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.