ML Aggarwal Class 12 Maths Solutions Section B Chapter 1 Vectors MCQs

Continuous practice using Class 12 ISC Maths Solutions Chapter 1 Vectors MCQs can lead to a stronger grasp of mathematical concepts.

ML Aggarwal Class 12 Maths Solutions Section B Chapter 1 Vectors MCQs

Choose the correct answer from the given four options in questions (1 to 52) :

Question 1.
The magnitude of the vector \(\) is
(a) 5 units
(b) 7 units
(c) 11 units
(d) 1 unit
Solution:
(b) 7 units

Let \(\vec{a}=6 \hat{i}-2 \hat{j}+3 \hat{k}\)
∴ \(|\vec{a}|=\sqrt{6^2+(-2)^2+3^2}\)
= \(\sqrt{36+4+9}\)
= \(\sqrt{49}\)
= 7 units

Question 2.
The vector hi the direction of the vector \(\hat{i}-2 \hat{j}+2 \hat{k}\) that has magnitude 9 units is
(a) \(\hat{i}-2 \hat{j}+2 \hat{k}\)
(b) \(\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{3}\)
(c) 3 \((\hat{i}-2 \hat{j}+2 \hat{k})\)
(d) 9 \((\hat{i}-2 \hat{j}+2 \hat{k})\)
Solution:
(c) 3 \((\hat{i}-2 \hat{j}+2 \hat{k})\)

Let \(\vec{a}=\hat{i}-2 \hat{j}+2 \hat{k}\) ;
\(|\vec{a}|=\sqrt{1^2+(-2)^2+2^2}\) = 3
∴ Required unit vector in the direction of \(\vec{a}\)
having magnitude 9 units = 9\(\hat{a}\)
= \(\frac{9 \vec{a}}{|\vec{a}|}\)
= \(\frac{9(\hat{i}-2 \hat{j}+2 \hat{k})}{\sqrt{1^2+(-2)^2+2^2}}\)
= \(3(\hat{i}-2 \hat{j}+2 \hat{k})\)

Question 3.
If \(\hat{a}\) is a non-zero vector and m is a non zero scalar, then m \(\hat{a}\) is a unit vector if
(a) m = ± 1
(b) m = |a|
(c) m = ± \(|\vec{a}|\)
(d) m = ± \(\frac{1}{|\vec{a}|}\)
Solution:
(d) m = ± \(\frac{1}{|\vec{a}|}\)

Now m \(\hat{a}\) is a unit vector
∴ |m \(\hat{a}\)| = 1
⇒ |m| |\(\hat{a}\)| = 1
⇒ |m| = ± \(\frac{1}{|\vec{a}|}\)
⇒ m = ± \(\frac{1}{|\vec{a}|}\).

Question 4.
If |\(\hat{a}\)| = 4 and – 3 ≤ k ≤ 2, then the range of |k \(\hat{a}\)| is
(a) [0, 8]
(b) [0, 12]
(c) [- 12, 8]
(d) [8, 12]
Solution:
(b) [0, 12]

Given – 3 ≤ k ≤ 2 < 3
⇒ |k| ≤ 3
⇒ 4 |k| ≤ 12
⇒ |\(\hat{a}\)| |k| ≤ 12
⇒ 0 ≤ \(|k \vec{a}|\) ≤ 12
⇒ \(|k \vec{a}|\) ∈ [0, 12]

Question 5.
The vector having initial and terminal points as (2, 5, 0) and (- 3, 7, 4) respectively is
(a) \(-5 \hat{i}+2 \hat{j}+4 \hat{k}\)
(b) \(5 \hat{i}-2 \hat{j}-4 \hat{k}\)
(c) \(5 \hat{i}+2 \hat{j}+4 \hat{k}\)
(d) \(-5 \hat{i}+2 \hat{j}-4 \hat{k}\)
Solution:
(a) \(-5 \hat{i}+2 \hat{j}+4 \hat{k}\)

P.V of initial point A = \(2 \hat{i}+5 \hat{j}+0 \hat{k}\)
P.V of terminal point B = \(-3 \hat{i}+7 \hat{j}+4 \hat{k}\)
\(\overrightarrow{\mathrm{AB}}\) = P.V of B – P.V of A
= \((-3 \hat{i}+7 \hat{j}+4 \hat{k})-(2 \hat{i}+5 \hat{j}+0 \hat{k})\)
= \(-5 \hat{i}+2 \hat{j}+4 \hat{k}\)

Question 6.
If the sides AB and AD of a parallelogram ABCD are represented by the vectors \(2 \hat{i}+4 \hat{j}-5 \hat{k}\) and \(\hat{i}+2 \hat{j}+3 \hat{k}\), then a unit vector being \(\overrightarrow{\mathbf{A C}}\) is
(a) \(\frac{1}{3}(3 \hat{i}-6 \hat{j}-2 \hat{k})\)
(b) \(\frac{1}{3}(3 \hat{i}+6 \hat{j}-2 \hat{k})\)
(c) \(\frac{1}{7}(3 \hat{i}-6 \hat{j}-2 \hat{k})\)
(d) \(\frac{1}{7}(3 \hat{i}+6 \hat{j}-2 \hat{k})\)
Solution:
(d) \(\frac{1}{7}(3 \hat{i}+6 \hat{j}-2 \hat{k})\)

Since ABCD is a parallelogram
Then \(\overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{BC}}\)
= \(\hat{i}+2 \hat{j}+3 \hat{k}\)

Question 7.
The position vector of the point which divides the line segment joining the points with position vectors \(2 \vec{a}-3 \vec{b}\) and \(\vec{a}+\vec{b}\) in the ratio 3 : 1 is
(a) \(\frac{3 \vec{a}-2 \vec{b}}{4}\)
(b) \(\frac{7 \vec{a}-8 \vec{b}}{4}\)
(c) \(\frac{3 \vec{a}}{4}\)
(d) \(\frac{5 \vec{a}}{4}\)
Solution:
(d) \(\frac{5 \vec{a}}{4}\)

Question 8.
If the position vector of the point A is \(\vec{a}+2 \vec{b}\) and a point P with position vector \(\vec{a}\) divides a line segment AB in the ratio 2 : 3, then the position vector of the point B ¡s
(a) \(2 \vec{a}-\vec{b}\)
(b) \(\vec{b}-2 \vec{a}\)
(c) \(\vec{a}-3 \vec{b}\)
(d) \(\vec{b}\)
Solution:
(c) \(\vec{a}-3 \vec{b}\)

Let the P.V of point B be \(\alpha \vec{a}+\beta \vec{b}\)
Then by section formula P.V. of P which divides line segment AB in the ratio 2 : 3 be given by

⇒ 2α = 5 – 3 = 2
⇒ α = 1
and 2β + 6 = 0
⇒ β = – 3
Thus, P.V. of B = \(\alpha \vec{a}+\beta \vec{b}\)
= \(\vec{a}-3 \vec{b}\)

Question 9.
The value of λ for which the vectors \(3 \hat{i}-6 \hat{j}+\hat{k}\) and \(2 \hat{i}-4 \hat{j}+\lambda \hat{k}\) are parallel is
(a) \(\frac{3}{2}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{5}{2}\)
(d) \(\frac{2}{5}\)
Solution:
(b) \(\frac{2}{3}\)

Let \(\vec{a}=3 \hat{i}-6 \hat{j}+\hat{k}\)
and \(\vec{b}=2 \hat{i}-4 \hat{j}+\lambda \hat{k}\)
Since \(\vec{a}\) is parallel to \(\vec{b}\)
∴ \(\vec{a}=k \vec{b}\)
⇒ \((3 \hat{i}-6 \hat{j}+\hat{k})=k(2 \hat{i}-4 \hat{j}+\lambda \hat{k})\)
⇒ 3 = 2k
and – 6 = – 4k
⇒ k = \(\frac{3}{2}\)
and 1 = λ k
⇒ 1 = \(\frac{3}{2}\) λ
⇒ λ = \(\frac{2}{3}\)

Question 10.
If \(\vec{a}\) = (1, – 1) and \(\vec{b}\) = (- 2, m) are collinear vector, then m is equal to
(a) 2
(b) 3
(c) 4
(d) – 2
Solution:
(a) 2

Given \(\vec{a}=\hat{i}-\hat{j}\)
and \(\vec{b}=-2 \hat{i}+m \hat{j}\)
Since \(\vec{a} \text { and } \vec{b}\) are collinear vectors.
∴ \(\vec{a}=k \vec{b}\)
⇒ \(\hat{i}-\hat{j}=k(-2 \hat{i}+m \hat{j})\)
⇒ 1 = – 2k
⇒ k = – \(\frac{1}{2}\)
and – 1 = mk
⇒ – 1 = m (- \(\frac{1}{2}\))
⇒ m = 2

Question 11.
If A, B, C, D and E are five coplanar points, then \(\overrightarrow{\mathrm{DA}}+\overrightarrow{\mathrm{DB}}+\overrightarrow{\mathrm{DC}}+\overrightarrow{\mathrm{AE}}+\overrightarrow{\mathrm{BE}}+\overrightarrow{\mathrm{CE}}\) is equal to
(a) \(\overrightarrow{\mathbf{D E}}\)
(b) 3 \(\overrightarrow{\mathbf{D E}}\)
(c) 2 \(\overrightarrow{\mathbf{D E}}\)
(d) 4 \(\overrightarrow{\mathbf{D E}}\)
Solution:
(b) 3 \(\overrightarrow{\mathbf{D E}}\)

\(\overrightarrow{\mathrm{DA}}+\overrightarrow{\mathrm{DB}}+\overrightarrow{\mathrm{DC}}+\overrightarrow{\mathrm{AE}}+\overrightarrow{\mathrm{BE}}+\overrightarrow{\mathrm{CE}}\)
= \((\overrightarrow{\mathrm{DA}}+\overrightarrow{\mathrm{AE}})+(\overrightarrow{\mathrm{DB}}+\overrightarrow{\mathrm{BE}})+(\overrightarrow{\mathrm{DC}}+\overrightarrow{\mathrm{CE}})\)
= \(\overrightarrow{\mathrm{DE}}+\overrightarrow{\mathrm{DE}}+\overrightarrow{\mathrm{DE}}=3 \overrightarrow{\mathrm{DE}}\)

Question 12.
If A, B and C are the vertices of a triangle with position vectors \(\vec{a}, \vec{b} \text { and } \vec{c}\) respectively and G is the centroid of ∆ABC, then \(\overrightarrow{\mathbf{G A}}+\overrightarrow{\mathbf{G B}}+\overrightarrow{\mathbf{G C}}\) is equal to
(a) \(\overrightarrow{0}\)
(b) \(\vec{b}+\vec{b}+\vec{c}\)
(c) \(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\)
(d) \(\frac{\vec{a}-\vec{b}-\vec{c}}{3}\)
Solution:
(a) \(\overrightarrow{0}\)

Question 13.
The angle between two vectors \(\vec{a} \text { and } \vec{b}\) with magnitudes √3 and 4, respectively and \(\vec{a} \cdot \vec{b}\) = 2√3 is
(a) \(\frac{2 \pi}{3}\)
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{3}\)
(d) \(\frac{\pi}{6}\)
Solution:
(c) \(\frac{\pi}{3}\)

Let θ be the angle between \(\vec{a} \text { and } \vec{b}\)
Given \(|\vec{a}|\) = √3 ;
\(|\vec{b}|\) = 4
and \(\vec{a} \cdot \vec{b}\) = 2√3
⇒ \(|\vec{a}||\vec{b}|\) cos θ = 2√3
⇒ √3 × 4 cos θ = 2√3
⇒ cos θ = \(\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{3}\)

Question 14.
The angle between the vectors \(\hat{i}-\hat{j}\) and \(\hat{j}-\hat{k}\) is
(a) – \(\frac{\pi}{3}\)
(b) \(\frac{2 \pi}{3}\)
(c) \(\frac{\pi}{6}\)
(d) \(\frac{5 \pi}{6}\)
Solution:
(b) \(\frac{2 \pi}{3}\)

Question 15.
The value of λ for which the vectors \(2 \hat{i}+\lambda \hat{j}+\hat{k}\) and \(\hat{i}+2 \hat{j}+3 \hat{k}\) are perpendicular is
(a) 0
(b) 1
(c) \(\frac{3}{2}\)
(d) – \(\frac{5}{2}\)
Solution:
(d) – \(\frac{5}{2}\)

Let \(\vec{a}=2 \hat{i}+\lambda \hat{j}+\hat{k}\)
and \(\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}\)
Since \(\vec{a} \perp \vec{b}\)
∴ \(\vec{a} \cdot \vec{b}\) = 0
⇒ \((2 \hat{i}+\lambda \hat{j}+\hat{k}) \cdot(\hat{i}+2 \hat{j}+3 \hat{k})\) = 0
⇒ 2 × 1 + λ × 2 + 1 × 3 = 0
⇒ 2λ = – 5
⇒ λ = – \(\frac{5}{2}\)

Question 16.
If the angle between the vectors \(\hat{i}+\hat{k}\) and \(\hat{i}+\hat{j}+\lambda \hat{k}\) is \(\frac{\pi}{3}\), then the values of Xare
(a) 0, 2
(b) 0, – 2
(c) 0, – 4
(d) 2, – 2
Solution:
(c) 0, – 4

⇒ 2 (2 + λ²) = 4 (1 + λ)²
⇒ 2 + λ² = 2λ² + 2 + 4λ
⇒ λ² + 4λ = 0
⇒ λ = 0, – 4

Question 17.
If points A, B and C with position vectors \(2 \hat{i}-\hat{j}+\hat{k}\), \(\hat{i}-3 \hat{j}-5 \hat{k}\) and \(\alpha \hat{i}-3 \hat{j}+\hat{k}\) respectively are the vertices of a right angled triangle with ∠C = \(\frac{\pi}{2}\), then the values of α are
(a) 2, – 1
(b) 2, 1
(c) – 2, 1
(d) – 2, – 1
Solution:
(b) 2, 1

Question 18.
If \(\vec{a} \text { and } \vec{b}\) are unit vectors, then the angle between \(\vec{a} \text { and } \vec{b}\) for \(\sqrt{3} \vec{a}-\vec{b}\) to be a unit vector is
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:

Question 19.
If θ is the angle between two vectors \(\vec{a} \text { and } \vec{b}\), then \(\vec{a} \cdot \vec{b}\) ≥ 0 only when
(a) 0 < θ < \(\frac{\pi}{2}\)
(b) 0 ≤ θ ≤ \(\frac{\pi}{2}\)
(c) 0 < θ < π
(d) 0 ≤ θ ≤ π
Solution:
(b) 0 ≤ θ ≤ \(\frac{\pi}{2}\)

Now \(\vec{a} \cdot \vec{b}\) ≥ 0
⇒ \(|\vec{a}||\vec{b}|\) cos θ ≥ 0
⇒ cos θ ≥ 0
⇒ 0 \(\frac{\pi}{2}\) or \(\frac{3 \pi}{2}\) ≤ θ ≤ 2π

Question 20.
The projection of the vector \(\hat{i}+\hat{j}+\hat{k}\) along vector \(\hat{j}\) is
(a) 1
(b) 2
(c) – 1
(d) – 2
Solution:
(a) 1

Let \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\)
and \(\vec{b}=\hat{j}\)
∴ projection of \(\vec{a} \text { along } \vec{b}\) = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\)
= \(\frac{\hat{i}+\hat{j}+\hat{k}) \cdot \hat{j}}{|\hat{j}|}\)
= 1

Question 21.
The projection of the vector \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k}\) along the vector \(\vec{b}=\hat{i}+2 \hat{j}+2 \hat{k}\) is
(a) 2
(b) √6
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{3}\)
Solution:
(c) \(\frac{2}{3}\)

Question 22.
If \(\vec{a}=2 \hat{i}+\hat{j}+2 \hat{k}\) and \(\vec{b}=5 \hat{i}-3 \hat{j}+\hat{k}\), then the perpendicular of \(\vec{b} \text { on } \vec{a}\) is
(a) 3
(b) 4
(c) 5
(d) 1
Solution:
(a) 3

Question 23.
If \(|\vec{a}|\) = 3 and \(|\vec{b}|\) = 4, then the value of λ for which \(\vec{a}+\lambda \vec{b}\) and \(\vec{a}-\lambda \vec{b}\) are perpendicular is
(a) \(\frac{9}{16}\)
(b) \(\frac{3}{4}\)
(c) \(\frac{3}{2}\)
(d) \(\frac{4}{3}\)
Solution:
(b) \(\frac{3}{4}\)

Question 24.
\((\vec{a} \cdot \hat{i})^2+(\vec{a} \cdot \hat{j})^2+(\vec{a} \cdot \hat{k})^2\) is equal to
(a) 1
(b) \(\vec{a}\)
(c) – \(\vec{a}\)
(d) \(|\vec{a}|^2\)
Solution:
(d) \(|\vec{a}|^2\)

Question 25.
If \(\vec{a} \cdot \hat{i}=\vec{a} \cdot(\hat{i}+\hat{j})=\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 1, then \(\vec{a}\) is equal to
(a) \(\overrightarrow{0}\)
(b) \(\hat{i}\)
(c) \(\hat{j}\)
(d) \(\hat{i}+\hat{j}+\hat{k}\)
Solution:
(b) \(\hat{i}\)

Let \(\vec{a}=x \hat{i}+y \hat{j}+z \hat{k}\)
Now \(\vec{a} \cdot \hat{i}\) = 1
⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot \hat{i}\) = 1
⇒ x = 1
[∵ \(\hat{j} \cdot \hat{i}\) = 0 = \(\hat{k} \cdot \hat{k}\)]
\(\vec{a} \cdot(\hat{i}+\hat{j})\) = 1
⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{j})\) = 1
[∵ \(\hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{k}\)]
⇒ x + y = 1
⇒ 1 + y = 1
⇒ y = 0
and \(\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 1
⇒ \((x \hat{j}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})\) = 1
⇒ x + y + z = 1
⇒ 1 + 0 + z = 1
⇒ z = 0
Thus, \(\vec{a}=\hat{i}\)

Question 26.
If \(|\vec{a} \times \vec{b}|^2+|\vec{a} \cdot \vec{b}|^2\) = 144 and \(|\vec{a}| \) = 4, then \(|\vec{b}|\) is equal to
(a) 12
(b) 3
(c) 4
(d) – 3
Solution:
(b) 3

Question 27.
If \(|\vec{a}|\) = 3, \(|\vec{b}|\) = 4 and \(|\vec{a} \times \vec{b}|\) = 10, then \(|\vec{a} \cdot \vec{b}|^2\) is equal to
(a) 22
(b) 88
(c) 44
(d) none of these
Solution:
(c) 44

Given \(|\vec{a}|\) = 3 ;
\(|\vec{b}|\) = 4
and \(|\vec{a} \times \vec{b}|\) = 10
⇒ 10 = \(|\vec{a}|\) \(|\vec{b}|\) sin θ
⇒ 10 = 3 × 4 sin θ
⇒ sin θ = \(\frac{5}{6}\)
∴ \(|\vec{a} \cdot \vec{b}|^2\) = (\(|\vec{a}||\vec{b}|\) cos θ)²
= \(|\vec{a}|^2|\vec{b}|^2\) cos² θ
= 3² × 4² × (1 – sin² θ)
= 9 × 16 × (1 – \(\frac{25}{36}\))
= 144 × \(\frac{11}{36}\)
= 44.

Question 28.
If \(|\vec{a}|^2\) = 10, \(|\vec{b}|\) = 2 and \(\vec{a} \cdot \vec{b}\) = 12, then \(|\vec{a} \times \vec{b}|\) is equal to
(a) 4
(b) 10
(c) – 16
(d) 16
Solution:
(d) 16

Given \(|\vec{a}|^2\) = 10 ;
\(|\vec{b}|\) = 2
and \(\vec{a} \cdot \vec{b}\) = 12
⇒ 10 × 2 × cos θ = 12
⇒ cos θ = \(\frac{12}{20}=\frac{3}{5}\)
∴ sin θ = \(\sqrt{1-\cos ^2 \theta}\)
= \(\sqrt{1-\frac{9}{25}}=\frac{4}{5}\)
Thus, \(|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta\)
= 10 × 2 × \(\frac{4}{5}\)
= 16.

Question 29.
If \(|\vec{a} \times \vec{b}|\) = 4 and \(|\vec{a} \cdot \vec{b}|\) = 2, then \(|\vec{a}|^2\) , \(|\vec{b}|^2\) is equal to
(a) 6
(b) 8
(c) 20
(d) none of these
Solution:
(c) 20

Given \(|\vec{a} \times \vec{b}|\) = 4
⇒ \(|\vec{a}|\) \(|\vec{b}|\) sin θ = 4 ………………(1)
and \(|\vec{a} \cdot \vec{b}|\) = 2
⇒ \(|\vec{a}|\) \(|\vec{b}|\) cos θ = 2 …………………(2)
On squaring and adding eqn. (1) and (2) ;
\(|\vec{a}|^2|\vec{b}|^2\) [sin² θ + cos² θ] = 16 + 4
⇒ \(|\vec{a}|^2|\vec{b}|^2\) = 20

Question 30.
If \(|\vec{a}|\) = 8, \(|\vec{b}|\) = 3 and \(|\vec{a} \times \vec{b}|\) = 12, then the value of \(\vec{a} \cdot \vec{b}\) is
(a) 12√3 only
(b) – 12 √3 only
(c) 12√3 or – 12√3 only
(d) none of these
Solution:
(c) 12√3 or – 12√3 only

Question 31.
If \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\), \(|\vec{a}|=\sqrt{37}\), \(|\vec{b}|\) = 3 and \(|\vec{c}|\) = 4, then the angle between \(\vec{b} \text { and } \vec{c}\) is
(a) 90°
(b) 60°
(c) 45°
(d) 30°
Solution:
(b) 60°

Question 32.
The number of unit vectors perpendicualr to the vectors \(\vec{a}=2 \hat{i}+\hat{j}+2 \hat{k}\) and \(\vec{b}=\hat{j}+\hat{k}\) is
(a) one
(b) two
(c) three
(d) infinite
Solution:
(b) two

Given \(\vec{a}=2 \hat{i}+\hat{j}+2 \hat{k}\)
and \(\vec{b}=\hat{j}+\hat{k}\)
∴ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 1 & 2 \\
0 & 1 & 1
\end{array}\right|\)
= \(\hat{i}(1-2)-\hat{j}(2-0)+\hat{k}(2-0)\)
= \(-\hat{i}-2 \hat{j}+2 \hat{k}\)
and \(|\vec{a} \times \vec{b}|=\sqrt{1+4+4}\) = 3
Thus required unit vector ⊥\(\vec{a} \text { and } \vec{b}\) = ± \(\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\)
= ± \(\frac{(-\hat{i}-2 \hat{j}+2 \hat{k})}{3}\)
Hence the required no. of unit vectors ⊥to given vectors.

Question 33.
Unit vectors perpendicualr to the vectors \(\hat{i}-\hat{j}\) and \(\hat{i}+\hat{j}\) is
(a) \(\hat{\boldsymbol{k}}\)
(b) – \(\hat{\boldsymbol{k}}\)
(c) \(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\)
(d) \(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\)
Solution:
either (a) \(\hat{\boldsymbol{k}}\) or (b) – \(\hat{\boldsymbol{k}}\)

Question 34.
Unit vectors perpendicualr to the plane of vectors \(\vec{a}=2 \hat{i}-6 \hat{j}-3 \hat{k}\) and \(\vec{b}=4 \hat{i}+3 \hat{j}-\hat{k}\) are
(a) ± \(\frac{4 \hat{i}+3 \hat{j}-\hat{k}}{\sqrt{26}}\)
(b) ± \(\frac{2 \hat{i}-6 \hat{j}-3 \hat{k}}{7}\)
(c) ± \(\frac{2 \hat{i}-3 \hat{j}+6 \hat{k}}{7}\)
(d) ± \(\frac{3 \hat{i}-2 \hat{j}+6 \hat{k}}{7}\)
Solution:
(d) ± \(\frac{3 \hat{i}-2 \hat{j}+6 \hat{k}}{7}\)

Question 35.
A vector of magnitude 5 and perpendicular to \(\hat{i}-2 \hat{j}+\hat{k}\) and \(2 \hat{i}+\hat{j}-3 \hat{k}\) is
(a) \(\frac{5 \sqrt{3}(\hat{i}+\hat{j}+\hat{k})}{3}\)
(b) \(\frac{5 \sqrt{3}(\hat{i}-\hat{j}+\hat{k})}{3}\)
(c) \(\frac{5 \sqrt{3}(\hat{i}-\hat{j}-\hat{k})}{3}\)
(d) \(\frac{5 \sqrt{3}(\hat{i}+\hat{j}-\hat{k})}{3}\)
Solution:
(a) \(\frac{5 \sqrt{3}(\hat{i}+\hat{j}+\hat{k})}{3}\)

Question 36.
The area (in sq. units) of a parallelogram whose adjacent sides are given by the vectors \(\hat{i}+\hat{k}\) and \(2 \hat{i}+\hat{j}+\hat{k}\) is
(a) √2
(b) √3
(c) 3
(d) 4
Solution:
(b) √3

\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 0 & 1 \\
2 & 1 & 1
\end{array}\right|\)
= \((0-1) \hat{i}-\hat{j}(1-2)+\hat{k}(1-0)\)
= \(-\hat{i}+\hat{j}+\hat{k}\)
∴ area of ||gm = \(|\vec{a} \times \vec{b}|\)
= \(\sqrt{1+1+1}\)
= \(\sqrt{3}\) sq. units

Question 37.
If \(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=2 \hat{i}-4 \hat{j}+4 \hat{k}\), then area of ∆ABC is
(a) 3 sq. units
(b) 6 sq. units
(c) 9 sq. units
(d) 16 sq. units
Solution:
(a) 3 sq. units

Given \(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=2 \hat{i}-4 \hat{j}+4 \hat{k}\)
∴ area of ∆ABC = \(\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|\)
= \(\frac{1}{2} \sqrt{2^2+(-4)^2+(4)^2}\)
= \(\frac{1}{2} \sqrt{4+16+16}\)
= 3 sq. units

Question 38.
The vectors from origin to the points A and B are \(\vec{a}=2 \hat{i}-3 \hat{j}+2 \hat{k}\) and \(\vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}\), respectively, then the area (in sq. units) of ∆ABC is
(a) \(\sqrt{340}\)
(b) 5
(c) \(\sqrt{229}\)
(d) \(\frac{1}{2} \sqrt{229}\)
Solution:
(d) \(\frac{1}{2} \sqrt{229}\)

\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -3 & 2 \\
2 & 3 & 1
\end{array}\right|\)
= \(\hat{i}(-3-6)-\hat{j}(2-4)+\hat{k}(6+6)\)
= \(-9 \hat{i}+2 \hat{j}+12 \hat{k}\)
∴ area of ∆OAB = \(\frac{1}{2}|\vec{a} \times \vec{b}|\)
= \(\frac{1}{2} \sqrt{81+4+144}\)
= \(\frac{\sqrt{229}}{2}\) sq. units

Question 39.
The area (in sq. units) of the triangle having vertices with position vectors \(\hat{i}-2 \hat{j}+3 \hat{k}\), \(-2 \hat{i}+3 \hat{j}-\hat{k}\) and \(4 \hat{i}-7 \hat{j}+7 \hat{k}\) is
(a) 36
(b) 0
(c) 39
(d) 11
Solution:
(b) 0

Let the vertices of ∆ABC with position vectors \(\hat{i}-2 \hat{j}+3 \hat{k}\), \(-2 \hat{i}+3 \hat{j}-\hat{k}\) and \(4 \hat{i}-7 \hat{j}+7 \hat{k}\) are A, B and C respectively.

Question 40.
If \(\vec{a}, \vec{b} \text { and } \vec{c}\) are unit vectors such that \(\vec{a}+\vec{b}+\vec{c}\) = 0, then the value of \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}\) is
(a) 1
(b) \(\frac{3}{2}\)
(c) – \(\frac{3}{2}\)
(d) none of these
Solution:
(c) – \(\frac{3}{2}\)

Question 41.
If \(\vec{a}, \vec{b}, \vec{c}\) are three vectors such that \(\vec{a}+\vec{b}+\vec{c}\) = 0 and \(\vec{a}\) = 2, \(\vec{b}\) = 3, \(\vec{c}\) = 5, then the value of \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}\) is
(a) 0
(b) 1
(c) 38
(d) – 19
Solution:
(d) – 19

Given latex]\vec{a}[/latex] = 2 ;
\(\vec{b}\) = 3
and \(\vec{c}\) = 5
Now \(\vec{a}+\vec{b}+\vec{c}\) = 0
⇒ \((\vec{a}+\vec{b}+\vec{c})^2\) = 0
⇒ \((\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})\) = 0
⇒ \(\vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{c}+2[\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}]\) = 0
⇒ \(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2[\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}]\) = 0
⇒ 4 + 9 + 25 + 2 \([\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}]\) = 0
⇒ \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}\) = – 19

Question 42.
The volume (in cubioc units) of the parallelopiped whose conterminous edges are given by the vectors \(\hat{i}-\hat{j}+\hat{k}\), \(2 \hat{i}-4 \hat{j}+5 \hat{k}\) and \(3 \hat{i}-5 \hat{j}+2 \hat{k}\) is
(a) 5
(b) 11
(c) 6
(d) 8
Solution:
(d) 8

Let \(\vec{a}=\hat{i}-\hat{j}+\hat{k}\) ;
\(\vec{b}=2 \hat{i}-4 \hat{j}+5 \hat{k}\) ;
\(\vec{c}=3 \hat{i}-5 \hat{j}+2 \hat{k}\)
∴ \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=\left|\begin{array}{rrr}
1 & -1 & 1 \\
2 & -4 & 5 \\
3 & -5 & 2
\end{array}\right|\) ;
Expanding along R₁
= 1 (- 8 + 25) + 1 (4 – 15) + 1 (- 10 + 12)
= 17 – 11 + 2 = 8
∴ Volume of || piped = \(\left|\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]\right|\)
= 8 cubic units

Question 43.
If \(\vec{a}=2 \hat{i}-3 \hat{j}+2 \hat{k}\), \(\vec{b}=2 \hat{i}-4 \hat{k}\) and \(\vec{c}=\hat{i}+\lambda \hat{j}+\mathbf{3} \hat{k}\) are coplanar, then the value of λ is
(a) \(\frac{5}{2}\)
(b) \(\frac{5}{3}\)
(c) \(\frac{3}{5}\)
(d) – \(\frac{15}{6}\)
Solution:
(d) – \(\frac{15}{6}\)

Since \(\vec{a}, \vec{b} \text { and } \vec{c}\) are coplanar
∴ \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]\) = 0
⇒ \(\left|\begin{array}{rrr}
2 & -3 & 2 \\
2 & 0 & -4 \\
1 & \lambda & 3
\end{array}\right|\) = 0 ;
expanding along R₁
⇒ 2 (0 + 4λ) + 3 (6 + 4) + 2 (2λ – 0) = 0
12λ = – 30
λ = – \(\frac{15}{6}\)

Question 44.
If \(\vec{a}\) is perpendicualr to \(\vec{b} \text { and } \vec{c}\), \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 3 and \(|\vec{c}|\) = 4 and angle between \(\vec{b} \text { and } \vec{c}\) is \(\frac{2 \pi}{3}\), then \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]\) is equal to
(a) 4√3
(b) 6√3
(c) 12√3
(d) 18√3
Solution:
(c) 12√3

Question 45.
If \(|\vec{a}|=|\vec{b}|\) = 1 and \(|\vec{a} \times \vec{b}|\) = 1, then
(a) \(\vec{a} \perp \vec{b}\)
(b) \(\vec{a} \| \vec{b}\)
(c) \(\vec{a} \cdot \vec{b}\) = 1
(d) none of these
Solution:
(a) \(\vec{a} \perp \vec{b}\)

Given \(|\vec{a} \times \vec{b}|\) = 1
⇒ \(|\vec{a}||\vec{b}|\) sin θ = 1
⇒ 1 × 1 sin θ = 1
⇒ sin θ = 1
⇒ θ = \(\frac{\pi}{2}\)
∴ \(\vec{a} \perp \vec{b}\)

Question 46.
If \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\), \(\vec{a} \cdot \vec{b}\) = 1 and \(\vec{a} \times \vec{b}=\hat{j}-\hat{k}\), then the vector \(\vec{b}\) is
(a) \(\hat{i}-\hat{j}+\hat{k}\)
(b) \(2 \hat{i}-\hat{k}\)
(c) 2 \(\hat{i}\)
(d) \(\hat{i}\)
Solution:
(d) \(\hat{i}\)

∴ γ – β = 0
⇒ γ = β ……………….(3)
and – γ + α = 1 ………………….(4)
Also, β – α = – 1 …………………..(5)
From (2) ;
α + 2β = 1 ……………..(6)
From (5) and (6) ;
3 β = 0
β = 0
∴ α = 1
∴ from (2) ;
γ = 0
Thus from (1) ;
\(\vec{b}=\hat{i}\).

Question 47.
If \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 7 and \(\vec{a} \times \vec{b}=3 \hat{i}+2 \hat{j}-6 \hat{k}\) then the angle between \(\vec{a} \text { and } \vec{b}\) is
(a) \(\frac{\pi}{6}\)
(b) \(\frac{\pi}{4}\)
(c) \(\frac{\pi}{3}\)
(d) \(\frac{2 \pi}{3}\)
Solution:
(a) \(\frac{\pi}{6}\)

Given \(\vec{a} \times \vec{b}=3 \hat{i}+2 \hat{j}-6 \hat{k}\)
⇒ \(|\vec{a} \times \vec{b}|=|3 \hat{i}+2 \hat{j}-6 \hat{k}|\)
⇒ \(|\vec{a}||\vec{b}|\) sin θ = \(\sqrt{9+4+36}\) = 7
where θ be an angle between \(\vec{a} \text { and } \vec{b}\)
⇒ 2 × 7 × sin θ = 7
⇒ sin θ = \(\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{6}\)

Question 48.
If \((\vec{a}+\vec{b}) \perp \vec{b}\) and \((\vec{a}+2 \vec{b}) \perp \vec{a}\), then
(a) \(|\vec{a}|=|\vec{b}|\)
(b) \(2 |\vec{a}|=|\vec{b}|\)
(c) \(|\vec{a}|=2|\vec{b}|\)
(d) \(|\vec{a}|=\sqrt{2}|\vec{b}|\)
Solution:
(d) \(|\vec{a}|=\sqrt{2}|\vec{b}|\)

Question 49.
If ABCD is a rhombus whose diagonals intersects at E, then \(\overrightarrow{\mathrm{EA}}+\overrightarrow{\mathrm{EB}}+\overrightarrow{\mathrm{EC}}+\overrightarrow{\mathrm{ED}}\) equals
(a) \(\overrightarrow{0}\)
(b) \(\overrightarrow{AD}\)
(c) \(\overrightarrow{BC}\)
(d) 2 \(\overrightarrow{AD}\)
Solution:
(a) \(\overrightarrow{0}\)

Question 50.
If \(\hat{i}, \hat{j}, \hat{k}\) are unit vectors along three mutually perpendicualr directions, then
(a) \(\hat{i} \cdot \hat{j}\) = 1
(b) \(\hat{i} \times \hat{j}\) = 1
(c) \(\hat{i} \cdot \hat{k}\) = 1
(d) \(\hat{i} \times \hat{k}\) = 0
Solution:
(c) \(\hat{i} \cdot \hat{k}\) = 1

Since \(\hat{i}, \hat{j}, \hat{k}\) are unit vectors along three mutually ⊥ directions.
∴ \(\hat{i} \cdot \hat{i}=1=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}\)
and \(\hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{i}\) = 0

Question 51.
The area of a triangle formed by vertices O, A and B where \(\overrightarrow{\mathrm{OA}}=\hat{i}+2 \hat{j}+3 \hat{k}\) and \(\overrightarrow{\mathrm{OB}}=-3 \hat{i}-2 \hat{j}+\hat{k}\) is
(a) 3√5 sq. units
(b) 5√5 sq. units
(c) 6√5 sq. units
(d) 4 sq. units
Solution:
(a) 3√5 sq. units

Question 52.
The vectors \(3 \hat{i}-\hat{j}+2 \hat{k}\), \(2 \hat{i}+\hat{j}+3 \hat{k}\) and \(\hat{i}+\lambda \hat{j}-\hat{k}\) are coplanar if the value of λ is
(a) – 2
(b) 0
(c) 2
(d) any real number
Solution:
(a) – 2

Let \(\vec{a}=3 \hat{i}-\hat{j}+2 \hat{k}\) ;
\(\vec{b}=2 \hat{i}+\hat{j}+3 \hat{k}\)
and \(\vec{c}=\hat{i}+\lambda \hat{j}-\hat{k}\)
Since it is given that \(\vec{a}, \vec{b} \text { and } \vec{c}\) are coplanar.
∴ \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]\) = 0
⇒ \(\left|\begin{array}{rrr}
3 & -1 & 2 \\
2 & 1 & 3 \\
1 & \lambda & -1
\end{array}\right|\) = 0 ;
Expanding along R₁
⇒ 3 (- 1 – 3λ) + 1 (- 2 – 3) + 2 (2λ – 1) = 0
⇒ – 5λ – 10 = 0
⇒ λ = – 2.