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ML Aggarwal Class 12 Maths Solutions Section B Chapter 1 Vectors Ex 1.3

Very short answer type questions (1 to 13) :

Question 1.
(i) If $$\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}$$ and $$\vec{b}=3 \hat{i}+5 \hat{j}-2 \hat{k}$$, then find $$|\vec{a} \times \vec{b}|$$. (NCERT)
(ii) Find the magnitude of $$\vec{a}$$, where $$\vec{a}=(\hat{i}+3 \hat{j}-2 \hat{k}) \times(-\hat{i}+3 \hat{k})$$.
(iii) If $$\vec{a}=3 \hat{i}-5 \hat{j}$$, $$\vec{b}=6 \hat{i}+3 \hat{j}$$ are two vectors and $$\vec{c}$$ is a vector such that $$\vec{c}=\vec{a} \times \vec{b}$$, then find $$|\vec{a}|:|\vec{b}|:|\vec{c}|$$.
Solution:
(i) ∴ $$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2 \end{array}\right|$$
= $$\hat{i}(-2-15)-\hat{j}(-4-9)+\hat{k}(10-3)$$
= $$-17 \hat{i}+13 \hat{j}+7 \hat{k}$$
Thus $$|\vec{a} \times \vec{b}|=\sqrt{(-17)^2+(13)^2+7^2}$$
= $$=\sqrt{289+169+49}=\sqrt{507}$$

(ii) $$\vec{a}=(\hat{i}+3 \hat{j}-2 \hat{k}) \times(-\hat{i}+3 \hat{k})$$
∴ $$\vec{a}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -2 \\ -1 & 0 & 3 \end{array}\right|$$
= $$\hat{i}(9-0)-\hat{j}(3-2)+\hat{k}(0+3)$$
= $$9 \hat{i}-\hat{j}+3 \hat{k}$$
∴ $$|\vec{a}|=\sqrt{9^2+(-1)^2+3^2}$$
= $$\sqrt{81+1+9}=\sqrt{91}$$

(iii) Given $$\vec{a}=3 \hat{i}-5 \hat{j}$$ ;
$$\vec{b}=6 \hat{i}+3 \hat{j}$$
∴ $$\vec{c}=\vec{a} \times \vec{b}$$
= $$\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & -5 & 0 \\ 6 & 3 & 0 \end{array}\right|$$
= $$\hat{i}(0-0)-\hat{j}(0-0)+\hat{k}(9+30)$$
= 39 $$\hat{k}$$
Thus $$|\vec{c}|=|39 \hat{k}|$$
= 39 × 1 = 39
$$|\vec{a}|=\sqrt{3^2+(-5)^2}$$
= $$\sqrt{9+25}=\sqrt{34}$$
$$|\vec{b}|=\sqrt{6^2+3^2}$$
= $$\sqrt{36+9}=\sqrt{45}$$
Thus, $$|\vec{a}|:|\vec{b}|:|\vec{c}|=\sqrt{34}: \sqrt{45}: 39$$.

Question 1 (old).
(ii) If $$\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}$$ and $$\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}$$, then find $$|\vec{a} \times \vec{b}|$$.
Solution:
Given $$\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}$$
and $$\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}$$
∴ $$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{array}\right|$$
= $$\hat{i}(0)-\hat{j}(2-21)+\hat{k}(-2+21)$$
= $$19 \hat{j}+19 \hat{k}$$
∴ $$|\vec{a} \times \vec{b}|=\sqrt{(19)^2+(19)^2}=19 \sqrt{2}$$

Question 2.
(i) Find the angle between two vectors $$\vec{a} \text { and } \vec{b}$$ with magnitudes 1 and 2 respectively when $$|\vec{a} \times \vec{b}|=\sqrt{3}$$.
(ii) If vectors $$\vec{a} \text { and } \vec{b}$$ are such that $$|\vec{a}|=3,|\vec{b}|=\frac{2}{3}$$ and $$\vec{a} \times \vec{b}$$ is a unit vector then find the angle between $$\vec{a} \text { and } \vec{b}$$.
(iii) If $$|\vec{a}|$$ = 2, $$|\vec{b}|$$ = 7 and $$\vec{a} \times \vec{b}=3 \hat{i}+2 \hat{j}+6 \hat{k}$$, find the angle between $$\vec{a} \text { and } \vec{b}$$.
Solution:
(i) Given $$|\vec{a}|$$ = 1 ;
$$|\vec{b}|$$ = 2
also, $$\sqrt{3}=|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}|$$ sin θ
[∵ $$|\hat{\eta}|$$ = 1]
where θ be the angle between $$\vec{a} \text { and } \vec{b}$$
⇒ √3 = 1 × 2 × sin θ
⇒ sin θ = $$\frac{\sqrt{3}}{2}$$
⇒ θ = $$\frac{\pi}{3}$$

(ii) Given $$|\vec{a}|$$ =3 ;
$$|\vec{b}|$$ = $$\frac{2}{3}$$
and $$\vec{a} \times \vec{b}$$ is a unit vector
or $$\vec{a} \times \vec{b}$$ = 1
⇒ $$|\vec{a}||\vec{b}|$$ sin θ = 1,
where θ be the angle between $$\vec{a} \text { and } \vec{b}$$
⇒ 3 × $$\frac{2}{3}$$ sin θ = 1
sin θ = $$\frac{1}{2}$$
θ = $$\frac{\pi}{6}$$

(iii) Given $$|\vec{a}|$$ = 2 ;
$$|\vec{b}|$$ = 7
and $$\vec{a} \times \vec{b}=3 \hat{i}+2 \hat{j}+6 \hat{k}$$
= $$\sqrt{9+4+36}$$
= $$\sqrt{49}$$ = 7
Let θ be the angle between $$\vec{a} \text { and } \vec{b}$$
Then sin θ = $$\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}$$
∴ sin θ = $$\frac{7}{2 \times 7}=\frac{1}{2}$$
θ = 30°, 150°

Question 3.
Find the angle between two vectors $$\vec{a} \text { and } \vec{b}$$ if $$|\vec{a} \times \vec{b}|=\vec{a} \cdot \vec{b}$$. (NCERT)
Solution:
Let θ be the angle between $$\vec{a} \text { and } \vec{b}$$
Then $$|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}|$$ sin θ
[∵ $$|\hat{\eta}|$$ = 1]
and $$|\vec{a} \cdot \vec{b}|=|\vec{a}||\vec{b}|$$ cos θ
Since $$|\vec{a} \times \vec{b}|=|\vec{a} \cdot \vec{b}|$$ (given)
⇒ $$|\vec{a}||\vec{b}|$$ sin θ = $$|\vec{a}||\vec{b}|$$ cos θ
⇒ tan θ = 1
⇒ θ = $$\frac{\pi}{4}$$

Question 4.
(i) Find $$\vec{a} \cdot \vec{b}$$ if $$|\vec{a}|$$ = 2, $$|\vec{b}|$$ = 5 and $$|\vec{a} \times \vec{b}|$$ = 8.
(ii) If $$|\vec{a}|$$ = 4 ; $$|\vec{b}|$$ = 2 and angle between $$|\vec{a}|$$ and $$|\vec{b}|$$ is $$\frac{\pi}{3}$$, find $$(\vec{a} \times \vec{b})^2$$.
(iii) If vectors $$\vec{a} \text { and } \vec{b}$$ are such that $$|\vec{a}|$$ = $$\frac{1}{2}$$, $$|\vec{b}|=\frac{4}{\sqrt{3}}$$ and $$|\vec{a} \times \vec{b}|=\frac{1}{\sqrt{3}}$$ then find $$|\vec{a} \cdot \vec{b}|$$.
Solution:
(i) Given, $$|\vec{a}|$$ = 2,
$$|\vec{b}|$$ = 5
and $$|\vec{a} \times \vec{b}|$$ = 8
By Lagranges identity, we have
$$|\vec{a} \times \vec{b}|^2=|\vec{a}||\vec{b}|^2-(\vec{a} \cdot \vec{b})^2$$
⇒ 64 = 4 × 25 – $$(\vec{a} \cdot \vec{b})^2$$
⇒ $$(\vec{a} \cdot \vec{b})^2$$ = 36
⇒ $$\vec{a} \cdot \vec{b}$$ = 6 (Taking +ve sign)

(ii) Given $$|\vec{a}|$$ = 4 ;
$$|\vec{b}|$$ = 2
since angle between $$|\vec{a}|$$ and $$|\vec{b}|$$ be $$\frac{\pi}{3}$$

(iii) Given $$|\vec{a}|$$ = $$\frac{1}{2}$$ ;
$$|\vec{b}|=\frac{4}{\sqrt{3}}$$
and $$|\vec{a} \times \vec{b}|=\frac{1}{\sqrt{3}}$$

Question 4 (old).
(ii) If $$|\vec{a}|$$ = 5, $$|\vec{b}|$$ = 13 and $$|\vec{a} \times \vec{b}|$$ = 25, then find $$\vec{a} \cdot \vec{b}$$.
Solution:
Given $$|\vec{a}|$$ = 5,
$$|\vec{b}|$$ = 13
and $$|\vec{a} \times \vec{b}|$$ = 25
If θ be the angle between $$\vec{a} \text { and } \vec{b}$$
sin θ = $$\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}$$
= $$\frac{25}{5 \times 13}=\frac{5}{13}$$
∴ cos θ = $$\sqrt{1-\frac{25}{169}}=\frac{12}{13}$$
∴ $$\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|$$ cos θ
= 5 × 13 × $$\frac{12}{13}$$ = 60.

Question 5.
(i) Find the value of λ if $$(2 \hat{i}+6 \hat{j}+14 \hat{k}) \times(\hat{i}-\lambda \hat{j}+7 \hat{k})=\overrightarrow{0}$$.
(ii) Find the value of p if $$(2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+3 \hat{j}+p \hat{k})=\overrightarrow{0}$$.
(iii) Find λ and µ $$(2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}$$. (NCERT)
Solution:
(i) Let $$\vec{a}=2 \hat{i}+6 \hat{j}+14 \hat{k}$$ ;
$$\vec{b}=(\hat{i}-\lambda \hat{j}+7 \hat{k})$$
∴ $$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 14 \\ 1 & -\lambda & 7 \end{array}\right|$$
= $$\hat{i}(42+14 \lambda)-\hat{j}(14-14)+\hat{k}(-2 \lambda-6)$$
= $$\hat{i}(42+14 \lambda)+0 \hat{j}+(-2 \lambda-6) \hat{k}$$
It is given that $$\vec{a} \times \vec{b}=\overrightarrow{0}$$
⇒ $$(42+14 \lambda) \hat{i}+0 \hat{j}+(-2 \lambda-6) \hat{k}$$ = $$0 \hat{i}+0 \hat{j}+0 \hat{k}$$
∴ 42 + 14λ = 0
⇒ λ = – 3
and – 2λ – 6 = 0
λ = 3

(ii) Let $$\vec{a}=2 \hat{i}+6 \hat{j}+27 \hat{k}$$ ;
$$\vec{b}=\hat{i}+3 \hat{j}+p \hat{k}$$

(iii) Let $$\vec{a}=2 \hat{i}+6 \hat{j}+27 \hat{k}$$
and $$\vec{b}=\hat{i}+\lambda \hat{j}+\mu \hat{k}$$
∴ $$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda & \mu \end{array}\right|$$
= $$\hat{i}(6 \mu-27 \lambda)-\hat{j}(2 \mu-27)+\hat{k}(2 \lambda-6)$$
also $$\vec{a} \times \vec{b}=\overrightarrow{0}$$
∴ $$(6 \mu-27 \lambda) \hat{i}-\hat{j}(2 \mu-27)+\hat{k}(2 \lambda-6)=\overrightarrow{0}=0 \hat{i}+0 \hat{j}+0 \hat{k}$$ ……………….(1)
Thus 6µ – 27λ = 0 ………………………(1)
– (2µ – 27) = 0
⇒ µ = $$\frac{27}{2}$$
and 2λ – 6 = 0
⇒ λ = 3
Also, λ = 3, µ = $$\frac{27}{2}$$ satisfies eqn. (1).

Question 6.
(i) Find a unit vector perpendicular to the two vectors $$\hat{i}+2 \hat{j}-\hat{k}$$ and $$2 \hat{i}+3 \hat{j}+\hat{k}$$. (ISC 2004)
(ii) Find a unit vector perpendicular to $$\vec{a} \text { and } \vec{b}$$, where $$\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}$$ and $$\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}$$.
Solution:
(i) Let $$\vec{a}=\hat{i}+2 \hat{j}-\hat{k}$$ ;
$$\vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}$$

(ii) Given $$\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}$$
and $$\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}$$
∴ $$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{array}\right|$$
= $$\hat{i}(-14+14)-\hat{j}(2-21)+\hat{k}(-2+21)$$
= $$0 \hat{i}+19 \hat{j}+19 \hat{k}$$
Thus unit vector = $$\pm \frac{\vec{a} \times \vec{b}}{|\vec{a}+\vec{b}|}$$
= $$\pm \frac{19(\hat{j}+\hat{k})}{\sqrt{361+361}}$$
= $$\pm \frac{(\hat{j}+\hat{k})}{\sqrt{2}}$$

(ii) Find a unit vector perpendicular to the plane of $$\vec{a} \text { and } \vec{b}$$, where $$\vec{a}=3 \hat{i}+2 \hat{j}+5 \hat{k}$$ and $$\vec{b}=\hat{i}-3 \hat{j}+\hat{k}$$.
Solution:
Given latex]\vec{a}=3 \hat{i}+2 \hat{j}+5 \hat{k}[/latex]
and $$\vec{b}=\hat{i}-3 \hat{j}+\hat{k}$$

Question 7.
(i) What conclusion can you draw about vectors $$\vec{a} \text { and } \vec{b}$$ when $$\vec{a} \times \vec{b}=\overrightarrow{0}$$ and $$\vec{a} \cdot \vec{b}$$ = 0? (NCERT)
(ii) If either $$\vec{a}$$ = 0 or $$\vec{b}$$ = 0, then $$\vec{a} \times \vec{b}=\overrightarrow{0}$$. Is the converse true? Justify your answer with an example. (NCERT)
Solution:
(i) When $$\vec{a} \times \vec{b}=\overrightarrow{0}$$
⇒ $$\vec{a}$$ = 0
or $$\vec{a} \| \vec{b}$$
or $$\vec{a} \cdot \vec{b}$$ = 0
⇒ $$\vec{a}=\overrightarrow{0}$$
or $$\vec{b}=\overrightarrow{0}$$
or $$\vec{a} \perp \vec{b}$$
Now a vector can never be parallel and perpendicular simultaneously.
Thus $$\vec{a}=\overrightarrow{0}$$ or $$\vec{b}=\overrightarrow{0}$$ = 0.

(ii) Given $$\vec{a}=\overrightarrow{0}$$
or $$\vec{b}=\overrightarrow{0}$$
∴ $$\vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \hat{n}$$ = 0
Let $$\vec{a}=\hat{i}$$ ;
b = 2 $$\hat{i}$$
Here 
but neither $$\vec{a}=\overrightarrow{0}$$ nor $$\vec{b}=\overrightarrow{0}$$.

Question 7 (old).
(i) show that $$\vec{a} \times \vec{b}=\vec{a} \times \vec{c}$$ may not imply that $$\vec{b}=\vec{c}$$.
(ii) If $$\vec{b}=\vec{c}$$, is $$\vec{a} \times \vec{b}=\vec{c} \times \vec{a}$$ ?
Solution:
(i)

(ii) Not necessary ;

Question 8.
Find λ such that $$\vec{a}=\hat{i}+\lambda \hat{j}+3 \hat{k}$$ and $$\vec{b}=3 \hat{i}+2 \hat{j}+9 \hat{k}$$ are parallel.
Solution:
Given $$\vec{a}=\hat{i}+\lambda \hat{j}+3 \hat{k}$$
and $$\vec{b}=3 \hat{i}+2 \hat{j}+9 \hat{k}$$
Since $$\vec{a} \text { and } \vec{b}$$ are parallel.
∴ $$\vec{a} \times \vec{b}=\overrightarrow{0}$$ …………….(1)
Here, $$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & \lambda & 3 \\ 3 & 2 & 9 \end{array}\right|$$
= $$\hat{i}(9 \lambda-6)-\hat{j}(9-9)+\hat{k}(2-3 \lambda)$$
= $$(9 \lambda-6) \hat{i}+(2-3 \lambda) \hat{k}$$
∴ from (1) ; we have
$$(9 \lambda-6) \hat{i}+(2-3 \lambda) \hat{k}$$
= $$\overrightarrow{0}=0 \hat{i}+0 \hat{j}+0 \hat{k}$$
∴ 9λ – 6 = 0
⇒ λ = $$\frac{2}{3}$$
and 2 – 3λ = 0
⇒ λ = $$\frac{2}{3}$$

Question 9.
(i) $$(\hat{i} \times \hat{j}) \cdot \hat{k}+\hat{i} \cdot \hat{j}$$
(ii) $$(\hat{j} \times \hat{k}) \cdot \hat{i}+(\hat{i} \times \hat{k}) \cdot \hat{j}+(\hat{i} \times \hat{j}) \cdot \hat{k}$$
Solution:
Since $$\hat{i}, \hat{j}, \hat{k}$$ forms right handed orthogonal system
∴ $$\hat{i} \times \hat{j}=\hat{k}$$
$$\hat{j} \times \hat{k}=\hat{i}$$
$$\hat{k} \times \hat{i}=\hat{j}$$
and $$\hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}$$ = 0

(i) $$(\hat{i} \times \hat{j}) \cdot \hat{k}+\hat{i} \cdot \hat{j}$$
$$\hat{k} \cdot \hat{k}+\hat{i} \cdot \hat{j}$$
= 1 + 0 = 1
[∵ $$\hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}$$ = 1
and $$\hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{i}$$ = 0]

(ii) $$(\hat{j} \times \hat{k}) \cdot \hat{i}+(\hat{i} \times \hat{k}) \cdot \hat{j}+(\hat{i} \times \hat{j}) \cdot \hat{k}$$
$$\hat{i} \cdot \hat{i}+(-\hat{j}) \cdot \hat{j}+(\hat{k} \cdot \hat{k})$$
= 1 – 1 + 1 = 0

Question 10.
Find the area of the parallelogram whose adjacent sides are represented by the vectors
(i) $$3 \hat{i}+\hat{j}+4 \hat{k} \text { and } \hat{i}-\hat{j}+\hat{k}$$ (NCERT)
(ii) $$2 \hat{i}-3 \hat{k} \text { and } 4 \hat{j}+2 \hat{k}$$
Solution:
(i) Let $$\vec{a}=3 \hat{i}+\hat{j}+4 \hat{k}$$
and $$\vec{b}=\hat{i}-\hat{j}+\hat{k}$$

(ii) Let $$\vec{a}=2 \hat{i}-3 \hat{k}$$
and $$\vec{b}=4 \hat{j}+2 \hat{k}$$

Question 11.
Find the area of a triangle whose two sides are represented by the vectors $$3 \hat{i}+\hat{j}+4 \hat{k}$$ and $$\hat{i}-\hat{j}+\hat{k}$$.
Solution:
Let $$\vec{a}=3 \hat{i}+\hat{j}+4 \hat{k}$$
and $$\vec{b}=\hat{i}-\hat{j}+\hat{k}$$
∴ $$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{array}\right|$$
= $$\hat{i}(1+4)-\hat{j}(3-4)+\hat{k}(-3-1)$$

∴ $$\vec{a} \times \vec{b}=5 \hat{i}+\hat{j}-4 \hat{k}$$
Thus, $$|\vec{a} \times \vec{b}|=\sqrt{5^2+1^2+(-4)^2}$$
= $$\sqrt{25+1+16}=\sqrt{42}$$

Therefore, area of triangle = $$\frac{1}{2}|\vec{a} \times \vec{b}|$$
= $$\frac{1}{2} \sqrt{42}$$ sq. units

Question 12.
(i) If $$\vec{a}=4 \hat{i}+3 \hat{j}+2 \hat{k}$$ and $$\vec{b}=3 \hat{i}+2 \hat{k}$$, then find $$|\vec{b} \times 2 \vec{a}|$$.
(ii) If $$\vec{a}=\hat{i}+\hat{j}-3 \hat{k}$$ and $$\vec{b}=\hat{j}+2 \hat{k}$$, then find $$|2 \vec{b} \times \vec{a}|$$.
Solution:
(i) Given, $$\vec{a}=4 \hat{i}+3 \hat{j}+2 \hat{k}$$
and $$\vec{b}=3 \hat{i}+2 \hat{k}$$

(ii) Given, $$\vec{a}=\hat{i}+\hat{j}-3 \hat{k}$$ ;
$$\vec{b}=\hat{j}+2 \hat{k}$$

∴ $$2 \vec{b}=2 \hat{j}+4 \hat{k}$$
$$2 \vec{b} \times \vec{a}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & 4 \\ 1 & 1 & -3 \end{array}\right|$$
= $$\hat{i}(-6-4)-\hat{j}(-4)+\hat{k}(-2)$$
= $$-10 \hat{i}+4 \hat{j}-2 \hat{k}$$

∴ $$|2 \vec{b} \times \vec{a}|=\sqrt{(-10)^2+(4)^2+(-2)^2}$$
= $$\sqrt{100+16+4}$$
= $$\sqrt{120}=2 \sqrt{30}$$

Question 13.
(i) If $$\vec{a}=3 \hat{i}-\hat{j}-2 \hat{k}$$ and $$\vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}$$, then find $$(\vec{a}+2 \vec{b}) \times(2 \vec{a}-\vec{b})$$.
(ii) Taking $$\vec{a}=2 \hat{i}-3 \hat{j}-\hat{k}$$ and $$\vec{b}=\hat{i}+4 \hat{j}-2 \hat{k}$$ verify that $$\vec{a} \times \vec{b} \neq \vec{b} \times \vec{a}$$.
Solution:
(i) Given $$\vec{a}=3 \hat{i}-\hat{j}-2 \hat{k}$$ ;
$$\vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}$$
$$\vec{a}+2 \vec{b}=(3 \hat{i}-\hat{j}-2 \hat{k})+2(2 \hat{i}+3 \hat{j}+\hat{k})$$
= $$7 \hat{i}+5 \hat{j}+0 \hat{k}$$

$$2 \vec{a}-\vec{b}=2(3 \hat{i}-\hat{j}-2 \hat{k})-(2 \hat{i}+3 \hat{j}+\hat{k})$$
$$4 \hat{i}-5 \hat{j}-5 \hat{k}$$

Thus $$(\vec{a}+2 \vec{b}) \times(2 \vec{a}-\vec{b})$$
= $$\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 7 & 5 & 0 \\ 4 & -5 & -5 \end{array}\right|$$
= $$\hat{i}(-25-0)-\hat{j}(-35-0)+\hat{k}(-35-20)$$
= $$-25 \hat{i}+35 \hat{j}-55 \hat{k}$$

(ii) Given $$\vec{a}=2 \hat{i}-3 \hat{j}-\hat{k}$$
and $$\vec{b}=\hat{i}+4 \hat{j}-2 \hat{k}$$

∴ $$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & -1 \\ 1 & 4 & -2 \end{array}\right|$$
= $$\hat{i}(6+4)-\hat{j}(-4+1)+\hat{k}(8+3)$$
= $$10 \hat{i}+3 \hat{j}+11 \hat{k}$$

∴ $$\vec{b} \times \vec{a}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & -2 \\ 2 & -3 & -1 \end{array}\right|$$
= $$\hat{i}(-4-6)-\hat{j}(-1+4)+\hat{k}(-3-8)$$
= $$-10 \hat{i}-3 \hat{j}-11 \hat{k}$$

∴ $$\vec{a} \times \vec{b} \neq \vec{b} \times \vec{a}$$

Question 14.
If θ is the angle between two vectors $$\hat{i}-2 \hat{j}+3 \hat{k}$$ and $$3 \hat{i}-2 \hat{j}+\hat{k}$$, find sin θ.
Solution:
Let $$\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}$$
and $$\vec{b}=3 \hat{i}-2 \hat{j}+\hat{k}$$

Question 15.
If $$\vec{a}=3 \hat{i}+\hat{j}+2 \hat{k}$$ and $$\vec{b}=2 \hat{i}-2 \hat{j}+4 \hat{k}$$, then find :
(i) the sine of the angle between $$\vec{a} \text { and } \vec{b}$$. (NCERT Exemplar)
(ii) a unit vector perpendicular to both the vectors $$\vec{a} \text { and } \vec{b}$$.
Solution:
(i) Given $$\vec{a}=3 \hat{i}+\hat{j}+2 \hat{k}$$
and $$\vec{b}=2 \hat{i}-2 \hat{j}+4 \hat{k}$$
Let θ be the angle between $$\vec{a} \text { and } \vec{b}$$

(ii) Here,
$$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 2 & -2 & 4 \end{array}\right|$$
= $$8 i-8 j-8 k$$
$$|\vec{a} \times \vec{b}|=\sqrt{8^2+(-8)^2+(-8)^2}$$
= 8√3
Thus required unit vector ⊥to both $$\vec{a} \text { and } \vec{b}$$
= $$\pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \pm \frac{8(\hat{i}-\hat{j}-\hat{k})}{8 \sqrt{3}}$$
= ± $$\frac{1}{\sqrt{3}}(\hat{i}-\hat{j}-\hat{k})$$

Question 15 (old).
(i) If $$\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$$, $$\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$$ and $$\vec{c}=2 \hat{i}+3 \hat{j}$$, find $$(\vec{a} \times \vec{b}) \times \vec{c}$$ and $$\vec{a} \times(\vec{b} \times \vec{c})$$, and verify that these are not the same.
(ii) If $$\vec{a}, \vec{b} \text { and } \vec{c}$$ represent the position vectors of the points with coordinates (2, – 10, 2), (3, 1, 2) and (2, 1, 3) respectively, find the value of $$\vec{a} \times(\vec{b} \times \vec{c})$$. (ISC 2009)
Solution:
(i) Given $$\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$$,
$$\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$$
and $$\vec{c}=2 \hat{i}+3 \hat{j}$$

$$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 2 & -1 \end{array}\right|$$
= $$\hat{i}(1-2)-\hat{j}(-2-1)+\hat{k}(4+1)$$
= $$-\hat{i}+3 \hat{j}+5 \hat{k}$$

(ii) Given $$\vec{a}=2 \hat{i}-10 \hat{j}+2 \hat{k}$$,
$$\vec{b}=3 \hat{i}+\hat{j}+2 \hat{k}$$
and $$\vec{c}=2 \hat{i}+\hat{j}+3 \hat{k}$$

∴ $$\vec{b} \times \vec{c}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 2 & 1 & 3 \end{array}\right|$$
= $$\hat{i}(3-2)-\hat{j}(9-4)+\hat{k}(3-2)$$
= $$\hat{i}-5 \hat{j}+\hat{k}$$

∴ $$\vec{a} \times(\vec{b} \times \vec{c})=\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 2 & -10 & 2 \\ 1 & -5 & 1 \end{array}\right|$$
= $$\hat{i}(-10+10)-\hat{j}(2-2)+\hat{k}(-10+10)$$
= $$0 \hat{i}+0 \hat{j}+0 \hat{k}$$
= $$\overrightarrow{0}$$

Question 16.
(i) Find a vector of magnitude 6 units which is perpendicular to both the vectors $$2 \hat{i}-\hat{j}+2 \hat{k}$$ and $$4 \hat{i}-\hat{j}+3 \hat{k}$$ (NCERT Exemplar)
(ii) Find a vector of magnitude 19 which is perpendicular to both the vectors $$-\hat{j}+\hat{k}$$ and $$4 \hat{i}-\hat{j}+8 \hat{k}$$.
Solution:
(i) Let $$\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}$$
and $$\vec{b}=4 \hat{i}-\hat{j}+3 \hat{k}$$

(ii) Let $$\vec{a}=4 \hat{i}-\hat{j}+8 \hat{k}$$
and $$\vec{b}=-\hat{j}+\hat{k}$$

Question 17.
Find a vector of magnitude 6, perpendicular to each of the vectors $$(\vec{a}+\vec{b})$$ and $$(\vec{a}-\vec{b})$$ where $$\vec{a}=\hat{i}+\hat{j}+\hat{k}$$ and $$\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}$$.
Solution:
Given, $$\vec{a}=\hat{i}+\hat{j}+\hat{k}$$
and $$\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}$$

Question 18.
If $$\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}$$ and $$\vec{b}=2 \hat{i}+3 \hat{j}-5 \hat{k}$$, find $$\vec{a} \times \vec{b}$$. Verify that $$\vec{a}$$ and $$\vec{a} \times \vec{b}$$ are perpendicular to each other.
Solution:
Given $$\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}$$
and $$\vec{b}=2 \hat{i}+3 \hat{j}-5 \hat{k}$$
$$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{array}\right|$$
= $$\hat{i}(10-9)-\hat{j}(-5-6)+\hat{k}(3+4)$$
= $$\hat{i}+11 \hat{j}+7 \hat{k}$$

Now $$\vec{a} \cdot(\vec{a} \times \vec{b})$$
= $$(\hat{i}-2 \hat{j}+3 \hat{k}) \cdot(\hat{i}+11 \hat{j}+7 \hat{k})$$
= 1 (1) – 2 (11) + 3 (7) = 0
∴ $$\vec{a} \perp \vec{a} \times \vec{b}$$

Question 18 (old).
Find a unit vector perpendicular to the vectors $$4 \hat{i}+3 \hat{j}+\hat{k}$$ and $$2 \hat{i}-\hat{j}+2 \hat{k}$$. Determine the sine of the angle between these two vectors. (ISC 2007)
Solution:
Given $$\vec{a}=4 \hat{i}+3 \hat{j}+\hat{k}$$
and $$\vec{b}=2 \hat{i}-\hat{j}+2 \hat{k}$$

Question 19.
(i) If $$\vec{a}=\hat{i}-\hat{j}$$, $$\vec{b}=3 \hat{j}-\hat{k}$$ and $$\vec{c}=7 \hat{i}-\hat{k}$$, find a vector $$\vec{d}$$ which is perpendicular to both $$\vec{a} \text { and } \vec{b}$$ and $$\vec{c} \cdot \vec{d}$$ = 1.
(ii) If $$\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}$$, $$\vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$$, and $$\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$$, find a vector $$\vec{d}$$ which is perpendicular to both the vectors $$\vec{a} \text { and } \vec{b}$$ and $$\vec{c} \cdot \vec{d}$$ = 27.
(iii) If $$\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}$$, $$\vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$$ and $$\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$$. Find a vector $$\vec{d$$ which is perpendicular to both $$\vec{a} \text { and } \vec{b}$$ and $$\vec{c} \cdot \vec{d}$$ = 18.
(iv) If $$\vec{a}=4 \hat{i}+5 \hat{j}-\hat{k}$$, $$\vec{b}=\hat{i}-4 \hat{j}+5 \hat{k}$$ and $$\vec{c}=3 \hat{i}+\hat{j}-\hat{k}$$. Find a vector $$\vec{d$$ which is perpendicular to both $$\vec{c} \text { and } \vec{b}$$ and $$\vec{d} \cdot \vec{a}$$ = 21.
Solution:
(i) Given $$\vec{a}=\hat{i}-\hat{j}$$,
$$\vec{b}=3 \hat{j}-\hat{k}$$
and $$\vec{c}=7 \hat{i}-\hat{k}$$

(ii) Given $$\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}$$,
$$\vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$$,
and $$\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$$
Let the required vector be $$\vec{d}=x \hat{i}+y \hat{j}+z \hat{k}$$
such that $$\vec{d}$$ is perpendicular to $$\vec{a} \text { and } \vec{b}$$
∴ $$\vec{d} \cdot \vec{b}$$ = 0
⇒ $$(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+4 \hat{j}+2 \hat{k})$$ = 0
⇒ x + 4y + 7z = 0 …………………(1)
Also $$\vec{d} \cdot \vec{b}$$ = 0
⇒ $$(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+7 \hat{k})$$ = 0
⇒ 3x – 2y + 7z = 0 ………………..(2)
also given $$\vec{c} \cdot \vec{d}$$ = 27
⇒ $$(2 \hat{i}-\hat{j}+4 \hat{k}) \cdot(x \hat{i}+y \hat{j}+z \hat{k})$$
⇒ 2x – y + 4z = 0 ………………..(3)
Multiplying eqn. (2) by 2 and adding to eqn. (1) ; we get
7x + 16z = 0 …………………..(4)
Multiplying eqn. (3) by 2 ; we have
4x – 2y + 8z = 54 …………………….(5)
eqn. (5) – eqn. (2) gives ;
x + z = 54 ……………….(6)
Multiplying eqn. (6) by (7) – eqn. (4) ; gives
⇒ 7z – 16z = + 54 × 7
⇒ – 9z = + 54 × 7
⇒ z = – 42
∴ from eqn. (6) ;
x = 96
∴ from eqn. (3) ;
y = 192 – 168 – 27 = – 3
Thus required vector be $$\vec{d}=96 \hat{i}-3 \hat{j}-42 \hat{k}$$
i.e., $$\vec{d}=3(32 \hat{i}-\hat{j}-14 \hat{k})$$

(iii) Given $$\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}$$,
$$\vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$$
and $$\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$$
Let the required vector be
$$\vec{d}=x \hat{i}+y \hat{j}+z \hat{k}$$ …………………..(1)

Since $$\vec{d}$$ is perpendicular to $$\vec{a} \text { and } \vec{b}$$
∴ $$\vec{d} \cdot \vec{b}$$ = 0
⇒ $$(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+4 \hat{j}+2 \hat{k})$$ = 0
⇒ x + 4y + 2z = 0 ………………….(2)

Also $$\vec{d} \cdot \vec{b}$$ = 0
⇒ $$(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+7 \hat{k})$$ = 0
⇒ 3x – 2y + 7z = 0 ……………………..(3)

given $$\vec{c} \cdot \vec{d}$$ = 18
⇒ $$(2 \hat{i}-\hat{j}+4 \hat{k}) \cdot(x \hat{i}+y \hat{j}+z \hat{k})$$ = 18
⇒ 2x – y + 4z = 18 ……………………..(4)
Multiplying eqn. (3) by (2) and adding to eqn. (2) ; we have
7x + 16z = 0 ……………………(5)
Multiplying eqn. (4) by (2) – eqn. (3) ; we have
x + z = 36 ……………………..(6)
Multiply eqn. (6) by 7 – eqn. (5) ; we have
– 9z = 36 × 7
⇒ z = – 28
∴ from (6) ;
x = 64
∴ from (4) ;
y = 128 – 112 – 18 = – 2
∴ from (1) ;
$$\vec{d}=64 \hat{i}-2 \hat{j}-28 \hat{k}$$
= $$2(32 \hat{i}-\hat{j}-14 \hat{k})$$

(iv) Given $$\vec{a}=4 \hat{i}+5 \hat{j}-\hat{k}$$,
$$\vec{b}=\hat{i}-4 \hat{j}+5 \hat{k}$$
and $$\vec{c}=3 \hat{i}+\hat{j}-\hat{k}$$
Let $$\vec{d}=x \hat{i}+y \hat{j}+z \hat{k}$$
Since $$\vec{d}$$ is perpendicular to $$\vec{c}$$
⇒ $$\vec{d} \text { and } \vec{c}$$ = 0
⇒ $$(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(3 \hat{i}+\hat{j}-\hat{k})$$ = 0
⇒ 3x + y – z = 0 ………………..(1)
Also $$\vec{d}$$ is perpendicular to $$\vec{b$$
⇒ $$\vec{d} \text { and } \vec{b}$$ = 0
i.e. $$(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}-4 \hat{j}+5 \hat{k})$$ = 0
⇒ x – 4y + 5z = 0 …………………..(2)
Also,  = 21
⇒ $$(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(4 \hat{i}+5 \hat{j}-\hat{k})$$ = 21
⇒ 4x + 5y – z = 21 ……………………(3)
Multiply eqn. (1) by 5 and adding to eqn. (2) ; we get
16x + y = 0 ………………….(4)
Multiply eqn. (3) by 5 and adding to eqn. (2) ; we get
21x + 21y = 105
⇒ x + y = 5 ……………………(5)
On solving eqn. (4) and (5) ; we have
x = – $$\frac{1}{3}$$
and y = $$\frac{16}{3}$$

Question 20.
Define $$\vec{a} \times \vec{b}$$ and prove that $$|\vec{a} \times \vec{b}|=(\vec{a} \cdot \vec{b})$$ tan θ, where θ is the angle between $$\vec{a} \text { and } \vec{b}$$.
Solution:
The vector product  is a vector quantity whose magnitude is $$|\vec{a}||\vec{b}|$$ sin θ where θ is the angle between $$\vec{a} \text { and } \vec{b}$$ and whose direction is perpendicular to plane of $$\vec{a} \text { and } \vec{b}$$ in such a way that $$\vec{a}, \vec{b}$$ and this direction forms a right handed system.
∴ $$\vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \hat{n}$$
where $$\hat{n$$ be the unit vector ⊥ to the plane of $$\vec{a} \text { and } \vec{b}$$ s.t. $$\vec{a}, \vec{b}, \hat{n}$$ forms a right handed system.
Since $$|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}|$$ ……………………..(1)
We know that $$\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$$
∴ $$|\vec{a}||\vec{b}|=\frac{\vec{a} \cdot \vec{b}}{\cos \theta}$$
∴ from (1) ;
$$|\vec{a} \times \vec{b}|=\frac{\vec{a} \cdot \vec{b}}{\cos \theta}$$ sin θ
∴ $$|\vec{a} \times \vec{b}|=(\vec{a} \cdot \vec{b})$$ tan θ

Question 21.
In ∆ OAB, $$\overrightarrow{\mathrm{OA}}=3 \hat{i}+2 \hat{j}-\hat{k}$$ and $$\overrightarrow{\mathrm{OB}}=\hat{i}+3 \hat{j}+\hat{k}$$. Find the area of the triangle OAB.
Solution:
Given $$\overrightarrow{\mathrm{OA}}=3 \hat{i}+2 \hat{j}-\hat{k}$$
and $$\overrightarrow{\mathrm{OB}}=\hat{i}+3 \hat{j}+\hat{k}$$

Question 22.
Using vectors, find the area of the triangle whose vertices are : A (3, – 1, 2), B (1, – 1, – 3), C (4, – 3, 1). (ISC 2020)
Solution:

Question 23.
(i) Find the area of a parallelogram whose diagonals are determined by the vectors $$\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}$$ and $$\vec{b}=\hat{i}-3 \hat{j}+4 \hat{k}$$. (ISC 2009)
(ii) The vectors $$-2 \hat{i}+4 \hat{j}+4 \hat{k}$$ and $$-4 \hat{i}-2 \hat{k}$$ represent the diagonals BD and AC of a parallelogram ABCD. Find the area of the parallelogram. (ISC 2006)
Solution:
(i) Given $$\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}$$
and $$\vec{b}=\hat{i}-3 \hat{j}+4 \hat{k}$$
Now $$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{array}\right|$$
= $$\hat{i}(4-6)-\hat{j}(12+2)+\hat{k}(-9-1)$$
= $$-2 \hat{i}-14 \hat{j}-10 \hat{k}$$
∴ Required area of ||gm = $$\frac{1}{2}|\vec{a} \times \vec{b}|$$
= $$\frac{1}{2} \sqrt{4+196+100}=5 \sqrt{3}$$

(ii) Given $$\vec{a}=-2 \hat{i}+4 \hat{j}+4 \hat{k}$$
and $$\vec{b}=-4 \hat{i}-2 \hat{k}$$
Now, $$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -2 & 4 & 4 \\ -4 & 0 & -2 \end{array}\right|$$
= $$\hat{i}(-8)-\hat{j}(4+16)+\hat{k}(16)$$
= $$-8 \hat{i}-20 \hat{j}+16 \hat{k}$$
= $$4(-2 \hat{i}-5 \hat{j}+4 \hat{k})$$
∴ Required area of ||gm = $$\frac{1}{2}|\vec{a} \times \vec{b}|$$
= $$\frac{\sqrt{64+400+256}}{2}$$
= $$\frac{\sqrt{720}}{2}$$
= $$\frac{12 \sqrt{5}}{2}$$
= 6√5

Question 24.
The two adjacent sides of a parallelogram are represented by the vectors $$2 \hat{i}-4 \hat{j}-5 \hat{k}$$ and $$2 \hat{i}+2 \hat{j}+3 \hat{k}$$.
Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram.
Solution:
Given, $$\vec{a}=2 \hat{i}-4 \hat{j}-5 \hat{k}$$
and $$\vec{b}=2 \hat{i}+2 \hat{j}+3 \hat{k}$$

Question 25.
Find the area of the triangle formed by the points A (1, 2, 3), B(2, -1 , 4) and C (4, 5, – 1).
Solution:
Given P.V. of A = $$\hat{i}+2 \hat{j}+3 \hat{k}$$ ;
P.V. of B = $$2 \hat{i}-\hat{j}+4 \hat{k}$$
and P.V. of C = $$4 \hat{i}+5 \hat{j}-\hat{k}$$
∴ $$\overrightarrow{\mathrm{AB}}$$ = P.V. of B – P.V. of A
= $$(2 \hat{i}-\hat{j}+4 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})$$
= $$\hat{i}-3 \hat{j}+\hat{k}$$

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

∴ $$\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{array}\right|$$
= $$\hat{i}(12-3)-\hat{j}(-4-3)+\hat{k}(3+9)$$
= $$9 \hat{i}+7 \hat{j}+12 \hat{k}$$

∴ $$|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{9^2+7^2+12^2}=\sqrt{274}$$
Thus required area of ∆ABC = $$\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|$$
= $$\frac{\sqrt{274}}{2}$$