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

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

Question 1.
Show that the points with position vectors \(2 \hat{i}+6 \hat{j}+3 \hat{k}\), \(\hat{i}+2 \hat{j}+\hat{k}\) and \(3 \hat{i}+10 \hat{j}+5 \hat{k}\) are collinear.
Solutions:
Let A, B and C are the points whose vectors are
\(2 \hat{i}+6 \hat{j}+3 \hat{k}\) ; \(\hat{i}+2 \hat{j}+\hat{k}\) and \(3 \hat{i}+10 \hat{j}+5 \hat{k}\)
∴ \(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V. of A
= \((\hat{i}+2 \hat{j}+\hat{k})-(2 \hat{i}+6 \hat{j}+3 \hat{k})\)
= \(-\hat{i}-4 \hat{j}-2 \hat{k}\)
\(\overrightarrow{\mathrm{AC}}\) = P.V. of C – P.V. of A
= \(3 \hat{i}+10 \hat{j}+5 \hat{k}-(2 \hat{i}+6 \hat{j}+3 \hat{k})\)
= \(\hat{i}+4 \hat{j}+2 \hat{k}\)
⇒ \(\overrightarrow{\mathrm{AC}}=-\overrightarrow{\mathrm{AB}}\)
⇒ vectors \(\overrightarrow{\mathrm{AC}} \text { and } \overrightarrow{\mathrm{AB}}\) are collinear.
i.e. vectors \(\overrightarrow{\mathrm{AC}} \text { and } \overrightarrow{\mathrm{AB}}\) have same or parallel supports.
Also \(\overrightarrow{\mathrm{AC}} \text { and } \overrightarrow{\mathrm{AB}}\) are coinitial vectors and have same supports.
Thus A, B, C are collinear.

Question 2.
Let \(\vec{a}=\hat{i}+2 \hat{j}\), \(\vec{b}=-2 \hat{i}+\hat{j}\) and \(\vec{c}=4 \hat{i}+3 \hat{j}\). Find scalars x and y such that \(\vec{c}=x \vec{a}+y \vec{b}\).
Solution:
Given \(\vec{a}=\hat{i}+2 \hat{j}\) ;
\(\vec{b}=-2 \hat{i}+\hat{j}\) ;
\(\vec{c}=4 \hat{i}+3 \hat{j}\)
Given \(\vec{c}=x \vec{a}+y \vec{b}\)
⇒ \(4 \hat{i}+3 \hat{j}=x(\hat{i}+2 \hat{j})+y(-2 \hat{i}+\hat{j})\)
⇒ \(4 \hat{i}+3 \hat{j}=(x-2 y) \hat{i}+(2 x+y) \hat{j})\)
[∵ two vectors are equal iff their corresponding components are equal]
x – 2y = 4
2x + y = 3
Multiplying eqn. (2) by 2 and adding to eqn. (1) ; we have
5x = 10
⇒ x = 2
∴ from (1);
y = – 1.

Question 3.
Show that the triangle ABC whose vertices A, B, C are \(7 \hat{j}+10 \hat{k}\), \(-\hat{i}+6 \hat{j}+6 \hat{k}\) and \(-4 \hat{i}+9 \hat{j}+6 \hat{k}\) is isosceles and right-angled. Also find the length of the median from
Solution:
Given P.V of A = \(7 \hat{j}+10 \hat{k}\) ;
P.V. of B = \(-\hat{i}+6 \hat{j}+6 \hat{k}\)
P.V. of C = \(-4 \hat{i}+9 \hat{j}+6 \hat{k}\)
Thus, the sides of triangle ABC are represented by \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{BC}} \text { and } \overrightarrow{\mathrm{CA}}\).

Question 4.
If position vectors of points A and B are \(-\hat{i}+\hat{j}+\hat{k}\) and \(-\hat{i}+2 \hat{j}-2 \hat{k}\) respectively, find the position vector of a point C which divides the join of A and B in the ratio 2 : 3
(i) internally
(ii) externally
Solution:
So P.V of point C which divides AB ¡n the ratio 2 : 3 internally

Question 5.
O is the origin and \(\overrightarrow{\mathrm{OA}}\) = (1, 2), \(\overrightarrow{\mathrm{OB}}\) = (3, 4) and \(\overrightarrow{\mathrm{OC}}\) = (x, – 2). Find by using vector method
(i) x so that points A, B, C are collinear.
(ii) x so that points A, B, C form a triangle right angled at B.
Solution:
Given \(\overrightarrow{\mathrm{OA}}=\hat{i}+2 \hat{j}\) ;
\(\overrightarrow{\mathrm{OB}}=3 \hat{i}+4 \hat{j}\) ;
\(\overrightarrow{\mathrm{OC}}=x \hat{i}-2 \hat{j}\)

(i) \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}\)
= \((3 \hat{i}+4 \hat{j})-(\hat{i}+2 \hat{j})\)
= \(2 \hat{i}+2 \hat{j}\)

\(\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}\)
= \((x \hat{i}-2 \hat{j})-(\hat{i}+2 \hat{j})\)
= \((x-1) \hat{i}-4 \hat{j}\)
Since AB and CD are collinear.
∴ \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{AC}}\) for some non-zero scalar λ.
\(2 \hat{i}+2 \hat{j}=\lambda[(x-1) \hat{i}-4 \hat{j}]\)
⇒ 2 = λ (x – 1) ;
2 = – 4λ
λ = – \(\frac{1}{2}\)
2 = – \(\frac{1}{2}\) (x – 1)
4 = – x + 1
⇒ x = – 3.

(ii) Since ∆ ABC be right angled at B.
∴ \(\overrightarrow{\mathrm{AB}} \perp \overrightarrow{\mathrm{BC}}\)
⇒ \(\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{BC}}\) = 0
⇒ \((2 \hat{i}+2 \hat{j}) \cdot(\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}})\) = 0
⇒ \((2 \hat{i}+2 \hat{j}) \cdot[(x-3) \hat{i}-6 \hat{j}]\) = 0
⇒ 2 (x – 3) + 2 (- 6) = 0
⇒ 2x – 18 = 0
⇒ x = 9.

Question 6.
If \(|\vec{a}|\) = 3 and – 2 ≤ k ≤ 2, then what can be said about the vector k \(\vec{a}\) ?
Solution:
Given \(|\vec{a}|\) = 3
\(|k \vec{a}|=|k||\vec{a}|\) = 3 |k|
Since – 2 ≤ k ≤ 2
⇒ 0 ≤ |k| ≤ 2
⇒ 0 ≤ |k| |\(\vec{a}\)| ≤ 2 |\(\vec{a}\)|
⇒ 0 ≤ |k \(\vec{a}\)| ≤ 6
[∵ \(|\vec{a}|\) = 3]
∴ k \(\vec{a}\) be a zero vector or collinear with \(\vec{a}\) s.t. 0 < |k \(\vec{a}\)| ≤ 6

Question 7.
If the position vectors of the three vertices A, B and C of a triangle are \(2 \hat{i}+3 \hat{j}+4 \hat{k}\), \(3 \hat{i}+4 \hat{j}+2 \hat{k}\) and \(4 \hat{i}+2 \hat{j}+3 \hat{k}\) respectively, then prove that ∆ ABC is equilateral.
Solution:
Given P.V of A = \(2 \hat{i}+3 \hat{j}+4 \hat{k}\)
P.V of B = \(3 \hat{i}+4 \hat{j}+2 \hat{k}\)
P.Vof C = \(4 \hat{i}+2 \hat{j}+3 \hat{k}\)
Thus the sides of ∆ ABC are represented by \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{BC}} \text { and } \overrightarrow{\mathrm{CA}}\)
Now, \(\overrightarrow{\mathrm{AB}}\) = P.V of B – P.V of A
= \((3 \hat{i}+4 \hat{j}+2 \hat{k})-(2 \hat{i}+3 \hat{j}+4 \hat{k})\)
= \(\hat{i}+\hat{j}-2 \hat{k}\)
\(\overrightarrow{\mathrm{BC}}\) = P.V of C – P.V of B
= \((4 \hat{i}+2 \hat{j}+3 \hat{k})-(3 \hat{i}+4 \hat{j}+2 \hat{k})\)
= \(\hat{i}-2 \hat{j}+\hat{k}\)

Question 8.
If \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\), \(\vec{b}=2 \hat{i}+3 \hat{j}\), \(\vec{c}=3 \hat{i}+5 \hat{j}-2 \hat{k}\), \(\vec{d}=\hat{k}-\hat{j}\), then prove that the vectors \((\vec{b}-\vec{a})\) and \((\vec{d}-\vec{c})\) are parallel. Also find the ratio of their radii.
Solution:
Given \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\) ;
\(\vec{b}=2 \hat{i}+3 \hat{j}\) ;
\(\vec{c}=3 \hat{i}+5 \hat{j}-2 \hat{k}\)
and \(\vec{d}=\hat{k}-\hat{j}\)

Question 9.
If A(2, 1, 3), B (2, – 3, 4) and C (- 1, 2, 7), find a point D such that \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{CD}}\) are parallel.
Solution:
Let the coordinates of point D be (α, β, γ)
such that \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{CD}}\) are parallel.
∴ \(\overrightarrow{\mathrm{AB}}=\lambda \overrightarrow{\mathrm{CD}}\)
for some scalar λ ≠ 0 …………..(1)
Here P.V of A = \(2 \hat{i}+\hat{j}+3 \hat{k}\) ;
P.V. of B = \(2 \hat{i}-3 \hat{j}+4 \hat{k}\)
and P.Vof C = \(-\hat{i}+2 \hat{j}+7 \hat{k}\)
∴ \(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V. of A
= \((2 \hat{i}-3 \hat{j}+4 \hat{k})-(2 \hat{i}+\hat{j}+3 \hat{k})\)
= \(-4 \hat{j}+\hat{k}\)

\(\overrightarrow{\mathrm{CD}}\) = P.V. of D – P.V. of C
= \((\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k})-(-\hat{i}+2 \hat{j}+7 \hat{k})\)
= \((\alpha+1) \hat{i}+(\beta-2) \hat{j}+(\gamma-7) \hat{k}\)

∴ from (1) ; we have
\(-4 \hat{j}+\hat{k}=\lambda[(\alpha+1) \hat{i}+(\beta-2) \hat{j}+(\gamma-7) \hat{k}]\)
∴ 0 = λ (α + 1) ……………(2)
– 4 = λ (β – 2) ……………….(3)
and 1 = λ (γ – 7) …………………(4)
from (2) ;
α + 1 = 0 (∵ λ ≠ 0)
⇒ α = – 1
from (3) ;
– \(\frac{4}{\lambda}\) = β – 2
⇒ β = 2 – \(\frac{4}{\lambda}\)
from (4) ;
\(\frac{1}{\lambda}\) = γ – 7
γ = \(\frac{1}{\lambda}\) + 7
On taking λ = 1 ;
α = – 1 ;
β = 2 – 4 = – 2
and γ = 1 + 7 = 8
Thus the required point D be D (- 1, – 2, 8).
For different values of λ, we get different points at which \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{CD}}\) are parallel.

Question 10.
(i) Find a unit vector in the direction of \(\vec{a}-\vec{b}\), where \(\vec{a}=-\hat{i}+\hat{j}+\hat{k}\) and \(\vec{b}=2 \hat{i}+\hat{j}-3 \hat{k}\).
(ii) If \(\vec{a}=\hat{i}+\hat{j}\), \(\vec{b}=\hat{j}+\hat{k}\) and \(\vec{c}=\hat{i}+\hat{k}\), find a unit vector in the direction of \(\vec{a}-2 \vec{b}+3 \vec{c}\).
Solution:
(i) Given \(\vec{a}=-\hat{i}+\hat{j}+\hat{k}\) ;
\(\vec{b}=2 \hat{i}+\hat{j}-3 \hat{k}\)

(ii)

Question 11.
If the vectors \(4 \hat{i}+\alpha \hat{j}+\hat{k}\) and \(\beta \hat{i}+2 \hat{j}+\hat{k}\) are parallel, find the values of α and β.
Solution:
Let \(\vec{a}=4 \hat{i}+\alpha \hat{j}+\hat{k}\)
and \(\vec{b}=\beta \hat{i}+2 \hat{j}+\hat{k}\)
Since \(\vec{a} \text { and } \vec{b}\) are parallel.
∴ \(\vec{a}=\lambda \vec{b}\) for some scalar λ ≠ 0
⇒ \((4 \hat{i}+\alpha \hat{j}+\hat{k})=\lambda(\beta \hat{i}+2 \hat{j}+\hat{k})\)
⇒ 4 = λβ ;
α = 2λ ;
1 = λ
∴ α = 2 × 1 = 2 ;
β = \(\frac{4}{1}\) = 4.

Question 12.
If \(\vec{a}=\vec{b}+\vec{c}\), then is it true that \(|\vec{a}|=|\vec{b}|+|\vec{c}|\) ? Justify your answer. (NCERT)
Solution:

Question 13.
If \(\vec{a}=\hat{i}-\hat{j}\) and \(\vec{b}=-\hat{j}+2 \hat{k}\), then find \((\vec{a}+\vec{b}) \cdot(\vec{a}-2 \vec{b})\).
Solution:
Given \(\vec{a}=\hat{i}-\hat{j}\) ;
\(\vec{b}=-\hat{j}+2 \hat{k}\)
\(\vec{a}+\vec{b}=(\hat{i}-\hat{j})+(-\hat{j}+2 \hat{k})\)
= \(\hat{i}-2 \hat{j}+2 \hat{k}\)
\(\vec{a}-2 \vec{b}=\hat{i}-\hat{j}-2(-\hat{j}+2 \hat{k})\)
= \(\hat{i}+\hat{j}-4 \hat{k}\)
\((\vec{a}+\vec{b}) \cdot(\vec{a}-2 \vec{b})\)
= \((\hat{i}-2 \hat{j}+2 \hat{k}) \cdot(\hat{i}+\hat{j}-4 \hat{k})\)
= 1 (1) – 2 (+ 1) + 2 (- 4)
= 1 – 2 – 8
= – 9.

Question 14.
(i) Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\). (NCERT)
(ii) A unit vector \(\vec{a}\) makes angles \(\frac{\pi}{4}\) and \(\frac{\pi}{3}\) with x-axís andy-axis respectively and an acute angle θ with z-axis, then find θ and the (scalar and vector) components of \(\vec{a}\) along the axes. (NCERT)
Solution:
(i) Let α, β, γ be the angle made by given vector with the ±ve direction of x-axis, y-axis and z-axis
Since the given vector is equally inclined to coordinate axes.
∴ α = β = γ
Let < l, m, n > be the d’cosines of the given vector.
∴ l = cos α;
m = cos α;
n = cos α
also, l² + m² + n² = 1
⇒ 3 cos² α = 1
⇒ cos α = \(\frac{1}{\sqrt{3}}\) [∵ 0 ≤ α ≤ π]
∴ direction cosines of vector < l, m, n >
i.e. < \(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\) >

(ii) Let l, m, n be the direction cosines of the vector \(\vec{a}\)
Then l = cos \(\frac{\pi}{4}\)
= \(\frac{1}{\sqrt{2}}\) ;
m = cos \(\frac{\pi}{3}\) = \(\frac{1}{2}\) ;
n = cos θ
Since vector \(\vec{a}\) makes angles \(\frac{\pi}{4}\), \(\frac{\pi}{3}\) and θ with coordinate axes.
Since l² + m² + n² = 1
⇒ \(\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{2}\right)^2\) + cos² θ = 1
⇒ cos² θ = 1 – \(\frac{3}{4}=\left(\frac{1}{2}\right)^2\)
⇒ cos θ = \(\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{3}\)
Since θ be acute angle.
∴ Scalar components of given vector are l, m, n
i.e. \(\frac{1}{\sqrt{2}}\), \(\frac{1}{2}\), cos \(\frac{\pi}{3}\) i.e. \(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}\)
and vector components of given vector are : \(\frac{\hat{i}}{\sqrt{2}}, \frac{\hat{j}}{2}, \frac{\hat{k}}{2}\).

Question 15.
Find unit vectors perpendicular to each of the vectors \(\hat{i}+\hat{j}-\hat{k}\) and \(-\hat{i}+2 \hat{j}+\hat{k}\).
Solution:
Let \(\vec{a}=\hat{i}+\hat{j}-\hat{k}\)
and \(\vec{b}=-\hat{i}+2 \hat{j}+\hat{k}\)
Since \(\vec{a} \times \vec{b}\) is ⊥ to both \(\vec{a} \text { and } \vec{b}\)

Question 15 (old).
If \(\vec{a}, \vec{b}, \vec{c}\) are mutually perpendicular vectors of equal magnitude, show that the vector \(\vec{a}+\vec{b}+\vec{c}\) is equally inclined to \(\vec{a}, \vec{b} \text { and } \vec{c}\).
Solution:
Since \(\vec{a}, \vec{b}, \vec{c}\) are mutually ⊥vector of equal magnitude.

⇒ a² = √3 a² cos α
⇒ cos α = \(\frac{1}{\sqrt{3}}\)
Now, \(\vec{b} \cdot(\vec{a}+\vec{b}+\vec{c})=|\vec{b}||\vec{a}+\vec{b}+\vec{c}|\) cos β
⇒ \(\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{b} \cdot \vec{c}=|\vec{b}||\vec{a}+\vec{b}+\vec{c}|\) cos β
⇒ a² = a . √3 a cos β
⇒ cos β = \(\frac{1}{\sqrt{3}}\)
Similarly cos γ = \(\frac{1}{\sqrt{3}}\)
Thus, cos α = cos β = cos γ = \(\frac{1}{\sqrt{3}}\)
⇒ α = β = γ
Therefore, the vector \(\vec{a}+\vec{b}+\vec{c}\) is equally inclined to \(\vec{a}, \vec{b} \text { and } \vec{c}\).

Question 16.
Prove that the triangle whose vertices A, B and C have position vectors \(2 \hat{i}+4 \hat{j}-\hat{k}\), \(4 \hat{i}+5 \hat{j}+\hat{k}\) and \(3 \hat{i}+6 \hat{j}-3 \hat{k}\) is an isosceles right-angled triangle.
Solution:
Given P.V. of A = \(2 \hat{i}+4 \hat{j}-\hat{k}\) ;
P.V. of B = \(4 \hat{i}+5 \hat{j}+\hat{k}\)
and P.V. of C = \(3 \hat{i}+6 \hat{j}-3 \hat{k}\)
\(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V. of A
= \((4 \hat{i}+5 \hat{j}+\hat{k})-(2 \hat{i}+4 \hat{j}-\hat{k})\)
= \(2 \hat{i}+\hat{j}+2 \hat{k}\)

\(\overrightarrow{\mathrm{CA}}\) = P.V. of A – P.V. of C
= \((2 \hat{i}+4 \hat{j}-\hat{k})-(3 \hat{i}+6 \hat{j}-3 \hat{k})\)
= \(-\hat{i}-2 \hat{j}+2 \hat{k}\)

Here,
\(\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{CA}}=(2 \hat{i}+\hat{j}+2 \hat{k}) \cdot(-\hat{i}-2 \hat{j}+2 \hat{k})\)
= 2 (- 1) + 1 (- 2) + 2 (2) = 0
∴ \(\overrightarrow{\mathrm{AB}} \perp \overrightarrow{\mathrm{CA}}\)
Thus, ∆ABC be a right angled ∆ and right angled at A.
Here \(|\overrightarrow{\mathrm{AB}}|=\sqrt{2^2+1^2+2^2}\)
= \(\sqrt{4+1+4}\) = 3
and \(|\overrightarrow{\mathrm{CA}}|=\sqrt{(-1)^2+(-2)^2+2^2}\)
= \(\sqrt{9}\) = 3
Thus, \(|\overrightarrow{\mathrm{AB}}|=|\overrightarrow{\mathrm{CA}}|\)
Hence, ∆ABC is right angled isosceles triangle.

Question 17.
Find the projection of \(\overrightarrow{\mathrm{AB}} \text { on } \overrightarrow{\mathrm{PQ}}\) where P, Q, A and B are the points (- 2, 1, 3), (0, 2, 5), (4, – 3, 0) and (7, – 5, – 1) respectively.
Solution:
Given P.V. of P = \(-2 \hat{i}+\hat{j}+3 \hat{k}\) ;
P.V. of Q = \(0 \hat{i}+2 \hat{j}+5 \hat{k}\)
P.V. of A = \(4 \hat{i}-3 \hat{j}\) ;
P.V. of B = \(7 \hat{i}-5 \hat{j}-\hat{k}\)
∴ \(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V. of A
= \((7 \hat{i}-5 \hat{j}-\hat{k})-(4 \hat{i}-3 \hat{j})\)
= \(3 \hat{i}-2 \hat{j}-\hat{k}\)

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

\(\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{PQ}}\) = \((3 \hat{i}-2 \hat{j}-\hat{k}) \cdot(2 \hat{i}+\hat{j}+2 \hat{k})\)
= 3 (2) – 2 (1) – 1 (2)
= 6 – 4 = 2

∴ \(|\overrightarrow{\mathrm{PQ}}|=\sqrt{2^2+1^2+2^2}\) = 3
∴ required projection of \(\overrightarrow{\mathrm{AB}}\) on \(\overrightarrow{\mathrm{PQ}}\) = \(\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{PQ}}}{|\overrightarrow{\mathrm{PQ}}|}=\frac{2}{3}\).

Question 18.
The resultant of two vectors \(\vec{a} \text { and } \vec{b}\) is perpendicular to \(\vec{a}\). If \(|\vec{b}|=\sqrt{2}|\vec{a}|\), show that the resultant of \(2 \vec{a} \text { and } \vec{b}\) is perpendicular to \(\vec{b}\).
Solution:
Since resultant of \(\vec{a} \text { and } \vec{b}\)
i.e. \(\vec{a}+\vec{b}\) is ⊥ to \(\vec{a}\).

Question 19.
Let \(\vec{a}=2 \hat{i}+3 \hat{j}\), \(\vec{b}=-\hat{i}+3 \hat{j}+\hat{k}\) and \(\vec{c}=\hat{i}+2 \hat{j}+5 \hat{k}\), find
(i) \(\vec{a} \times \vec{b}\)
(ii) \(\vec{b} \times \vec{b}\)
(iii) \((\vec{a}-\vec{b}) \times(\vec{c}-\vec{a})\)
Solution:
Given \(\vec{a}=2 \hat{i}+3 \hat{j}\) ;
\(\vec{b}=-\hat{i}+3 \hat{j}+\hat{k}\) ;
\(\vec{c}=\hat{i}+2 \hat{j}+5 \hat{k}\)

(i) \(\vec{a} \times \vec{b}\) = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & 0 & -1 \\
-1 & -1 & 5
\end{array}\right|\)
= \(\hat{i}(0-1)-\hat{j}(15-1)+\hat{k}(3-0)\)
= \(-\hat{i}-14 \hat{j}-3 \hat{k}\)

(ii) \(\vec{b} \times \vec{b}\)
= \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 3 & 1 \\
-1 & 3 & 1
\end{array}\right|\)
= 0

(iii)

Question 20.
If \(\vec{a}=2 \hat{i}-3 \hat{j}\) and \(\vec{b}=\hat{i}+\hat{j}-\hat{k}\), find \(\vec{a} \times \vec{b}\). Verify that \(\vec{a} \times \vec{b}\) is perpendicular to both the vectors \(\vec{a} \text { and } \vec{b}\).
Solution:
Given \(\vec{a}=2 \hat{i}-3 \hat{j}\) ;
\(\vec{b}=\hat{i}+\hat{j}-\hat{k}\)

Question 20 (old).
If \(|\vec{a}+\vec{b}|\) = 60, \(|\vec{a}-\vec{b}|\) = 40 and \(|\vec{a}|\) = 22, find \(|\vec{b}|\).
Solution:

Question 21.
The two adjacent sides of a parallelogram are represented by the vectors \(\vec{a}=2 \hat{i}-4 \hat{j}+5 \hat{k}\) and \(\vec{b}=\hat{i}-2 \hat{j}-3 \hat{k}\). Find unit vectors parallel to its diagonals. Also, find its area. (NCERT)
Solution:
Let \(\vec{a}=2 \hat{i}-4 \hat{j}+5 \hat{k}\)
and \(\vec{b}=\hat{i}-2 \hat{j}-3 \hat{k}\)
Thus one of the diagonal of ||gm lies along \(\vec{a}+\vec{b}\) and other along \(\vec{b}-\vec{a}\)

Question 22.
If the vectors \(\vec{a}, \vec{b} \text { and } \vec{c}\) are \(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\), \(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\) and \(c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\) respectively, then show that \(\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}\).
Solution:
Given \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\) ;
\(\vec{b}=b_1 \hat{j}+b_2 \hat{j}+b_3 \hat{k}\)
and \(\vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\)

Question 23.
If \(|\vec{a}|\) = 13, \(|\vec{b}|\) = 5 and \(\vec{a} \cdot \vec{b}\) = 60, find \(|\vec{a} \times \vec{b}|\).
Solution:
Given \(|\vec{a}|\) = 13,
\(|\vec{b}|\) = 5
and \(\vec{a} \cdot \vec{b}\) = 60
⇒ \(|\vec{a}||\vec{b}|\) cos θ = 60
⇒ 13 × 5 cos θ = 60
⇒ cos θ = \(\frac{60}{65}=\frac{12}{13}\) ;
∴ sin θ = \(\sqrt{1-\cos ^2 \theta}\)
= \(\sqrt{1-\frac{144}{169}}=\frac{5}{13}\)
Thus \(|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta\)
[∵ \(|\hat{n}|\) = 1]
= 13 × 5 × \(\frac{5}{13}\) = 25

Question 24.
If the points A, B, C and D have 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 \(-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}\) respectively, then show that ABCD is a rectangle. Also, find its area.
Solution:

∴ opposite sides are equal and parallel.
\(\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{AD}}=(2 \hat{i}) \cdot(-\hat{j})\)
= \(-2(\hat{i} \cdot \hat{j})\) = 0
∴ \(\overrightarrow{\mathrm{AB}} \perp \overrightarrow{\mathrm{AD}}\) ; \(\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{BC}}\) = 0
Also \(\overrightarrow{\mathrm{DC}} \cdot \overrightarrow{\mathrm{AD}}\) = 0 = \(\overrightarrow{\mathrm{DC}} \cdot \overrightarrow{\mathrm{BC}}\)
Hence ABCD be a rectangle.
∴ area of rectangle = \(|\overrightarrow{\mathrm{AB}}||\overrightarrow{\mathrm{AD}}|\)
= \(|2 \hat{i}||-\hat{j}|\)
= 2 × 1
= 2 sq. units

Question 25.
If the volume of the parallelopiped whose coterminus edges are represented by the vectors \(5 \hat{i}-4 \hat{j}+\hat{k}\), \(4 \hat{i}+3 \hat{j}+\lambda \hat{k}\) and \(\hat{i}-2 \hat{j}+7 \hat{k}\) is 216 cubic units, find the value of λ.
Solution:
We know that, volume of parallelopiped whose caterminus edges are represented by vectors \(5 \hat{i}-4 \hat{j}+\hat{k}\) is equal to \([\vec{a}, \vec{b}, \vec{c}]\).
Let \(\vec{a}=5 \hat{i}-4 \hat{j}+\hat{k}\) ;
\(\vec{b}=4 \hat{i}+3 \hat{j}+\lambda \hat{k}\)
and \(\vec{c}=\hat{i}-2 \hat{j}+7 \hat{k}\)
∴ Volume of parallelopiped = \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]\)
= \(\left|\begin{array}{rrr}
5 & -4 & 1 \\
4 & 3 & \lambda \\
1 & -2 & 7
\end{array}\right|\)
= 5 (21 + 2λ) + 4 (28 – λ) + 1 (- 8 – 3)
= (+ 206 + 6λ) cubic units
also given volume of parallelopiped = 216 cubic units
∴ 206 + 6λ = 216
⇒ 6λ = 10
⇒ λ = \(\frac{5}{3}\).

Question 26.
Using scalar triple product, prove that the points (- 1, 4, – 3), (3, 2, – 5), (- 3, 8, – 5) and (- 3, 2, 1) are coplanar.
Solution:
Given points A, B, C, D are coplanar if any one of triod of vectors are coplanar.
i.e. \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{AD}}\) ; \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{CD}}\) etc.
Now \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{AD}}\) are coplanar iff \(\left[\begin{array}{lll}
\overrightarrow{\mathrm{AB}} & \overrightarrow{\mathrm{AC}} & \overrightarrow{\mathrm{AD}}
\end{array}\right]\) = 0

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

\(\overrightarrow{\mathrm{AC}}\) = P.V. of C – P.V. of A
= \((-3 \hat{i}+8 \hat{j}-5 \hat{k})-(-\hat{i}+4 \hat{j}-3 \hat{k})\)
= \(-2 \hat{i}+4 \hat{j}-2 \hat{k}\)

\(\overrightarrow{\mathrm{AD}}\) = P.V. of D – P.V. of A
= \((-3 \hat{i}+2 \hat{j}+\hat{k})-(-\hat{i}+4 \hat{j}-3 \hat{k})\)
= \(-2 \hat{i}-2 \hat{j}+4 \hat{k}\)

Here, \([\overrightarrow{\mathrm{AB}} \overrightarrow{\mathrm{AC}} \overrightarrow{\mathrm{AD}}]\) = \(\left|\begin{array}{rrr}
4 & -2 & -2 \\
-2 & 4 & -2 \\
-2 & -2 & 4
\end{array}\right|\)
= 4 (16 – 4) + 2 (- 8 – 4) – 2 (4 + 8)
= 48 – 24 – 24 = 0
Thus \(\overrightarrow{\mathrm{AB}}\), \(\overrightarrow{\mathrm{AC}}\) and \(\overrightarrow{\mathrm{AD}}\) are coplanar vectors.
∴ given points are coplanar.

Question 27.
The position vectors of the points A, B, C and D are \(3 \hat{i}-2 \hat{j}-\hat{k}\), \(2 \hat{i}+3 \hat{j}-4 \hat{k}\), \(-\hat{i}+\hat{j}+2 \hat{k}\) and \(4 \hat{i}+5 \hat{j}+\lambda \hat{k}\) respectively. If the points A, B, C and D lie in a plane, then find the value of λ.
Solution:
Given P.V of A = \(3 \hat{i}-2 \hat{j}-\hat{k}\)
P.V of B = \(2 \hat{i}+3 \hat{j}-4 \hat{k}\)
P.V of C = \(-\hat{i}+\hat{j}+2 \hat{k}\) ;
P.V of D = \(4 \hat{i}+5 \hat{j}+\lambda \hat{k}\)
∴ \(\overrightarrow{\mathrm{AB}}\) = P.V of B – P.V of A
= \((2 \hat{i}+3 \hat{j}-4 \hat{k})-(3 \hat{i}-2 \hat{j}-\hat{k})\)
= \(-\hat{i}+5 \hat{j}-3 \hat{k}\)

\(\overrightarrow{\mathrm{AC}}\) = P.V of C – P.V of A
= \((-\hat{i}+\hat{j}+2 \hat{k})-(3 \hat{i}-2 \hat{j}-\hat{k})\)
= \(-4 \hat{i}+3 \hat{j}+3 \hat{k}\)

\(\overrightarrow{\mathrm{AD}}\) = P.V of D – P.V of A
= \((4 \hat{i}+5 \hat{j}+\lambda \hat{k})-(3 \hat{i}-2 \hat{j}-\hat{k})\)
= \(\hat{i}+7 \hat{j}+(\lambda+1) \hat{k}\)
Since, the points A, B, C and D are lies in a plane.
∴ \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}} \text { and } \overrightarrow{\mathrm{AD}}\) are coplanar.
⇒ \([\overrightarrow{\mathrm{AB}} \overrightarrow{\mathrm{AC}} \overrightarrow{\mathrm{AD}}]\) = 0
⇒ \(\left|\begin{array}{rrc}
-1 & 5 & -3 \\
-4 & 3 & 3 \\
1 & 7 & \lambda+1
\end{array}\right|\) = 0 ;
expanding along R₁
– 1 (3λ + 3 – 21) – 5 (- 4λ – 4 – 3) – 3 (- 28 – 3) = 0
⇒ – 3λ + 18 + 20λ + 25 + 93 = 0
⇒ 17λ + 146 = 0
⇒ λ = – \(\frac{146}{17}\).