\vec{0}\)

â‡’ [âˆµ \(\vec{a}\) âŠ¥ \(\vec{b}\) and \(\vec{a}\) || \(\vec{b}\) can never hold simultaneously.

Question 7.

Let the vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are given as a_{1}\(\hat {i} \) + a_{2}\(\hat {j} \) + a_{3}\(\hat {k} \),

b_{1}\(\hat {i} \) + b_{2}\(\hat {j} \) + b_{3}\(\hat {k} \) and c_{1}\(\hat {i} \) + c_{2}\(\hat {j} \) + c_{3}\(\hat {k} \).

Thus, show that \(\vec{a}\) Ã— (\(\vec{b}\) + \(\vec{c}\)) = \(\vec{a}\) Ã— \(\vec{b}\) + \(\vec{a}\) Ã— \(\vec{c}\).

Solution:

\(\vec{b}\) + \(\vec{c}\) = (b_{1}\(\hat {i} \) + b_{2}\(\hat {j} \) + b_{3}\(\hat {k} \)) + (c_{1}\(\hat {i} \) + c_{2}\(\hat {j} \) + c_{3}\(\hat {k} \))

= (b_{1} + c_{1})\(\hat {i} \) + (b_{2} + c_{2})\(\hat {j} \) + (b_{3} + c_{3})\(\hat {k} \)

From (1) and (2), we conclude that

\(\vec{a}\) Ã— (\(\vec{b}\) + \(\vec{c}\)) = \(\vec{a}\) Ã— \(\vec{b}\) + \(\vec{a}\) Ã— \(\vec{c}\).

Question 8.

If either \(\vec{a}\) = \(\vec{0}\) or \(\vec{b}\) = \(\vec{0}\), then \(\vec{a}\) Ã— \(\vec{b}\) = \(\vec{0}\). Is the converse true?

Justify your answer with an example.

Solution:

We have:

\(\vec{a}\) = \(\vec{0}\) â‡’ |\(\vec{a}\)| = 0

âˆ´ \(\vec{a}\) Ã— \(\vec{b}\) = |\(\vec{a}\)| |\(\vec{b}\)| sin Î¸ \(\hat {n} \),

where Î¸ is the angle between \(\vec{a}\) and \(\vec{b}\).

= 0 Ã— |\(\vec{b}\)| sin Î¸ = 0

Similarly, when \(\vec{b}\) = \(\vec{0}\) â‡’ |\(\vec{b}\)| = 0, \(\vec{a}\) Ã— \(\vec{b}\) = 0

Converse : Let \(\vec{a}\) = a_{1}\(\hat {i} \) + a_{2}\(\hat {j} \) + a_{3}\(\hat {k} \)

and \(\vec{b}\) = pa_{1}\(\hat {i} \) + pa_{2}\(\hat {j} \) + pa_{3}\(\hat {k} \).

i.e. \(\vec{a}\) and \(\vec{b}\) are parallel, So Î¸ = 0.

and |\(\vec{a}\)| â‰ 0, |\(\vec{b}\)| â‰ 0.

But \(\vec{a}\) Ã— \(\vec{b}\) = |\(\vec{a}\)| |\(\vec{b}\)| sin Î¸ \(\hat {n} \) = 0

Thus, \(\vec{a}\) Ã— \(\vec{b}\) = 0, even when |\(\vec{a}\)| â‰ 0, |\(\vec{b}\)| â‰ 0.

Question 9.

Find the area of the traiangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 3).

Solution:

The position vectors of vertices of âˆ† ABC are (1, 1, 2), (2, 3, 5) and (1, 5, 5).

âˆ´ \(\overrightarrow{\mathrm{AB}}\) = \(\overrightarrow{\mathrm{OB}}\) – \(\overrightarrow{\mathrm{OA}}\)

= (2\(\hat {i} \) + 3\(\hat {j} \) + 5\(\hat {k} \)) – (\(\hat {i} \) + \(\hat {j} \) + 2\(\hat {k} \))

= \(\hat {i} \) + 2\(\hat {j} \) + 3\(\hat {k} \).

\(\overrightarrow{\mathrm{AC}}\) = \(\overrightarrow{\mathrm{OC}}\) – \(\overrightarrow{\mathrm{OA}}\) = (\(\hat {i} \) + 5\(\hat {j} \) + 5\(\hat {k} \)) – (\(\hat {i} \) + \(\hat {j} \) + 2\(\hat {k} \))

= 0\(\hat {i} \) + 4\(\hat {j} \) + 3\(\hat {k} \) = 4\(\hat {j} \) + 3\(\hat {k} \).

Question 10.

Find the area of parallelogram whose adjacent sides are determined by the vectors \(\vec{a}\) = \(\hat {i} \) – \(\hat {j} \) + 3\(\hat {k} \)

and \(\vec{b}\) = 2\(\hat {i} \) – 7\(\hat {j} \) + \(\hat {k} \).

Solution:

Area of the parallelogram whose sides are vectors \(\vec{a}\) and \(\vec{b}\)

Question 11.

Let the vectors \(\vec{a}\) and \(\vec{b}\) are such that |\(\vec{a}\)| = 3 and |\(\vec{b}\)| = \(\frac{\sqrt{2}}{3}\), then \(\vec{a}\) Ã— \(\vec{b}\) is a unit vector, if the angle between \(\vec{a}\) and \(\vec{b}\) is

(A) \(\frac{Ï€}{6}\)

(B) \(\frac{Ï€}{4}\)

(C) \(\frac{Ï€}{3}\)

(D) \(\frac{Ï€}{2}\)

Solution:

Also, |\(\vec{a}\) Ã— \(\vec{b}\)| is a unit vector.

âˆ´ Part (B) is the correct answer.

Question 12.

Area of rectangle having vertices A, B, C and D with position vectors

(- \(\hat {i} \) + \(\frac{1}{2}\)\(\hat {j} \) + 4\(\hat {k} \)), (\(\hat {i} \) + \(\frac{1}{2}\)\(\hat {j} \) + 4\(\hat {k} \)), (\(\hat {i} \) – \(\frac{1}{2}\) \(\hat {j} \) + 4\(\hat {k} \))

and (- \(\hat {i} \) – \(\frac{1}{2}\)\(\hat {j} \) + 4\(\hat {k} \)) is

Solution:

The position vectors of A and B are – \(\hat {i} \) + \(\frac{1}{2}\)\(\hat {j} \) + 4\(\hat {k} \)

and \(\hat {i} \) + \(\frac{1}{2}\)\(\hat {j} \) + 4\(\hat {k} \).

âˆ´ \(\overrightarrow{\mathrm{AB}}\) = [1 – (- 1)]\(\hat {i} \) + 0.\(\hat {j} \) + 0.\(\hat {k} \)

âˆ´ |\(\overrightarrow{\mathrm{AB}}\)| = 2

The postion vectors of A and D are – \(\hat {i} \) + \(\frac{1}{2}\)\(\hat {j} \) + 4\(\hat {k} \) and – \(\hat {i} \) – \(\frac{1}{2}\)\(\hat {j} \) + 4\(\hat {k} \).

âˆ´ |\(\overrightarrow{\mathrm{AD}}\)| = (- \(\hat {i} \) + \(\hat {i} \)) + (- \(\frac{1}{2}\) – \(\frac{1}{2}\)) \(\hat {j} \) + 4\(\hat {k} \).

âˆ´ |\(\overrightarrow{\mathrm{AD}}\)| = 1.

Area of rectangle ABCD = |\(\overrightarrow{\mathrm{AB}}\)| |\(\overrightarrow{\mathrm{AD}}\)| = 2 Ã— 1 = 2.

âˆ´ Part (C) is the correct answer.