Effective ISC Maths Class 12 Solutions Chapter 1 Vectors Chapter Test can help bridge the gap between theory and application.

## 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 = – $$\frac{146}{17}$$.