OP Malhotra Class 12 Maths Solutions Chapter 22 Vectors (Continued) Ex 22(b)

Well-structured OP Malhotra ISC Class 12 Solutions Chapter 22 Vectors (Continued) Ex 22(b) facilitate a deeper understanding of mathematical principles.

S Chand Class 12 ICSE Maths Solutions Chapter 22 Vectors (Continued) Ex 22(b)

Question 1.
Find $\vec{a}$ × $\vec{b}$ when
(i) $\vec{a}$ = 3 $\hat{i}$ – $\hat{j}$ + $\hat{k}$,
$\vec{b}$ = 2 $\hat{i}$ + $\hat{j}$ – $\hat{k}$
(ii) $\vec{a}$ = 2 $\hat{i}$ – 3 $\hat{j}$ – $\hat{k}$,
$\vec{b}$ = $\hat{i}$ + 4 $\hat{j}$ – 2 $\hat{k}$
(iii) In (ii), verify that $\vec{a}$ × $\vec{b}$ ≠ $\vec{b}$ × $\vec{a}$
(iv) $\vec{a}$ = 2 $\hat{i}$ – $\hat{j}$ + $\hat{k}$, $\vec{b}$ = 3 $\hat{i}$ + 4 $\hat{j}$ – $\hat{k}$,
verify that $\vec{a}$ × $\vec{b}$ is perpendicular to both $\vec{a}$ and $\vec{b}$.
(v) Compute $\vec{a}$ × ($\vec{b}$ × $\vec{c}$),
if $\vec{a}$ = 7 $\hat{i}$ – 2 $\hat{j}$ + 3 $\hat{k}$,
$\vec{b}$ = 2 $\hat{i}$ + 8 $\hat{k}$, $\vec{c}$ = $\hat{i}$ + $\hat{j}$ + $\hat{k}$.
Answer:
(i) $\vec{a}$ = 3 $\hat{i}$ – $\hat{j}$ + $\hat{k}$ and $\vec{b}$ = 2 $\hat{i}$ + $\hat{j}$ – $\hat{k}$
∴ $\vec{a}$ × $\vec{b}$ = $|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
3 -1 1
2 1 -1
\end{array}|$
= $\hat{i}$(1 – 1) – $\hat{j}$(-3 – 2) + $\hat{k}$(3 + 2) = 0
$\hat{i}$ + 5 $\hat{j}$ + 5 $\hat{k}$

(ii) Given $\vec{a}$ = 2 $\hat{i}$ – 3 $\hat{j}$ – $\hat{k}$ and $\vec{b}$ = $\hat{i}$ + 4 $\hat{j}$ – 2 $\hat{k}$
∴ $\vec{a}$ × $\vec{b}$ = $|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
2 -3 -1
1 4 -2
\end{array}|$
= $\hat{i}$(6 + 4) – $\hat{j}$(-4 + 1) + $\hat{k}$(8 + 3)
= 10 $\hat{i}$ + 3 $\hat{j}$ + 11 $\hat{k}$

(iii) Now, $\vec{b}$ × $\vec{a}$
= $|\begin{array}{rrr}\hat{i} \hat{j} \hat{k} 1 4 -2 2 -3 -1\end{array}|$
= $\hat{i}$(-4 – 6) – $\hat{j}$(-1 + 4) + $\hat{k}$(-3 – 8)
= -10 $\hat{i}$ – 3 $\hat{j}$ – 11 $\hat{k}$
Thus $\vec{a}$ × $\vec{b}$ ≠ $\vec{b}$ × $\vec{a}$

(iv) Given $\vec{a}$ = 2 $\hat{i}$ – $\hat{j}$ + $\hat{k}$ and $\vec{b}$ = 3 $\hat{i}$ + 4 $\hat{j}$ – $\hat{k}$
∴ $\vec{a}$ × $\vec{b}$ = $|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
2 -1 1
3 4 -1
\end{array}|$
= $\hat{i}$(1 – 4) – $\hat{j}$(-2 – 3) + $\hat{k}$(8 + 3)
= -3 $\hat{i}$ + 5 $\hat{j}$ + 11 $\hat{k}$
($\vec{a}$ × $\vec{b}$) .$\vec{a}$ = (-3 $\hat{i}$ + 5 $\hat{j}$ + 11 $\hat{k}$).(2 $\hat{i}$ – $\hat{j}$ + $\hat{k}$)
= -3(2) + 5(-1) + 11(1) = -6 – 5 + 11 = 0
($\vec{a}$ × $\vec{b}$).$\vec{b}$ = (-3 $\hat{i}$ + 5 $\hat{j}$ + 11 $\hat{k}$).(3 $\hat{i}$ + 4 $\hat{j}$ – $\hat{k}$)
= (-3) 3 + 5(4) + 11(-1) = 0
Thus, $\vec{a}$ × $\vec{b}$ is ⊥ to $\vec{a}$ and $\vec{b}$.

(v) Given $\vec{a}$ = 7 $\hat{i}$ – 2 $\hat{j}$ + 3 $\hat{k}$;
$\vec{b}$ = 2 $\hat{i}$ + 8 $\hat{k}$ and $\vec{c}$ = $\hat{i}$ + $\hat{j}$ + $\hat{k}$
∴ $\vec{b}$ + $\vec{c}$ = 3 $\hat{i}$ + $\hat{j}$ + 9 $\hat{k}$
Now $\vec{a}$ × ($\vec{b}$ + $\vec{c}$)
= $|\begin{array}{rrr}\hat{i} \hat{j} \hat{k} 7 -2 3 3 1 9\end{array}|$
= $\hat{i}$(-18 – 3) – $\hat{j}$(63 – 9) + $\hat{k}$(7 + 6)
= -21 $\hat{i}$ – 54 $\hat{j}$ + 13 $\hat{k}$

Question 2.
Find |$\vec{a}$ × $\vec{b}$| when
(i) $\vec{a}$ = $\hat{i}$ + 3 $\hat{j}$ – 2 $\hat{k}$,
$\vec{b}$ = $-\hat{i}$ + 3 $\hat{k}$
(ii) $\vec{a}$ = 2 $\hat{i}$ + $\hat{k}$, $\vec{b}$ = $\hat{i}$ + $\hat{j}$ + $\hat{k}$
Answer:
(i) $\vec{a}$ = $\hat{i}$ + 3 $\hat{j}$ – 2 $\hat{k}$ and $\vec{b}$ = $-\hat{i}$ + 3 $\hat{k}$
∴ $\vec{a}$ × $\vec{b}$ = $|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
1 3 -2
-1 0 3
\end{array}|$
= $\hat{i}$(9) – $\hat{j}$(3 – 2) + $\hat{k}$(3)
= 9 $\hat{i}$ – $\hat{j}$ + 3 $\hat{k}$
Thus |$\vec{a}$ × $\vec{b}$| = $\sqrt{9^2+(-1)^2+3^2}$
= $\sqrt{91}$

(ii) Given $\vec{a}$ = 2 $\hat{i}$ + $\hat{k}$ and $\vec{b}$ = $\hat{i}$ + $\hat{j}$ + $\hat{k}$
∴ $\vec{a}$ × $\vec{b}$ = $|\begin{array}{ccc}
\hat{i} \hat{j} \hat{k}
2 0 1
1 1 1
\end{array}|$
= $\hat{i}$(0 – 1) – $\hat{j}$(2 – 1) + $\hat{k}$(2 – 0)
= $-\hat{i}$ – $\hat{j}$ + 2 $\hat{k}$
Thus, |$\vec{a}$ × $\vec{b}$| = $\sqrt{(-1)^2+(-1)^2+2^2}$
= $\sqrt{1+1+4}$ = $\sqrt{6}$

Question 3.
If $\vec{a}$ = 2 $\hat{i}$ + 3 $\hat{j}$, $\vec{b}$ = $-\hat{i}$ + 3 $\hat{j}$ + $\hat{k}$ and $\vec{c}$ = $\hat{i}$ + 2 $ \hat{j}$ + 5 $\hat{k}$ be three vectors, find
(i) $\vec{a}$ × $\vec{b}$
(ii) $\vec{b}$ × $\vec{c}$
(iii) ($\vec{a}$ × $\vec{b}$) × ($\vec{c}$ × $\vec{b}$)
Answer:
Given $\vec{a}$ = 2 $\hat{i}$ + 3 $\hat{j}$;
$\vec{b}$ = $-\hat{i}$ + 3 $\hat{j}$ + $\hat{k}$ and
$\vec{c}$ = $\hat{i}$ + 2 $\hat{j}$ + 5 $\hat{k}$

(i) ∴ $\vec{a}$ × $\vec{b}$ = $|\begin{array}{rrr}\hat{i} \hat{j} \hat{k} 2 3 0 -1 3 1\end{array}|$
= $\hat{i}$(3 – 0) – $\hat{j}$(2 – 0) + $\hat{k}$(6 + 3)
= 3 $\hat{i}$ – 2 $\hat{j}$ + 9 $\hat{k}$

(ii) $\vec{b}$ × $\vec{c}$ = $|\begin{array}{rrr}\hat{i} \hat{j} \hat{k} -1 3 1 1 2 5\end{array}|$
= $\hat{i}$(15 – 2) – $\hat{j}$(-5 – 1) + $\hat{k}$(-2 – 3)
= 13 $\hat{i}$ + 6 $\hat{j}$ – 5 $\hat{k}$

(iii) $\vec{a}$ – $\vec{b}$ = (2 $\hat{i}$ + 3 $\hat{j}$) – ($-\hat{i}$ + 3 $\hat{j}$ + $\hat{k}$)
= 3 $\hat{i}$ + 0 $\hat{j}$ – $\hat{k}$
$\vec{c}$ – $\vec{b}$ = ($\hat{i}$ + 2 $\hat{j}$ + 5 $\hat{k}$) – ($-\hat{i}$ + 3 $\hat{j}$ + $\hat{k}$)
= 2 $\hat{i}$ – $\hat{j}$ + 4 $\hat{k}$
∴($\vec{a}$ – $\vec{b}$) × ($\vec{c}$ – $\vec{b}$)
= $|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
3 0 -1
2 -1 4
\end{array}|$
= $\hat{i}$(0 – 1) – $\hat{j}$(12 + 2) + $\hat{k}$(-3 – 0)
= $-\hat{i}$ – 14 $\hat{j}$ – 3 $\hat{k}$

Question 4.
$\vec{a}$ = $\sqrt{26}$, $|\vec{b}|$ = 7 and |$\vec{a}$ × $\vec{b}$| = 35, find $\vec{a}$. $\vec{b}$.
Answer:
Given |$\vec{a}$| = $\sqrt{26}$,
|$\vec{b}$| = 7 and
|$\vec{a}$ × $\vec{b}$| = 35
⇒ |$\vec{a}$||$\vec{b}$|
sin θ = 35
⇒ $\sqrt{26}$× 7 × sinθ = 35
⇒ sinθ = $\frac{5}{\sqrt{26}}$
∴ cosθ = $\sqrt{1-\sin ^2 \theta}$
= $\sqrt{1-\frac{25}{26}}$
= $\frac{1}{\sqrt{26}}$
Thus, $\vec{a}$.$\vec{b}$ = |$\vec{a}$||$\vec{b}$|
cosθ = $\sqrt{26}$ × 7 × $\frac{1}{\sqrt{26}}$ = 7

Question 5.
If $\vec{a}$ = 3 $\hat{i}$ – $\hat{j}$ – 2 $\hat{k}$ and $\vec{b}$ = 2 $\hat{i}$ + 3 $\hat{j}$ + $\hat{k}$ find ($\vec{a}$ + 2 $\vec{b}$) × (2 $\vec{a}$ – $\vec{b}$).
Answer:
Given $\vec{a}$ = 3 $\hat{i}$ – $\hat{j}$ – 2 $\hat{k}$ and $\vec{b}$ = 2 $\hat{i}$ + 3 $\hat{j}$ + $\hat{k}$
∴ $\vec{a}$ + 2 $\vec{b}$
= 3 $\hat{i}$ – $\hat{j}$ – 2 $\hat{k}$ + 2(2 $\hat{i}$ + 3 $\hat{j}$ + $\hat{k}$)
= 7 $\hat{i}$ + 5 $\hat{j}$ + 0 $\hat{k}$
2 $\vec{a}$ – $\vec{b}$
= 2(3 $\hat{i}$ – $\hat{j}$ – 2 $\hat{k}$) – (2 $\hat{i}$ + 3 $\hat{j}$ + $\hat{k}$)
= 4 $\hat{i}$ – 5 $\hat{j}$ – 5 $\hat{k}$
Thus ($\vec{a}$ + 2 $\vec{b}$) × (2 $\vec{a}$ – $\vec{b}$)
= $|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} 7 & 5 & 0 4 & -5 & -5\end{array}|$
= $\hat{i}$(-25 – 0) – $\hat{j}$(-35 – 0) + $\hat{k}$(-35 – 20)
= -25 $\hat{i}$ + 35 $\hat{j}$ – 55 $\hat{k}$

Question 6.
Find a unit vector perpendicular to the plane of the vectors.
(i) -3 $\hat{i}$ + 4 $\hat{k}$ and 4 $\hat{i}$ + 3 $\hat{j}$
(ii) $\vec{a}$ = 2 $\hat{i}$ + $\hat{j}$ + $\hat{k}$ and $\vec{b}$ = $\hat{i}$ + 2 $\hat{j}$ + $\hat{k}$.
Answer:
(i) Let $\vec{a}$ = -3 $\hat{j}$ + 4 $\hat{k}$ and $\vec{b}$ = 4 $\hat{i}$ + 3 $\hat{j}$
∴ $\vec{a}$ × $\vec{b}$ = $|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
0 -3 4
4 3 0
\end{array}|$
= $\hat{i}$(0 – 12) – $\hat{j}$(0 – 16) + $\hat{k}$(0 + 12)
= -12 $\hat{i}$ + 16 $\hat{j}$ + 12 $\hat{k}$
|$\vec{a}$ × $\vec{b}$|
= $\sqrt{(-12)^2+(16)^2+(12)^2}$
= $\sqrt{144+256+144}$
= $\sqrt{544}$
= $\sqrt{16 × 34}$
= 4 $\sqrt{34}$
Thus required unit vector ⊥ to both $\vec{a}$ and $\vec{b}$
= $\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$
= $\frac{-12 \hat{i}+16 \hat{j}+12 \hat{k}}{4 \sqrt{34}}$
= $\frac{-3 \hat{i}}{\sqrt{34}}$ + $\frac{4 \hat{j}}{\sqrt{34}}+\frac{3 \hat{k}}{\sqrt{34}}$

(ii) Given $\vec{a}$ = 2 $\hat{i}$ + $\hat{j}$ + $\hat{k}$ and $\vec{b}$ = $\hat{i}$ + 2 $\hat{j}$ + $\hat{k}$
$\vec{a}$ × $\vec{b}$ =
$|\begin{array}{ccc}
\hat{i} \hat{j} \hat{k}
2 1 1
1 2 1
\end{array}|$ = $\hat{i}$(1 – 2) – $\hat{j}$(2 – 1) + $\hat{k}$(4 – 1)
= $-\hat{i}$ – $\hat{j}$ + 3 $\hat{k}$
and |$\vec{a}$ × $\vec{b}$|
= $\sqrt{(-1)^2+(-1)^2+3^2}$
= $\sqrt{11}$
Thus, unit vector ⊥ to the plane of $\vec{a}$ and $\vec{b}$
= $\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$
= $\frac{-\hat{i}-\hat{j}+3 \hat{k}}{\sqrt{11}}$
= $\frac{-1}{\sqrt{11}}(\hat{i}+\hat{j}-3 \hat{k})$

Question 7.
(i) Find the vector of magnitude 9, which is perpendicular to both the vectors
4 $\hat{i}$ – $\hat{j}$ + 3 $\hat{k}$ and 2 $\hat{i}$ + $\hat{j}$ – 2 $\hat{k}$
(ii) Find a vector whose length is 3 and which is perpendicular to both the vectors
$\vec{a}$ = 3 $\hat{i}$ + $\hat{j}$ + 3 $\hat{k}$ and -2 $\hat{i}$ + $\hat{j}$ – 2 $\hat{k}$
Answer:
(i) Let $\vec{a}$ = 4$\hat{i}$ – $\hat{j}$ + 3 $\hat{k}$ and
$\vec{b}$ = -2 $\hat{i}$ + $\hat{j}$ – 2 $\hat{k}$
∴ $\vec{a}$ × $\vec{b}$ = $|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
4 -1 3
-2 1 -2
\end{array}|$
= $\hat{i}$(2 – 3) – $\hat{j}$(-8 + 6) + $\hat{k}$(4 – 2)
= $-\hat{i}$ + 2 $\hat{j}$ + 2 $\hat{k}$
|$\vec{a}$ × $\vec{b}$|
= $\sqrt{(-1)^2+2^2+2^2}$ = 3
Thus the unit vector ⊥ to both $\vec{a}$ 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 required vector of magnitude 9 which is ⊥ to $\vec{a}$ and $\vec{b}$.
= 9($\frac{-\hat{i}+2 \hat{j}+2 \hat{k}}{3}$)
= -3 $\hat{i}$ + 6 $\hat{j}$ + 6$\hat{k}$

(ii) Given $\vec{a}$ = 3 $\hat{i}$ + $\hat{j}$ – 4 $\hat{k}$ and
$\vec{b}$ = 6 $\hat{i}$ + 5 $\hat{j}$ – 2 $\hat{k}$
∴ $\vec{a}$ × $\vec{b}$
= $|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
3 1 -4
6 5 -2
\end{array}|$
= $\hat{i}$(-2 + 20) – $\hat{j}$(-6 + 24) + $\hat{k}$(15 – 6)
= 18 $\hat{i}$ – 18 $\hat{j}$ + 9 $\hat{k}$
|$\vec{a}$ × $\vec{b}$|
= $\sqrt{(18)^2+(-18)^2+9^2}$
= $\sqrt{324+324+81}$
= $\sqrt{729}$ = 27
Thus required vector of magnitude 3 and ⊥ to both $\vec{a}$ and $\vec{b}$
= $\frac{3(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|}$
= $\frac{3(18 \hat{i}-18 \hat{j}+9 \hat{k})}{27}$
= 2 $\hat{i}$ – 2 $\hat{j}$ + $\hat{k}$

Question 8.
(i) If the position vectors of the three points A, B, C are respectively $\hat{i}$ + $\hat{j}$ + $\hat{k}$, $\hat{i}$ + 3 $\hat{j}$ – 4 $\hat{k}$ and 7 $\hat{i}$ + 4 $\hat{j}$ + 9 $\hat{k}$, find the unit vector perpendicular to the plane of the triangle ABC.
(ii) Find the unit vectors perpendicular to the plane A B C, when the position vectors of A, B and C are 2 $\hat{i}$ – $\hat{j}$ + $\hat{k}$, $\hat{i}$ + $\hat{j}$ + 2 $\hat{k}$ and $\hat{i}$ + 3 $\hat{k}$ respectively.
(iii) Find a unit vector perpendicular to the plane determiaed by the points P(1, -1, 2), Q(2, 0, -1) and R(0, 2, 1).
Answer:
(i) Given P.V. of A = $\hat{i}$ + $\hat{j}$ + $\hat{k}$;
P.V. of B = 2 $\hat{i}$ + 3 $\hat{j}$ – 4 $\hat{k}$ and
P.V. of C = 7 $\hat{i}$ + 4 $\hat{j}$ + 9 $\hat{k}$
∴ $\overrightarrow{\mathrm{BA}}$
= P.V. of A – P.V of B
= ($\hat{i}$ + $\hat{j}$ + $\hat{k}$) – (2 $\hat{i}$ + 3 $\hat{j}$ – 4 $\hat{k}$)
= $-\hat{i}$ – 2 $\hat{j}$ + 5 $\hat{k}$
$\overrightarrow{\mathrm{BC}}$
= P.V. of C – PV of B
= (7 $\hat{i}$ + 4 $\hat{j}$ + 9 $\hat{k}$) – (2 $\hat{i}$ + 3 $\hat{j}$ – 4 $\hat{k}$)
= 5 $\hat{i}$ – $\hat{j}$ + 13 $\hat{k}$
Thus $\overrightarrow{\mathrm{BA}}$ × $\overrightarrow{\mathrm{BC}}$ be the vector ⊥ to both $\overrightarrow{\mathrm{BA}}$ and $\overrightarrow{\mathrm{BC}}$
i.e. to the plane of ∆ABC.
Now $\overrightarrow{\mathrm{BA}}$ × $\overrightarrow{\mathrm{BC}}$
= $|\begin{array}{rrr}\hat{i} \hat{j} \hat{k} -1 -2 5 5 1 13\end{array}|$
= $\hat{i}$(-26 – 5) – $\hat{j}$(-13 – 25) + $\hat{k}$(-1 + 10)
= -31 $\hat{i}$ + 38 $\hat{j}$ + 9 $\hat{k}$
|$\overrightarrow{\mathrm{BA}}$ × $\overrightarrow{\mathrm{BC}}$|
= $\sqrt{(-31)^2+(38)^2+9^2}$
= $\sqrt{2486}$
Thus required unit vector
= $\frac{\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}}{|\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}|}$
= $\frac{-31 \hat{i}+38 \hat{j}+9 \hat{k}}{\sqrt{2486}}$

(ii) Given PV. of A
= 2 $\hat{i}$ – $\hat{j}$ + $\hat{k}$;
P.V. of B = $\hat{i}$ + $\hat{j}$ + 2 $\hat{k}$ and
P.V. of C = 2 $\hat{i}$ + 3 $\hat{k}$
∴ $\overrightarrow{\mathrm{BA}}$
= P.V. of A – P.V. of B
= (2 $\hat{i}$ – $\hat{j}$ + $\hat{k}$) – ($\hat{i}$ + $\hat{j}$ + 2 $\hat{k}$)
= $\hat{i}$ – 2 $\hat{j}$ – $\hat{k}$
$\overrightarrow{\mathrm{BC}}$ = P.V. of C – P.V. of B
= 2 $\hat{i}$ + 3 $\hat{k}$ – ($\hat{i}$ + $\hat{j}$ + 2 $\hat{k}$)
= $\hat{i}$ – $\hat{j}$ + $\hat{k}$
∴ $\overrightarrow{\mathrm{BA}}$ × $\overrightarrow{\mathrm{BC}}$
= $|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
1 -2 -1
1 -1 1
\end{array}|$
= $\hat{i}$(-2 – 1) – $\hat{j}$(1 + 1) + $\hat{k}$(-1 + 2)
= -3 $\hat{i}$ – 2 $\hat{j}$ + $\hat{k}$
|$\overrightarrow{\mathrm{BA}}$ × $\overrightarrow{\mathrm{BC}}$|
= $\sqrt{(-3)^2+(-2)^2+1^2}$
= $\sqrt{14}$
Since $\overrightarrow{\mathrm{BA}}$ × $\overrightarrow{\mathrm{BC}}$ be the vector ⊥ to both vector $\overrightarrow{\mathrm{BA}}$ and $\overrightarrow{\mathrm{BC}}$ i.e. ⊥ to the plane of ∆ABC
∴ required must vector
= $\frac{\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}}{|\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}|}$
= $\frac{1}{\sqrt{14}}(-3 \hat{i}-2 \hat{j}+\hat{k})$

(iii) Let $\vec{p}$, $\vec{q}$ and $\vec{r}$ respectively be the position vectors of point P, Q and R.
∴ $\vec{p}$ = $\hat{i}$ – $\hat{j}$ + 2 $\hat{k}$ ; $\vec{q}$
= 2 $\hat{i}$ + 0 $\hat{j}$ – $\hat{k}$; $\vec{r}$
= 0 $\hat{i}$ + 2 $\hat{j}$ + $\hat{k}$
$\overrightarrow{\mathrm{QR}}$ = $\vec{r}$ – $\vec{q}$
= 0 $\hat{i}$ + 2 $\hat{j}$ + $\hat{k}$ – 2 $\hat{i}$ – 0 $\hat{j}$ + $\hat{k}$
= -2 $\hat{i}$ + 2 $\hat{j}$ + 2 $\hat{k}$
$\overrightarrow{\mathrm{QP}}$ = $\vec{p}$ – $\vec{q}$
= ($\hat{i}$ – $\hat{j}$ + 2 $\hat{k}$) – (2 $\hat{i}$ – $\hat{k}$)
= $-\hat{i}$ – $\hat{j}$ + 3 $\hat{k}$
Now $\overrightarrow{\mathrm{QR}}$ × $\overrightarrow{\mathrm{QP}}$ is a vector ⊥ to both $\overrightarrow{\mathrm{QR}}$ and $\overrightarrow{\mathrm{QP}}$ i.e. ⊥ to the plane of $\overrightarrow{\mathrm{QR}}$ and $\overrightarrow{\mathrm{QP}}$ i.e. the plane $\mathrm{PQR}$.
∴ $\overrightarrow{\mathrm{QR}}$ × $\overrightarrow{\mathrm{QP}}$
= $|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
-2 2 2
-1 -1 3
\end{array}|$
= $\hat{i}$(6 + 2) – $\hat{j}$(-6 + 2) + $\hat{k}$(2 + 2)
= 8 $\hat{i}$ + 4 $\hat{j}$ + 4 $\hat{k}$
|$\overrightarrow{\mathrm{QR}}$ × $\overrightarrow{\mathrm{QP}}$|
= $\sqrt{8^2+4^2+4^2}$
= $\sqrt{64+16+16}$
= $\sqrt{96}$
= 4 $\sqrt{6}$
∴Required unit vector
= $\frac{\overrightarrow{\mathrm{QR}} × \overrightarrow{\mathrm{QP}}}{|\overrightarrow{\mathrm{QR}} × \overrightarrow{\mathrm{QP}}|}$
= $\frac{8 \hat{i}+4 \hat{j}+4 \hat{k}}{4 \sqrt{6}}$
= $\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}$

Question 9.
Find the sine of the angle between the vectors.
(i) $\vec{a}$ = 3 $\hat{i}$ – 4 $\hat{j}$ + 5 $\hat{k}$,
$\vec{b}$ = $\hat{i}$ – $\hat{j}$ + $\hat{k}$
(ii) $\hat{i}$ + $\hat{j}$ and $\hat{j}$ + $\hat{k}$
Answer:
(i) Given $\vec{a}$ = 3 $\hat{i}$ – 4 $\hat{j}$ + 5 $\hat{k}$ and $\vec{b}$ = $\hat{i}$ – $\hat{j}$ + $\hat{k}$
∴ $\vec{a}$ × $\vec{b}$
= $|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
3 -4 5
1 -1 1
\end{array}|$
= $\hat{i}$(-4 + 5) – $\hat{j}$(3 – 5) + $\hat{k}$(-3 + 4)
= $\hat{i}$ + 2 $\hat{j}$ + $\hat{k}$
⇒|$\vec{a}$ × $\vec{b}$| = $\sqrt{1^2+2^2+1^2}$ = $\sqrt{6}$
∴ |$\vec{a}$| = $\sqrt{3^2+(-4)^2+5^2}$ = $\sqrt{50}$ = 5 $\sqrt{2}$;
|$\vec{b}$| = $\sqrt{1^2+(-1)^2+1^2}$ = $\sqrt{3}$
Let θ be the angle between $\vec{a}$ and $\vec{b}$
∴ sin θ = $\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}$
= $\frac{\sqrt{6}}{5 \sqrt{2} × \sqrt{3}}$ = $\frac{1}{5}$

(ii) Let $\vec{a}$ = $\hat{i}$ + $\hat{j}$ and $\vec{b}$ = $\hat{j}$ + $\hat{k}$
∴ $\vec{a}$ × $\vec{b}$
= $|\begin{array}{ccc}
\hat{i} \hat{j} \hat{k}
1 1 0
0 1 1
\end{array}|$
= $\hat{i}$(1 – 0) – $\hat{j}$(1 – 0) + $\hat{k}$(1 – 0)
= $\hat{i}$ – $\hat{j}$ + $\hat{k}$
Thus, |$\vec{a}$ × $\vec{b}$|
= $\sqrt{1^2+(-1)^2+1^2}$
= $\sqrt{3}$
∴ |$\vec{a}$| = $\sqrt{1^2+1^2}$
= $\sqrt{2}$ ;
|$\vec{b}$| = $\sqrt{1^2+1^2}$ = $\sqrt{2}$
Let θ be the angle between $\vec{a}$ and $\vec{b}$
∴ sin θ = $\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}$
= $\frac{\sqrt{3}}{\sqrt{2} \times \sqrt{2}}$
= $\sqrt{\frac{3}{4}}$

Question 10.
Find the angle between two vectors $\vec{a}$ and $\vec{b}$, if |$\vec{a}$ × $\vec{b}$| = $\vec{a}$ . $\vec{b}$.
Answer:
Let θ be the required angle between $\vec{a}$ and $\vec{b}$ and also given |$\vec{a}$ × $\vec{b}$|
= $\vec{a}$.$\vec{b}$
⇒ |$\vec{a}$||$\vec{b}$|
sin θ = | $\vec{a}$|| $\vec{b}$|
cos θ ⇒ tan θ = 1 [∵|$\vec{a}$| ≠ 0 and |$\vec{b}$| ≠ 0]
⇒ θ = $\frac{\pi}{4}$

Question 11.
If |$\vec{a}$| = 2,|$\vec{b}$| = 7 and $\vec{a}$ × $\vec{b}$
= 3$ \hat{i}$ + 2 $\hat{j}$ + 6 $\hat{k}$, find the angle between $\vec{a}$ and $\vec{b}$.
Answer:
Given |$\vec{a}$| = 2 ;|$\vec{b}$| = 7 and $\vec{a}$ × $\vec{b}$ = 3 $\hat{i}$ + 2 $\hat{j}$ + 6 $\hat{k}$
Let θ be the required angle between $\vec{a}$ and $\vec{b}$
Then sin θ = $\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}$
= $\frac{|3 \hat{i}+2 \hat{j}+6 \hat{k}|}{2 \times 7}$
arrow sinθ = $\frac{\sqrt{3^2+2^2+6^2}}{2 \times 7}$ = $\frac{7}{14}$
= $\frac{1}{2}$
∴ θ = $\frac{\pi}{6}$

Question 12.
If θ is the angle between $\vec{a}$ and $\vec{b}$, then prove that tanθ = $\frac{|\vec{a} \times \vec{b}|}{\vec{a} \cdot \vec{b}}$.
Answer:
Given θ be the angle between $\vec{a}$ and $\vec{b}$
Then
$\vec{a}$.$\vec{b}$ = |$\vec{a}$||$\vec{b}$| cosθ
⇒ cosθ = $\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$
and
sin θ = $\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}$
on dividing eqn. (2) by eqn. (1); we have tanθ = $\frac{|\vec{a} \times \vec{b}|}{\vec{a} \cdot \vec{b}}$

Question 13.
If $\vec{a}$ = 4 $\hat{i}$ – $\hat{j}$ + $\hat{k}$, $\vec{b}$ = $\hat{i}$ + $\hat{j}$ + $\hat{k}$ and $\vec{c}$ = $\hat{i}$ – $\hat{j}$ + $\hat{k}$, verify that,
$\vec{a}$ × ($\vec{b}$ × $\vec{c}$) = ($\vec{a}$ × $\vec{b}$ + $\vec{a}$ × $\vec{c}$) .
Answer:
Given $\vec{a}$ = 4 $\hat{i}$ – $\hat{j}$ + $\hat{k}$;
$\vec{b}$ = $\hat{i}$ + $\hat{j}$ + $\hat{k}$ and
$\vec{c}$ = $\hat{i}$ – $\hat{j}$ + $\hat{k}$
∴ $\vec{b}$ + $\vec{c}$ = ($\hat{i}$ + $\hat{j}$ + $\hat{k}$) + ($\hat{i}$ – $\hat{j}$ + $\hat{k}$) = 2 $\hat{i}$ + 2 $\hat{k}$
L.H.S. = $\vec{a}$ × ($\vec{b}$ + $\vec{c}$)
= $|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
4 -1 1
2 0 2
\end{array}|$
= $\hat{i}$(-2 – 0) – $\hat{j}$(8 – 2) + $\hat{k}$(0 + 2)
= -2 $\hat{i}$ – 6 $\hat{j}$ + 2 $\hat{k}$
∴ $\vec{a}$ × $\vec{b}$ = $|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
4 -1 1
1 1 1
\end{array}|$
= $\hat{i}$(-1 – 1) – $\hat{j}$(4 – 1) + $\hat{k}$(4 + 1)
= -2 $\hat{i}$ – 3 $\hat{j}$ + 5 $\hat{k}$
and $\vec{a}$ × $\vec{c}$
= $|\begin{array}{rrr}\hat{i} \hat{j} \hat{k} 4 -1 1 1 -1 1\end{array}|$
= $\hat{i}$(-1 + 1) – $\hat{j}$(4 – 1) + $\hat{k}$(-4 + 1)
= 0 $\hat{i}$ – 3 $\hat{j}$ – 3 $\hat{k}$
∴ R.H.S. = $\vec{a}$ × $\vec{b}$ + $\vec{a}$ × $\vec{c}$
= (-2 $\hat{i}$ – 3 $\hat{j}$ + 5 $\hat{k}$) + (0 $\hat{i}$ – 3 $\hat{j}$ – 3 $\hat{k}$)
= -2 $\hat{i}$ – 6 $\hat{j}$ + 2 $\hat{k}$
From eqn. (1) and eqn. (2), we have
L.H.S. = R.H.S.

Question 14.
Find the area of the parallelogram whose adjacent sides are
(i) $\hat{i}$ – 3 $\hat{j}$ + $\hat{k}$ and $\hat{i}$ + $\hat{j}$ + $\hat{k}$
(ii) $\hat{i}$ + 2 $\hat{j}$ + 3 $\hat{k}$ and 3 $\hat{i}$ – 2 $\hat{j}$ + $\hat{k}$
(iii) 2 $\hat{i}$ + $\hat{j}$ + 3 $\hat{k}$ and $\hat{i}$ – $\hat{j}$
(iv) 2 $\hat{i}$ and 3 $\hat{j}$.
Answer:
OP Malhotra Class 12 Maths Solutions Chapter 22 Vectors (Continued) Ex 22(b) 1

Question 15.
Find the area of a parallelogram whose diagonals are 3 $\hat{i}$ + 4 $\hat{j}$ and $\hat{i}$ + $\hat{j}$ + $\hat{k}$.
Answer:
Let $\vec{a}$ = 3 $\hat{i}$ + 4 $\hat{j}$ and $\vec{b}$
= $\hat{i}$ + $\hat{j}$ + $\hat{k}$
∴ $\vec{a}$ × $\vec{b}$
= $|\begin{array}{ccc}
\hat{i} \hat{j} \hat{k}
3 4 0
1 1 1
\end{array}|$
= $\hat{i}$(4 – 0) – $\hat{j}$(3 – 0) + $\hat{k}$(3 – 4)
= 4 $\hat{i}$ – 3 $\hat{j}$ – $\hat{k}$
Thus |$\vec{a}$ × $\vec{b}$|
= $\sqrt{4^2+(-3)^2+(-1)^2}$
= $\sqrt{26}$
∴ Required area of parallelogram having diagonals along $\vec{a}$ and $\vec{b}$
= $\frac{1}{2}|\vec{a} \times \vec{b}|$ sq. units
= $\frac{1}{2} \sqrt{26}$ sq. units

Question 16.
If $\vec{a}$ = 2 $\hat{i}$ – 3 $\hat{j}$ + $\hat{k}$, $\vec{b}$ = $-\hat{i}$ + $\hat{k}$, $\vec{c}$ = 2 $\hat{j}$ – $\hat{k}$, find the area of the parallelogram having diagonals $\vec{a}$ + $\vec{b}$ and $\vec{b}$ + $\vec{c}$.
Answer:
Given $\vec{a}$ = 2 $\hat{i}$ – 3 $\hat{j}$ + $\hat{k}$;
$\vec{b}$ = $-\hat{i}$ + $\hat{k}$ and $\vec{c}$
= 2 $\hat{j}$ – $\hat{k}$
∴ $\vec{a}$ + $\vec{b}$ = (2 $\hat{i}$ – 3 $\hat{j}$ + $\hat{k}$) + ($-\hat{i}$ + $\hat{k}$)
= $\hat{i}$ – 3 $\hat{j}$ + 2 $\hat{k}$
$\vec{b}$ + $\vec{c}$ = ($-\hat{i}$ + $\hat{k}$) + (2 $\hat{j}$ – $\hat{k}$)
= $-\hat{i}$ + 2 $\hat{j}$ + 0 $\hat{k}$
Thus, ($\vec{a}$ + $\vec{b}$) × ($\vec{b}$ + $\vec{c}$)
= $|\begin{array}{rrr}\hat{i} \hat{j} \hat{k} 1 -3 2 -1 2 0\end{array}|$
= $\hat{i}$(0 – 4) – $\hat{j}$(0 + 2) + $\hat{k}$(2 – 3)
= -4 $\hat{i}$ – 2 $\hat{j}$ – $\hat{k}$
∴|($\vec{a}$ + $\vec{b}$) × ($\vec{b}$ + $\vec{c}$)|
= $\sqrt{(-4)^2+(-2)^2+(-1)^2}$ = $\sqrt{21}$
Thus required area of parallelogram = $\frac{1}{2}|(\vec{a}+\vec{b}) × (\vec{b}+\vec{c})|$
= $\frac{1}{2} \sqrt{21}$ sq. units.

Question 17.
If $\vec{p}$ and $\vec{q}$ are unit vectors forming an angle of 30°; find the area of the parallelogram having $\vec{a}$ = $\vec{p}$ + $\overrightarrow{2 q}$ and $\vec{b}$ = 2 $\vec{p}$ + $\vec{q}$ as its diagonals.
Answer:
Given $\vec{a}$ = $\vec{p}$ + 2 $\vec{q}$; $\vec{b}$ = 2 $\vec{p}$ + $\vec{q}$
∴ $\vec{a}$ × $\vec{b}$
= ($\vec{p}$ + 2 $\vec{q}$) × (2$\vec{p}$ + $\vec{q}$)
= 2($\vec{p}$ × $\vec{p}$) + $\vec{p}$ × $\vec{q}$ + 4($\vec{q}$ × $\vec{p}$) + 2($\vec{q}$ × $\vec{q}$)
= 2(0) + $\vec{p}$ × $\vec{q}$ – 4($\vec{p}$ × $\vec{q}$) + 2 × 0 [since $\vec{a}$ ×$\vec{a}$
= 0 and $\vec{a}$ × $\vec{b}$
= $-\vec{b}$ × $\vec{a}$
= -3($\vec{p}$ × $\vec{q}$)
Also it is given that |$\vec{p}$| = |$\vec{q}$| = 1 and
$\frac{\pi}{6}$ be the angle between $\vec{p}$ and $\vec{q}$.
∴ |$\vec{p}$ × $\vec{q}$| = |$\vec{p}$|.|$\vec{q}$|
sinθ = 1 × 1 × sin 30° = $\frac{1}{2}$
from (1); |$\vec{a}$ × $\vec{b}$|
= 3|$\vec{p}$ × $\vec{q}$|
= $\frac{3}{2}$
[using (2)]
Thus the required area of parallelogram having diagonals along $\vec{a}$ and $\vec{b}$
= $\frac{1}{2}$|$\vec{a}$ × $\vec{b}$|
= $\frac{3}{4}$ sq. units.

Question 18.
Find the area of the triangle whose adjacent sides are determined by the vectors.
(i) $\overrightarrow{\mathrm{OA}}$ = 3 $\hat{i}$ + 2 $\hat{j}$ – $\hat{k}$, $\overrightarrow{\mathrm{OB}}$ = $\hat{i}$ + 3 $\hat{j}$ + $\hat{k}$
(ii) 3 $\hat{i}$ + 4 $\hat{j}$ and -5 $\hat{i}$ + 7 $\hat{j}$
Answer:
(i) Given $\overrightarrow{\mathrm{OA}}$ = 3 $\hat{i}$ + 2 $\hat{j}$ – $\hat{k}$;
$\overrightarrow{\mathrm{OB}}$ = $\hat{i}$ + 3 $\hat{j}$ + $\hat{k}$
Now $\overrightarrow{\mathrm{OA}}$ × $\overrightarrow{\mathrm{OB}}$
= $|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} 3 & 2 & -1 1 & 3 & 1\end{array}|$
= $\hat{i}$(2 + 3) – $\hat{j}$(3 + 1) + $\hat{k}$(9 – 2)
= 5 $\hat{i}$ – 4 $\hat{j}$ + 7 $\hat{k}$
Thus area of triangle whose adjacent sides along the vector $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OB}}$
= $\frac{1}{2}$|$\overrightarrow{\mathrm{OA}}$ × $\overrightarrow{\mathrm{OB}}$|
= $\frac{1}{2}$ $\sqrt{5^2+(-4)^2+7^2}$ sq. units
= $\frac{1}{2}$ $\sqrt{90}$ sq. units
= $\frac{3}{2}$ $\sqrt{10}$ sq. units

(ii) Let $\vec{a}$ = 3 $\hat{i}$ + 4 $\hat{j}$ ;
$\vec{b}$ = -5 $\hat{i}$ + 7 $\hat{j}$
∴ $\vec{a}$ × $\vec{b}$
= $|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k}
3 & 4 & 0
-5 & 7 & 0
\end{array}|$
= $\hat{i}$(0 – 0) – $\hat{j}$(0 – 0) + $\hat{k}$(21 + 20)
= 0 $\hat{i}$ – 0 $\hat{j}$ + 41 $\hat{k}$
Thus required area of ∆ having sides determined by vectors $\vec{a}$ and $\vec{b}$
= $\frac{1}{2}$|$\vec{a}$ × $\vec{b}$| sq. units
= $\frac{1}{2}$ $\sqrt{0^2+0^2+(41)^2}$
= $\frac{41}{2}$ sq. units.

Question 19.
Find by vector method, the area of triangle A B C whose vertices are
(i) A(1, 3, 2), B(2, -1, 1) and C(-1, 2, 3)
(ii) A(1, 2, 3), B(2, 5, -1) and C(-1, 1, 2).
Answer:
Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be the position vectors of the vertices A, B and C of $\triangle \mathrm{ABC}$ with reference to O as origin.
∴ $\vec{a}$ = $\overrightarrow{\mathrm{OA}}$
= $\hat{i}$ + 3 $\hat{j}$ + 2 $\hat{k}$;
$\vec{b}$ = $\overrightarrow{\mathrm{OB}}$ = 2 $\hat{i}$ – $\hat{j}$ + $\hat{k}$;
$\vec{c}$ = $\overrightarrow{\mathrm{OC}}$ = $-\hat{i}$ + 2 $\hat{j}$ + 3 $\hat{k}$
Thus $\overrightarrow{\mathrm{AB}}$ = $\vec{b}$ – $\vec{a}$
= (2 $\hat{i}$ – $\hat{j}$ + $\hat{k}$) – ($\hat{i}$ + 3 $\hat{j}$ + 2 $\hat{k}$)
= $\hat{i}$ – 4 $\hat{j}$ – $\hat{k}$
$\overrightarrow{\mathrm{AC}}$ = $\vec{c}$ – $\vec{a}$
= ($-\hat{i}$ + 2 $\hat{j}$ + 3 $\hat{k}$) – ($\hat{i}$ + 3 $\hat{j}$ + 2 $\hat{k}$)
= -2 $\hat{i}$ – $\hat{j}$ + $\hat{k}$
∴ $\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}$ = $|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
1 -4 -1
-2 -1 1
\end{array}|$
= $\hat{i}$(-4 – 1) – $\hat{j}$(1 – 2) + $\hat{k}$(-1 – 8)
= -5 $\hat{i}$ + $\hat{j}$ – 9 $\hat{k}$
∴ $|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|$
= $\sqrt{(-5)^2+1^2+(-9)^2}$
= $\sqrt{25+1+81}$
= $\sqrt{107}$
Thus required area of ∆ABC = $\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|$
= $\frac{1}{2}$ $\sqrt{107}$ sq. units.

(ii) Let $\vec{a}$, $\vec{b}$, $\vec{c}$ are the position vectors of the vertices A, B and C of ∆ABC with reference to O as origin.
∴ $\vec{a}$ = $\overrightarrow{\mathrm{OA}}$
= $\hat{i}$ + 2 $\hat{j}$ + 3 $\hat{k}$;
$\vec{b}$ = 2 $\hat{i}$ + 5 $\hat{j}$ – $\hat{k}$;
$\vec{c}$ = $-\hat{i}$ + $\hat{j}$ + 2 $\hat{k}$
Thus $\overrightarrow{\mathrm{AB}}$ = $\vec{b}$ – $\vec{a}$ = (2 $\hat{i}$ + 5 $\hat{j}$ – $\hat{k}$) – ($\hat{i}$ + 2 $\hat{j}$ + 3 $\hat{k}$)
= $-\hat{i}$ + $\hat{j}$ + 2 $\hat{k}$
$\overrightarrow{\mathrm{AC}}$ = $\vec{c}$ – $\vec{a}$
= ($-\hat{i}$ + $\hat{j}$ + 2 $\hat{k}$) – ($\hat{i}$ + 2 $\hat{j}$ + 3 $\hat{k}$)
= 2 $\hat{i}$ – $\hat{j}$ – $\hat{k}$
∴ $\overrightarrow{\mathrm{AB}}$ × $\overrightarrow{\mathrm{AC}}$ = $|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
+1 3 -4
-2 -1 -1
\end{array}|$
= $\hat{i}$(-3 – 4) – $\hat{j}$(-1 – 8) + $\hat{k}$ (-1 + 6)
= -7 $\hat{i}$ + 9 $\hat{j}$ + 5 $\hat{k}$
⇒ |$\overrightarrow{\mathrm{AB}}$ × $\overrightarrow{\mathrm{AC}}$|
= $\sqrt{(-7)^2+9^2+5^2}$ = $\sqrt{155}$
∴ Required area of ∆ABC
= $\frac{1}{2}$|$\overrightarrow{\mathrm{AB}}$ × $\overrightarrow{\mathrm{AC}}$|
= $\frac{1}{2}$ $\sqrt{155}$ sq. units.

Question 20.
Show that the points whose position vectors are given below, are collinear.
(i) $\vec{a}$ – 2 $\vec{b}$ + 3 $\vec{c}$, 2 $\vec{a}$ + 3 $\vec{b}$ – 4 $\vec{c}$ and -7 $\vec{b}$ + 10 $\vec{c}$
(ii) 5 $\hat{i}$ + 6 $\hat{j}$ + 7 $\hat{k}$, 7 $\hat{i}$ – 8 $\hat{j}$ + 9 $\hat{k}$ and 3 $\hat{i}$ + 20 $\hat{j}$ + 5 $\hat{k}$
Answer:
(i) Let $\vec{α}$, $\vec{\beta}$, $\vec{\gamma}$ are the position vectors of given points
Then $\vec{α}$ = $\vec{a}$ – 2 $\vec{b}$ + 3 $\vec{c}$;
$\vec{\beta}$ = 2 $\vec{a}$ + 3 $\vec{b}$ – 4 $\vec{c}$ and
$\vec{\gamma}$ = -7 $\vec{b}$ + 10 $\vec{c}$
Here $\vec{α}$ × $\vec{\beta}$
= ($\vec{a}$ – 2 $\vec{b}$ + 3 $\vec{c}$) × (2 $\vec{a}$ + 3 $\vec{b}$ – 4 $\vec{c}$)
= 2($\vec{a}$ × $\vec{a}$) + 3($\vec{a}$ × $\vec{b}$) – 4($\vec{a}$ × $\vec{c}$) – 4($\vec{b}$ × $\vec{a}$) – 6($\vec{b}$ × $\vec{b}$) + 8($\vec{b}$ × $\vec{c}$) ……………. (1)
= 7($\vec{a}$ × $\vec{b}$) – $\vec{b}$ × $\vec{c}$ + 10($\vec{c}$ × $\vec{a}$)
$\vec{\beta}$ × $\vec{\gamma}$ = (2 $\vec{a}$ + 3 $\vec{b}$ – 4 $\vec{c}$) × (-7 $\vec{b}$ + 10 $\vec{c}$) + 9($\vec{c}$ × $\vec{b}$) – 12($\vec{c}$ × $\vec{b}$)
= -14($\vec{a}$ × $\vec{b}$) + 20($\vec{a}$ × $\vec{c}$) – 21($\vec{b}$ × $\vec{b}$) + 30($\vec{b}$ × $\vec{c}$) + 28($\vec{c}$ × $\vec{b}$) + 40($\vec{c}$ × $\vec{c}$)
= -14($\vec{a}$ × $\vec{b}$) + 2($\vec{b}$ × $\vec{c}$) – 20($\vec{c}$ × $\vec{a}$) ………………… (2)
$\vec{\gamma}$ × $\vec{α}$ = (-7 $\vec{a}$ + 10 $\vec{c}$) × ($\vec{a}$ – 2 $\vec{b}$ + 3 $\vec{c}$)
= -7($\vec{b}$ × $\vec{a}$) + 14($\vec{b}$ × $\vec{b}$) – 21($\vec{b}$ × $\vec{c}$) + 10($\vec{c}$ × $\vec{a}$) – 20($\vec{c}$ × $\vec{b}$) + 30($\vec{c}$ × $\vec{c}$)
= 7($\vec{a}$ × $\vec{b}$) – ($\vec{b}$ × $\vec{c}$) + 10($\vec{c}$ × $\vec{a}$)
on adding (1), (2) and (3); we have
$\vec{a}$ × $\vec{b}$ + $\vec{b}$ × $\vec{c}$ + $\vec{c}$ × $\vec{a}$ = $\overrightarrow{0}$
We know that $\vec{a}$, $\vec{b}$, $\vec{c}$ are collinear points if $\vec{a}$ × $\vec{b}$ +$\vec{b}$ × $\vec{c}$ + $\vec{c}$ × $\vec{a}$
= $\overrightarrow{0}$ Hence given points are collinear.

(ii) Let $\vec{a}$ = 5 $\hat{i}$ + 6 $\hat{j}$ + 7 $\hat{k}$;
$\vec{b}$ = 7 $\hat{i}$ – 8 $\hat{j}$ + 9 $\hat{k}$ and
$\vec{c}$ = 3 $\hat{i}$ + 20 $\hat{j}$ + 5 $\hat{k}$
Here
$\vec{a}$ × $\vec{j}$ = $|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
5 6 7
7 -8 9
\end{array}|$
= $\hat{i}$(54 + 56) – $\hat{j}$(45 – 49) + $\hat{k}$(-40 – 42)
= 110 $\hat{i}$ + 4 $\hat{j}$ – 82 $\hat{k}$
$\vec{b}$ × $\vec{c}$ = $|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
7 -8 9
3 20 5
\end{array}|$
= $\hat{i}$(-40 – 180) – $\hat{j}$(35 – 27) + $\hat{k}$(140 + 24)
= -220 $\hat{i}$ – 8 $\hat{j}$ + 164 $\hat{k}$
and
$\vec{c}$ × $\vec{a}$ = $|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
3 20 5
5 6 7
\end{array}|$
= $\hat{i}$(140 – 30) – $\hat{j}$(21 – 25) + $\hat{k}$(18 – 100)
= -110 $\hat{i}$ + 4 $\hat{j}$ – 82 $\hat{k}$
∴ $\vec{a}$ × $\vec{b}$ + $\vec{b}$ × $\vec{c}$ + $\vec{c}$ × $\vec{a}$ = $\overrightarrow{0}$
Thus given points are collinear.
[since area of δ formed by three points as vertices = $\frac{1}{2}$|$\vec{a}$ × $\vec{b}$ + $\vec{b}$ × $\vec{c}$ + $\vec{c}$ × $\vec{a}$| = 0 ]

Question 21.
Find λ such that $\vec{a}$ = $\hat{i}$ + λ $\hat{j}$ + 3 $\hat{k}$ and $\vec{b}$ = 3 $\hat{i}$ + 2 $\hat{j}$ + 9 $\hat{k}$ are parallel.
Answer:
Given $\vec{a}$ = $\hat{i}$ + λ $\hat{j}$ + 3 $\hat{k}$ and
$\vec{b}$ = 3 $\hat{i}$ + 2 $\hat{j}$ + 9 $\hat{k}$
Here $\vec{a}$ × $\vec{b}$
= $|\begin{array}{ccc}\hat{i} \hat{j} \hat{k} 1 λ 3 3 2 9\end{array}|$
= $\hat{i}$(9 λ – 6) – $\hat{j}$(9 – 9) + $\hat{k}$(2-3 λ)
= (9 λ – 6) $\hat{i}$ + 0 $\hat{j}$ + (2 – 3 λ) $\hat{k}$
Since $\vec{a}$ is parallel to $\vec{b}$
∴ $\vec{a}$ × $\vec{b}$
= $\overrightarrow{0}$ (9 λ – 6) $\hat{i}$ + 0 $\hat{j}$ + (2 – 3 λ) $\hat{k}$
= $\overrightarrow{0}$ = 0 $\hat{i}$ + 0 $\hat{j}$ + 0 $\hat{k}$
⇒ 9 λ = 6
⇒ λ = $\frac{2}{3}$

Question 22.
Find the values of a for which the vectors 3 $\hat{i}$ + 2 $\hat{j}$ + 9 $\hat{k}$ and $\hat{i}$ + a $\hat{j}$ + 3 $\hat{k}$ are
(i) perpendicular
(ii) parallel.
Answer:
(i) Let $\vec{α}$ = 3 $\hat{i}$ + 2 $\hat{j}$ + 9 $\hat{k}$ and $\vec{\beta}$ = $\hat{i}$ + a $\hat{j}$ + 3 $\hat{k}$
Now $\vec{α}$ ⊥ $\vec{\beta}$
∴ $\vec{α}$ . $\vec{\beta}$ = 0
⇒ (3 $\hat{i}$ + 2 $\hat{j}$ + 9 $\hat{k}$) .($\hat{i}$ + 9 $\hat{j}$ + 3 $\hat{k}$) = 0
⇒ 3(1) + 2(a) + 9(3) = 0
⇒ 2 a + 30 = 0
⇒ a = -15

(ii) Now $\vec{α}$ || $\vec{\beta}$
∴ $\vec{α}$ × $\vec{\beta}$
= $\overrightarrow{0}$
⇒ $|\begin{array}{lll}
\hat{i} \hat{j} \hat{k}
3 2 9
1 a 3
\end{array}|$ = 0
⇒ $\hat{i}$(6 – 9 a) + $\hat{j}$(9 – 9) + $\hat{k}$(3 a – 2)
= $\overrightarrow{0}$
= 0 $\hat{i}$ + 0 $\hat{j}$ + 0 $\hat{k}$
⇒ 6 – 9 a = 0
⇒ a = $\frac{2}{3}$

Question 23.
Find a unit vector parallel to the x y plane and perpendicular to the vector 4 $\hat{i}$ – 3 $\hat{j}$ + $\hat{k}$.
Answer:
We know that a vector || to XOY plane is of the form α $\hat{i}$ + β $\hat{j}$
(For XOY plane, z = 0 )
Now α $\hat{i}$ + β $\hat{j}$ is ⊥ to 4 $\hat{i}$ – 3 $\hat{j}$ + $\hat{k}$
∴ (c $\hat{i}$ + β $\hat{j}$).(4 $\hat{i}$ – 3 $\hat{j}$ + $\hat{k}$) = 0
⇒ 4 α – 3β = 0
⇒ β = $\frac{4 α}{3}$
∴$\vec{c}$ = α $\hat{i}$ + β $\hat{j}$
= α $\hat{i}$ + $\frac{4 α}{3}$ $\hat{j}$
= $\frac{α}{3}$(3 $\hat{i}$ + 4 $\hat{j}$)
Hence the required unit vector = $\frac{\frac{α}{3}(3 \hat{i}+4 \hat{j})}{|\frac{α}{3}| \sqrt{3^2+4^2}}$
= ± $\frac{1}{5}$(3 $\hat{i}$ + 4 $\hat{j}$)

Question 24.
By vector method obtain the perpendi- cular distance of the point (6,-4,4) from the line passing through the points (2,1,2) and (3,-1,4).
Answer:
OP Malhotra Class 12 Maths Solutions Chapter 22 Vectors (Continued) Ex 22(b) 2

Question 25.
Given, $\vec{a}$ = $\frac{1}{7}$(2 $\hat{i}$ + 3 $\hat{j}$ + 6 $\hat{k}$), $\vec{b}$ = $\frac{1}{7}$(3 $\hat{i}$ – 6 $\hat{j}$ + 2 $\hat{k}$), $\vec{c}$ = $\frac{1}{7}$(6 $\hat{i}$ + 2 $\hat{j}$ – 3 $\hat{k}$), $\hat{i}$, $\hat{j}$, $\hat{k}$ being a right handed orthogonal system of unit vectors in space, show that $\vec{a}$, $\vec{b}$, $\vec{c}$ is also another system of unit vectors mutually perpendicular.
Answer:
OP Malhotra Class 12 Maths Solutions Chapter 22 Vectors (Continued) Ex 22(b) 3

Question 26.
$\vec{a}$, $\vec{b}$, $\vec{c}$ represnets three sides $\overrightarrow{B C}$, $\overrightarrow{C A}$ and $\overrightarrow{A B}$ of a ∆A B C such that |$\vec{a}$| = 13,
|$\vec{b}$|= 14, |$\vec{c}$|= 15 Find < C.
Answer:
Given |$\vec{a}$| = 13 ;
|$\vec{b}$| = 14 and |$\vec{c}$| = 15
∴ a² = |$\vec{a}$|² = 169 ;
b² = |$\vec{b}$|² = 196 and c² = |$\vec{c}$|² = 225
Then by cosine’s formula, we have
cos C= $\frac{a^2+b^2-c^2}{2 a b}$
= $\frac{169+196-225}{2 \times 13 \times 14}$
∴ cos C = $\frac{365-225}{2 \times 13 \times 14}$
= $\frac{140}{2 \times 13 \times 14}$
= $\frac{5}{13}$
⇒ c = cos^-1 ($\frac{5}{13}$)

Question 27.
$\vec{a}$, $\vec{b}$, $\vec{c}$ are three vectors such that $\vec{a}$.$\vec{b}$ = $\vec{b}$ $ \vec{c}$ ; $\vec{a}$ × $\vec{b}$ = $\vec{a}$ × $\vec{c}$, $\vec{a}$ ≠ 0 prove that $\vec{b}$ ≠ $\vec{c}$.
Answer:
Given $\vec{a}$.$\vec{b}$ = $\vec{a}$.$\vec{c}$
⇒ $\vec{a}$.$\vec{b}$ – $\vec{a}$.$\vec{c}$ = $\overrightarrow{0}$
⇒ $\vec{a}$ .($\vec{b}$ – $\vec{c}$) = 0
⇒ $\vec{a}$ =$\overrightarrow{0}$ or $\vec{b}$ – $\vec{c}$
= $\overrightarrow{0}$ or $\vec{a}$ ⊥ $\vec{b}$ – $\vec{c}$
but $\vec{a}$ ≠ $\overrightarrow{0}$
∴ $\vec{b}$ – $\vec{c}$ = $\overrightarrow{0}$ or $\vec{a}$ ⊥$\vec{b}$ – $\vec{c}$
Also given $\vec{a}$ × $\vec{b}$ = $\vec{a}$ × $\vec{c}$
⇒ $\vec{a}$ × ($\vec{b}$ – $\vec{c}$) = $\overrightarrow{0}$
⇒ either $\vec{a}$ = $\overrightarrow{0}$ or $\vec{b}$ – $\vec{c}$
= $\overrightarrow{0}$ or $\vec{a}$ ||
$\vec{b}[latex] – [latex]\vec{c}$
since $\vec{a}$ ≠$\overrightarrow{0} $
∴ $\vec{b}$ – $\vec{c}$ = $\overrightarrow{0}$ or $\vec{a}$ || $\vec{b}$ – $\vec{c}$
Now $\vec{a}[latex] can never be ⊥ and || to [latex]\vec{b}$ – $\vec{c}$
simultaneously
[From eqn. (1) and eqn. (2)]
Thus, $\vec{b}$ – $\vec{c}$ = 0
⇒ $\vec{b}$ = $\vec{c}$

Question 28.
If $\vec{r}$ = x $\hat{i}$ + y $\hat{j}$ + z $\hat{k}$ then show that
(i) ($\vec{r}$ × $\hat{i}$)² = y² + z²
(ii) ($\vec{r}$ × $\hat{i}$) .($\vec{r}$ × $\hat{j}$)+x y=0
Answer:
Given $\vec{r}$ = x $\hat{i}$ + y $\hat{j}$ + z $\hat{k}$
(i) $\vec{r}$ × $\hat{i}$ = (x $\hat{i}$ + y $\hat{j}$ + z $\hat{k}$) × $\hat{i}$
= x($\hat{i}$ × $\hat{i}$) + y($\hat{j}$ × $\hat{i}$) + z($\hat{k}$ × $\hat{i}$)
= x(0) + y($-\hat{k}$) + z($\hat{j}$)
= +z $\hat{j}$ – y $\hat{k}$
∴ ($\vec{r}$ × $\hat{i}$)²
= |$\vec{r}$ × $\hat{i}$|²
= ($\sqrt{z^2+(-y)^2}$)²
= y² + z²

(ii) $\vec{r}$ × $\hat{j}$ = (x $\hat{i}$ + y $\hat{j}$ + z $\hat{k}$) × $\hat{j}$
= x($\hat{i}$ × $\hat{j}$) + y($\hat{j}$ × $\hat{j}$) + z($\hat{k}$ × $\hat{j}$)
= x $\hat{k}$ + y(0) +z($-\hat{i}$)
= x $\hat{k}$ – z $\hat{i}$
Thus, ($\vec{r}$ × $\hat{i}$).($\vec{r}$ × $\hat{j}$)
= (z $\hat{j}$ – y $\hat{k}$).(-z $\hat{i}$ – x $\hat{k}$)
= 0(-z) + z(0) – y(x)
⇒ ($\vec{r}$ × $\hat{i}$).($\vec{r}$ × $\hat{j}$) + x y = 0

Question 29.
For any vector $\vec{a}$, prove that |$\vec{a}$ × $\hat{i}$|² + |$\vec{a}$ × $\hat{j}$|² + |$\vec{a}$ × $\hat{k}$|² = 2|$\vec{a}$|²
Answer:

OP Malhotra Class 12 Maths Solutions Chapter 22 Vectors (Continued) Ex 22(b) 4

OP Malhotra Class 12 Maths Solutions Chapter 22 Vectors (Continued) Ex 22(b)

S Chand Class 12 ICSE Maths Solutions Chapter 22 Vectors (Continued) Ex 22(b)

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