# GSEB Solutions Class 12 Maths Chapter 10 Vector Algebra Ex 10.3

Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions 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 $$\sqrt{3}$$ and 2 respectively,
and such that $$\vec{a}$$.$$\vec{b}$$ = $$\sqrt{6}$$.
Solution:
We know that the angle θ between two vectors $$\vec{a}$$ and $$\vec{b}$$ is given by

Question 2.
Find the angle between the vectors $$\hat {i}$$ – 2$$\hat {j}$$ + 3$$\hat {k}$$ and 3$$\hat {i}$$ – 2$$\hat {j}$$ + $$\hat {k}$$.
Solution:
Let $$\vec{a}$$ = $$\hat {i}$$ – 2$$\hat {j}$$ + 3$$\hat {k}$$ and $$\vec{b}$$ = 3$$\hat {i}$$ – 2$$\hat {j}$$ + $$\hat {k}$$
are the given vectors and let θ be the angle between them. Then,
cos θ = $$\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$$ ……………………. (1)
Now, $$\vec{a}$$.$$\vec{b}$$ = ($$\hat {i}$$ – 2$$\hat {j}$$ + 3$$\hat {k}$$).(3$$\hat {i}$$ – 2$$\hat {j}$$ + $$\hat {k}$$)
= (1)(3) + (- 2)(- 2) + (3) (1)
= 3 + 4 + 3 = 10

Question 3.
Find the projection of vector $$\hat {i}$$ – $$\hat {j}$$ on the vector $$\hat {i}$$ + $$\hat {j}$$.
Solution:
Let $$\vec{a}$$ = $$\hat {i}$$ – $$\hat {j}$$ and $$\vec{b}$$ = $$\hat {i}$$ + $$\hat {j}$$
Then, projection of $$\vec{a}$$ and $$\vec{b}$$ = $$\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$ ………………….. (1)
Now, $$\vec{a}$$.$$\vec{b}$$ = ($$\hat {i}$$ – $$\hat {j}$$). ($$\hat {i}$$ + $$\hat {j}$$)
= (1) (1) + (- 1)(1) = 1 – 1 = 0
and |$$\vec{b}$$| = $$\sqrt{1^{2}+1^{2}}$$ = $$\sqrt{1+1}$$ = $$\sqrt{2}$$
From (1), we have:
∴ Projection of $$\hat {i}$$ – $$\hat {j}$$ on the vector $$\hat {i}$$ + $$\hat {j}$$
= $$\frac{(\hat{i}-\hat{j}) \cdot(\hat{i}+\hat{j})}{|\hat{i}+\hat{j}|}$$
= $$\frac{0}{\sqrt{2}}$$
= 0.

Question 4.
Find the projection of the vector $$\hat {i}$$ + 3$$\hat {j}$$ + 7$$\hat {k}$$ on the vector 7$$\hat {i}$$ – $$\hat {j}$$ + 8$$\hat {k}$$.
Solution:
Let $$\vec{a}$$ = $$\hat {i}$$ + 3$$\hat {j}$$ + 7$$\hat {k}$$
and $$\vec{b}$$ = 7$$\hat {i}$$ – $$\hat {j}$$ + 8$$\hat {k}$$.
Then, projection of $$\vec{a}$$ on $$\vec{b}$$ = $$\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$ ……………………. (1)
Now, $$\vec{a}$$.$$\vec{b}$$ = ($$\hat {i}$$ + 3$$\hat {j}$$ + 7$$\hat {k}$$).(7$$\hat {i}$$ – $$\hat {j}$$ + 8$$\hat {k}$$)
= (1)(7) + (3)(- 1) + (7)(8) = 7 – 3 + 56 = 60
and |$$\vec{b}$$| = $$\sqrt{7^{2}+(-1)^{2}+8^{2}}$$ = $$\sqrt{49+1+64}$$ = $$\sqrt{114}$$
From (1), we have:
Projection of $$\vec{a}$$ on $$\vec{b}$$ = $$\frac{60}{\sqrt{114}}$$.

Question 5.
Show that each of the given three vectors is a unit vector:
$$\frac{1}{7}$$(2$$\hat {i}$$ + 3$$\hat {j}$$ + 6$$\hat {k}$$),
$$\frac{1}{7}$$(3$$\hat {i}$$ + 6$$\hat {j}$$ + 2$$\hat {k}$$), $$\frac{1}{7}$$(6$$\hat {i}$$ + 2$$\hat {j}$$ – 3$$\hat {k}$$).
Solution:
Let $$\vec{a}$$ = $$\frac{1}{7}$$(2$$\hat {i}$$ + 3$$\hat {j}$$ + 6$$\hat {k}$$),
$$\vec{b}$$ = $$\frac{1}{7}$$(3$$\hat {i}$$ + 6$$\hat {j}$$ + 2$$\hat {k}$$), and $$\vec{c}$$ = $$\frac{1}{7}$$(6$$\hat {i}$$ + 2$$\hat {j}$$ – 3$$\hat {k}$$).

Hence, $$\vec{a}$$, $$\vec{b}$$ and $$\vec{c}$$ are three mutually perpendicular unit vectors.

Question 6.
Find |$$\vec{a}$$| and |$$\vec{b}$$|, if ($$\vec{a}$$ + $$\vec{b}$$).($$\vec{a}$$ – $$\vec{b}$$) = 8 and |$$\vec{a}$$| = 8|$$\vec{b}$$|.
Solution:
Given ($$\vec{a}$$ + $$\vec{b}$$).($$\vec{a}$$ – $$\vec{b}$$) = 8
⇒ $$\vec{a}$$.($$\vec{a}$$ – $$\vec{b}$$ + $$\vec{b}$$.($$\vec{a}$$ – $$\vec{b}$$) = 8
⇒ $$\vec{a}$$.$$\vec{a}$$ – $$\vec{a}$$.$$\vec{b}$$ + $$\vec{b}$$.$$\vec{a}$$ – $$\vec{b}$$.$$\vec{b}$$ = 8
⇒ |$$\vec{a}$$|2 – |$$\vec{b}$$|2 = 8 [∵ $$\vec{a}$$.$$\vec{b}$$ = $$\vec{b}$$.$$\vec{a}$$]
⇒ 64 |$$\vec{b}$$|2 – |$$\vec{b}$$|2 = 8 [∵ |$$\vec{a}$$| = 8|$$\vec{b}$$|

Question 7.
Evaluate the product (3$$\vec{a}$$ – 5$$\vec{b}$$).(2$$\vec{a}$$ + 7$$\vec{b}$$).
Solution:
(3$$\vec{a}$$ – 5$$\vec{b}$$).(2$$\vec{a}$$ + 7$$\vec{b}$$)
= 3$$\vec{a}$$.(2$$\vec{a}$$ + 7$$\vec{b}$$) – 5$$\vec{b}$$.(2$$\vec{a}$$ + 7$$\vec{b}$$)
= 6 $$\vec{a}$$. $$\vec{a}$$ + 21 $$\vec{a}$$.$$\vec{b}$$ – 10 $$\vec{b}$$. $$\vec{a}$$ – 35 $$\vec{b}$$.$$\vec{b}$$
= 6|$$\vec{a}$$|2 + 21$$\vec{a}$$.$$\vec{b}$$ – 10$$\vec{b}$$.$$\vec{a}$$ – 35 |$$\vec{b}$$|2
[∵ $$\vec{a}$$.$$\vec{b}$$ = $$\vec{b}$$.$$\vec{a}$$]
= 6|$$\vec{a}$$|2 + 11$$\vec{a}$$.$$\vec{b}$$ – 35 |$$\vec{b}$$|2.

Question 8.
Find the magnitude of two vectors $$\vec{a}$$ and $$\vec{b}$$ having the same magnitude such that the angle between them is 60° and their scalar product is $$\frac{1}{2}$$.
Solution:
We know that $$\vec{a}$$.$$\vec{b}$$ = |$$\vec{a}$$| |$$\vec{b}$$| cos θ
Here, $$\vec{a}$$.$$\vec{b}$$ = $$\frac{1}{2}$$, |$$\vec{a}$$| = |$$\vec{b}$$| and θ = 60°.
$$\frac{1}{2}$$ = |$$\vec{a}$$| |$$\vec{b}$$| cos 60°
⇒ $$\frac{1}{2}$$ = |$$\vec{a}$$|2 ($$\frac{1}{2}$$)
⇒ |$$\vec{a}$$|2 = 1 [∵ |$$\vec{a}$$| = |$$\vec{b}$$|, given]
|$$\vec{a}$$| = 1
|$$\vec{b}$$| = |$$\vec{a}$$| = 1.
Thus, |$$\vec{a}$$| = 1 and |$$\vec{b}$$| = 1.

Question 9.
Find |$$\vec{x}$$|, if for a unit vector $$\vec{a}$$, ($$\vec{x}$$ – $$\vec{a}$$). ($$\vec{x}$$ + $$\vec{a}$$) = 12.
Solution:
($$\vec{x}$$ – $$\vec{a}$$) ($$\vec{x}$$ + $$\vec{a}$$) = $$\vec{x}$$($$\vec{x}$$ + $$\vec{a}$$) – $$\vec{a}$$($$\vec{x}$$ + $$\vec{a}$$)
= |$$\vec{x}$$|2 + $$\vec{x}$$.$$\vec{a}$$ – $$\vec{a}$$.$$\vec{x}$$ – |$$\vec{a}$$|2
= |$$\vec{x}$$|2 + $$\vec{x}$$.$$\vec{a}$$ – $$\vec{x}$$ – $$\vec{a}$$ – |$$\vec{a}$$|2
[∵$$\vec{a}$$.$$\vec{x}$$ = $$\vec{x}$$.$$\vec{a}$$]
Now, |$$\vec{a}$$| = 1, ($$\vec{x}$$ + $$\vec{a}$$)($$\vec{x}$$ + $$\vec{a}$$) = 12
∴ |$$\vec{x}$$|2 – |$$\vec{a}$$|2 = 12
or |$$\vec{x}$$|2 = 12 + 1 [∵ |$$\vec{a}$$| = 1]
∴ |$$\vec{x}$$| = $$\sqrt{13}$$.

Question 10.
If $$\vec{a}$$.$$\vec{a}$$ = 0 and $$\vec{a}$$.$$\vec{b}$$ = 0, then what can be concluded about vector $$\vec{b}$$?
Solution:
$$\vec{a}$$ = 2$$\hat {i}$$ + 2$$\hat {j}$$ + 3$$\hat {k}$$
and $$\vec{b}$$ = – $$\hat {i}$$ + 2$$\hat {j}$$ + $$\hat {k}$$
∴ $$\vec{a}$$ + λ$$\vec{b}$$ = (2$$\hat {i}$$ + 2$$\hat {j}$$ + 3$$\hat {k}$$) + λ(- $$\hat {i}$$ + 2$$\hat {j}$$ + $$\hat {k}$$)
= (2 – λ)$$\hat {i}$$ + (2 + 2λ)$$\hat {j}$$ + (3 + λ)$$\hat {k}$$
$$\vec{c}$$ = 3$$\hat {i}$$ + $$\hat {j}$$
$$\vec{a}$$ + λ$$\vec{b}$$. $$\vec{c}$$ = 0
or [(2 – λ)$$\hat {i}$$ + (2 + 2λ)$$\hat {j}$$ + (3 + λ)$$\hat {k}$$
$$\vec{c}$$ = 3$$\hat {i}$$ + $$\hat {j}$$
$$\vec{a}$$ + λ$$\vec{b}$$ is perpendicular to $$\vec{c}$$.
⇒ ($$\vec{a}$$ + λ$$\vec{b}$$).$$\vec{c}$$ = 0
or [(2 – λ)$$\hat {i}$$ + (2 + 2λ)$$\hat {j}$$ + (3 + λ)$$\hat {k}$$].[3$$\hat {i}$$ + $$\hat {j}$$] = 0
⇒ 3(2 – λ) + (2 + 2λ) = 0
[Since a1$$\hat {i}$$ + b1$$\hat {j}$$ + c1$$\hat {k}$$).(a2$$\hat {i}$$ + b2$$\hat {j}$$ + c2$$\hat {k}$$) = (a1a2 + b1b2 + c1c2)]
⇒ 6 – 3λ + 2 + 2λ = 0 λ = 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 $$\vec{a}$$ and $$\vec{b}$$.
Solution:
Let $$\vec{p}$$ = |$$\vec{a}$$|$$\vec{b}$$ + |$$\vec{b}$$|$$\vec{a}$$
$$\vec{q}$$ = |$$\vec{a}$$|$$\vec{b}$$ – |$$\vec{b}$$|$$\vec{a}$$
$$\vec{p}$$.$$\vec{q}$$ = [|$$\vec{a}$$|$$\vec{b}$$ + |$$\vec{b}$$|$$\vec{a}$$. [|$$\vec{a}$$|$$\vec{b}$$ – |$$\vec{b}$$|$$\vec{a}$$]
= [$$\vec{a}$$|$$\vec{b}$$.[|$$\vec{a}$$|$$\vec{b}$$ – |$$\vec{b}$$|$$\vec{a}$$] + |$$\vec{b}$$$$\vec{a}$$.[|$$\vec{a}$$$$\vec{b}$$ – |$$\vec{b}$$|$$\vec{a}$$]
= |$$\vec{a}$$|2$$\vec{b}$$.$$\vec{b}$$ – |$$\vec{a}$$||$$\vec{b}$$| $$\vec{b}$$.$$\vec{a}$$ + |$$\vec{b}$$| |$$\vec{a}$$| $$\vec{a}$$.$$\vec{b}$$ – |$$\vec{b}$$|2 $$\vec{a}$$.$$\vec{a}$$
But $$\vec{b}$$.$$\vec{a}$$ = $$\vec{a}$$.$$\vec{b}$$,
$$\vec{b}$$.$$\vec{b}$$ = |$$\vec{b}$$|2 and $$\vec{a}$$.$$\vec{a}$$ = |$$\vec{a}$$|2.
∴ $$\vec{p}$$.$$\vec{q}$$ = |$$\vec{a}$$|2|$$\vec{b}$$|2 – |$$\vec{a}$$| |$$\vec{b}$$| $$\vec{a}$$.$$\vec{b}$$ + |$$\vec{a}$$| |$$\vec{b}$$| $$\vec{a}$$.$$\vec{b}$$ – |$$\vec{b}$$|2 |$$\vec{a}$$|2
= |$$\vec{a}$$|2 |$$\vec{b}$$|2 – |$$\vec{a}$$|2|$$\vec{b}$$|2 = 0
∴ $$\vec{p}$$.$$\vec{q}$$ = 0 ⇒ $$\vec{p}$$ ⊥ $$\vec{q}$$
Hence, |$$\vec{a}$$|$$\vec{b}$$ + |$$\vec{b}$$|$$\vec{a}$$ and |$$\vec{a}$$|$$\vec{b}$$ – |$$\vec{b}$$|$$\vec{a}$$ are perpendicular to each other.

Question 12.
Solution:
We have $$\vec{a}$$.$$\vec{a}$$ = 0 and $$\vec{a}$$.$$\vec{b}$$ = 0.
These equations are satisfied for $$\vec{a}$$ = 0.
⇒ $$\vec{b}$$ is any vector.

Question 13.
If $$\vec{a}$$, $$\vec{b}$$ and $$\vec{c}$$ are unit vectors such that $$\vec{a}$$ + $$\vec{b}$$ + $$\vec{c}$$ = 0, then find the value of $$\vec{a}$$.$$\vec{b}$$ + $$\vec{b}$$.$$\vec{c}$$ + $$\vec{c}$$.$$\vec{a}$$.
Solution:
We have:
$$\vec{a}$$ + $$\vec{b}$$ + $$\vec{c}$$ = 0
⇒ $$\vec{a}$$ + $$\vec{b}$$ = – $$\vec{c}$$
⇒ $$\vec{a}$$.($$\vec{a}$$ + $$\vec{b}$$) = $$\vec{a}$$.(- $$\vec{c}$$)
⇒ $$\vec{a}$$.$$\vec{a}$$ + $$\vec{a}$$.$$\vec{b}$$ = – $$\vec{a}$$.$$\vec{c}$$
⇒ |$$\vec{a}$$|2 $$\vec{a}$$.$$\vec{b}$$ = – $$\vec{a}$$.$$\vec{c}$$
⇒ |$$\vec{a}$$|2 $$\vec{a}$$.$$\vec{b}$$ + $$\vec{c}$$.$$\vec{a}$$ = 0
⇒ 1 + $$\vec{a}$$.$$\vec{b}$$ + $$\vec{c}$$.$$\vec{a}$$ = 0 ……………. (1) [∵ |$$\vec{a}$$| = 1]
Again, $$\vec{a}$$ + $$\vec{b}$$ + $$\vec{c}$$
⇒ $$\vec{a}$$ + $$\vec{b}$$ = – $$\vec{c}$$
⇒ $$\vec{b}$$.($$\vec{a}$$ + $$\vec{b}$$) = $$\vec{b}$$. (- $$\vec{c}$$)
⇒ $$\vec{a}$$.$$\vec{b}$$ + |$$\vec{b}$$|2 + $$\vec{b}$$.$$\vec{c}$$ = 0 ……….. (2) [∵ |$$\vec{b}$$| = 1]
Futher, $$\vec{a}$$ + $$\vec{b}$$ + $$\vec{c}$$ = 0
⇒ $$\vec{a}$$ + $$\vec{b}$$ = – $$\vec{c}$$
⇒ $$\vec{c}$$.($$\vec{a}$$ + $$\vec{b}$$) = $$\vec{c}$$.(- $$\vec{c}$$)
⇒ $$\vec{c}$$.$$\vec{a}$$ + $$\vec{b}$$.$$\vec{c}$$ + |$$\vec{c}$$|2 = 0
⇒ $$\vec{c}$$.$$\vec{a}$$ + $$\vec{b}$$.$$\vec{c}$$ + 1 = 0
Adding (1), (2) and (3), we have:
2($$\vec{a}$$.$$\vec{b}$$ + $$\vec{b}$$.$$\vec{c}$$ + $$\vec{c}$$.$$\vec{a}$$) + 3 = 0
∴ $$\vec{a}$$.$$\vec{b}$$ + $$\vec{b}$$.$$\vec{c}$$ + $$\vec{c}$$.$$\vec{a}$$ = – $$\frac{3}{2}$$.

Question 14.
If either $$\vec{a}$$ = 0 or $$\vec{b}$$ = 0, then $$\vec{a}$$.$$\vec{b}$$ = 0. But the converse need not be true. Justify your answer with an example.
Solution:
Let $$\vec{a}$$ = $$\hat {i}$$ – 2$$\hat {j}$$ + $$\hat {k}$$ and $$\vec{b}$$ = $$\hat {i}$$ + 3$$\hat {j}$$ + 5$$\hat {k}$$.
Thus, |$$\vec{a}$$| = $$\sqrt{1^{2}+(-2)^{2}+1^{2}}$$ = $$\sqrt{6}$$
|$$\vec{b}$$| = $$\sqrt{1^{2}+(3)^{2}+5^{2}}$$ = $$\sqrt{1+9+25}$$ = $$\sqrt{35}$$
⇒ a ≠ 0, $$\vec{b}$$ ≠ 0
But $$\vec{a}$$.$$\vec{b}$$ = ($$\hat {i}$$ – 2$$\hat {j}$$ + $$\hat {k}$$).($$\hat {i}$$ + 3$$\hat {j}$$ + 5$$\hat {k}$$)
[We know: (a1$$\hat {i}$$ + b1$$\hat {j}$$ + c1$$\hat {k}$$). (a2$$\hat {i}$$ + b2$$\hat {j}$$ + c2$$\hat {k}$$)
= (a1a2 + b1b2 + c1c2)]
∴ $$\vec{a}$$.$$\vec{b}$$ = 1 . 1 – 2 × 3 + 1 × 5 = 1 – 6 + 5 = 0
⇒ $$\vec{a}$$.$$\vec{b}$$ = 0 though $$\vec{a}$$ ≠ 0, $$\vec{b}$$ ≠ 0.

Question 15.
If the vertices A, B and C of a triangle ABC are (1, 2, 3), (- 1, 0, 0) and (0, 1, 2) respectively, then find ∠ABC.
Solution:
Let O be the origin. Thus,
$$\overrightarrow{\mathrm{OA}}$$ = $$\hat {i}$$ + 2$$\hat {j}$$ + 3$$\hat {k}$$, $$\overrightarrow{\mathrm{OB}}$$ = – $$\hat {i}$$
and $$\overrightarrow{\mathrm{OC}}$$ = $$\hat {j}$$ + 2$$\hat {k}$$.
∴ $$\overrightarrow{\mathrm{BC}}$$ = $$\overrightarrow{\mathrm{OC}}$$ – $$\overrightarrow{\mathrm{OB}}$$
= $$\hat {i}$$ + $$\hat {j}$$ + 2$$\hat {k}$$
and $$\overrightarrow{\mathrm{BA}}$$ = $$\overrightarrow{\mathrm{OA}}$$ – $$\overrightarrow{\mathrm{OB}}$$ = ($$\hat {i}$$ + 2$$\hat {j}$$ + 3$$\hat {k}$$) – (- $$\hat {i}$$)
= 2$$\hat {i}$$ + 2$$\hat {j}$$ + 2$$\hat {k}$$

Question 16.
Show that the poins A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1) are collinear.
Solution:
The position vectors of points A, B and C are $$\hat {i}$$ + 2$$\hat {j}$$ + 7$$\hat {k}$$, 2$$\hat {i}$$ + 6$$\hat {j}$$ + 3$$\hat {k}$$
and 3$$\hat {i}$$ + 10$$\hat {j}$$ – $$\hat {k}$$ respectively.
∴ $$\overrightarrow{\mathrm{AB}}$$ = $$\overrightarrow{\mathrm{OB}}$$ – $$\overrightarrow{\mathrm{OA}}$$
= (2$$\hat {i}$$ + 6$$\hat {j}$$ + 3$$\hat {k}$$) – ($$\hat {i}$$ + 2$$\hat {j}$$ + 7$$\hat {k}$$) = $$\hat {i}$$ + 4$$\hat {j}$$ – 4$$\hat {k}$$
$$\overrightarrow{\mathrm{BC}}$$ = $$\overrightarrow{\mathrm{OC}}$$ – $$\overrightarrow{\mathrm{OB}}$$
= (3$$\hat {i}$$ + 10$$\hat {j}$$ – $$\hat {k}$$) – (2$$\hat {i}$$ + 6$$\hat {j}$$ + 3$$\hat {k}$$)
= $$\hat {i}$$ + 4$$\hat {j}$$ – 4$$\hat {k}$$
and $$\overrightarrow{\mathrm{AC}}$$ = $$\overrightarrow{\mathrm{OC}}$$ – $$\overrightarrow{\mathrm{OA}}$$
= (3$$\hat {i}$$ + 10$$\hat {j}$$ – $$\hat {k}$$) – ($$\hat {i}$$ + 2$$\hat {j}$$ + 7$$\hat {k}$$)
= 2$$\hat {i}$$ + 8$$\hat {j}$$ – 8$$\hat {k}$$

Thus, AB + BC = AC
Hence, A, B and C are collinear.

Question 17.
Show that the vectors 2$$\hat {i}$$ – $$\hat {j}$$ + $$\hat {k}$$, $$\hat {i}$$ – 3$$\hat {j}$$ – 5$$\hat {k}$$
and 3$$\hat {i}$$ – 4$$\hat {j}$$ – 4$$\hat {k}$$ form the vertices of a right angled triangle.
Solution:
The position vectors of the points A, B and C are 2$$\hat {i}$$ – $$\hat {j}$$ + $$\hat {k}$$, $$\hat {i}$$ – 3$$\hat {j}$$ – 5$$\hat {k}$$
and 3$$\hat {i}$$ – 4$$\hat {j}$$ – 4$$\hat {k}$$ respectively.
∴ $$\overrightarrow{\mathrm{AB}}$$ = $$\overrightarrow{\mathrm{OB}}$$ – $$\overrightarrow{\mathrm{OA}}$$
= ($$\hat {i}$$ – 3$$\hat {j}$$ – 5$$\hat {k}$$) – (2$$\hat {i}$$ – $$\hat {j}$$ + $$\hat {k}$$) = $$\hat {i}$$ – 2$$\hat {j}$$ – 6$$\hat {k}$$
So, |$$\overrightarrow{\mathrm{AB}}$$| = $$\sqrt{(-1)^{2}+(-2)^{2}+(-6)^{2}}$$ = $$\sqrt{1+4+36}$$ = $$\sqrt{41}$$
∴ |$$\overrightarrow{\mathrm{AB}}$$|2 = 41.
$$\overrightarrow{\mathrm{BC}}$$ = $$\overrightarrow{\mathrm{OC}}$$ – $$\overrightarrow{\mathrm{OB}}$$
= (3$$\hat {i}$$ – 4$$\hat {j}$$ – 4$$\hat {k}$$) – $$\hat {i}$$ – 3$$\hat {j}$$ – 5$$\hat {k}$$) = 2$$\hat {i}$$ – $$\hat {j}$$ + $$\hat {k}$$
∴ |$$\overrightarrow{\mathrm{BC}}$$|2 = BC2 = 22 + (- 1)2 + 12 = 4 + 1 + 1 = 6.
$$\overrightarrow{\mathrm{AC}}$$ = $$\overrightarrow{\mathrm{OC}}$$ – $$\overrightarrow{\mathrm{OA}}$$
= (3$$\hat {i}$$ – 4$$\hat {j}$$ – 4$$\hat {k}$$) – (2$$\hat {i}$$ – $$\hat {j}$$ + $$\hat {k}$$) = $$\hat {i}$$ – 3$$\hat {j}$$ – 5$$\hat {k}$$
∴ |$$\overrightarrow{\mathrm{AC}}$$|2 = AC2 = 12 + (- 3)2 + (- 5)2
= 1 + 9 + 25 = 35.
Now, BC2 + AC2
= 6 + 35 = 41
= AB2
i.e., BC2 + AC2 = AB2
⇒ Triangle ABC is a right angled triangle, right angled at C.

Question 18.
If $$\vec{a}$$ is a non-zero vector of magnitude a and λ, a non-zero scalar, then λ$$\vec{a}$$ is a unit vector, if
(A) λ = 1
(B) λ = – 1
(C) a = |λ|
(D) a = $$\frac{1}{|\lambda|}$$
Solution:
$$\vec{a}$$ is a non-zero vector of magnitude a
⇒ |$$\vec{a}$$| = a
Now, λ$$\vec{a}$$ is a unit vector, if
|λ$$\vec{a}$$| = 1 or |λ| |$$\vec{a}$$| = 1
or |λ| a = 1
∴ a = $$\frac{1}{|\lambda|}$$
∴ Part (D) is the correct answer.