#### Need solution for rd sharma maths class 12 chapter 24 vector or cross product Exercises Multiple choice questions question 11 maths textbook solution

Answer: $\frac{2}{\sqrt{7}}$

Given: $\theta$ is the angle between the vectors $2\hat{i}-2\hat{j}+4\hat{k}$ & $3\hat{i}+\hat{j}+2\hat{k}$, then $\sin \theta$

Hint: Using $\sin \theta =\frac{\overrightarrow{a}\times \overrightarrow{b}}{\left | \overrightarrow{a}\times \overrightarrow{b} \right |}$

Solution:

$\vec{a}=2 \hat{i}-2 \hat{j}+4 \hat{k} ; \vec{b}=3 \hat{i}+\hat{j}+2 \hat{k} \\ \Rightarrow|\vec{a}|=\sqrt{2^{2}+(-2)^{2}+(4)^{2}}=\sqrt{4+4+16}=\sqrt{24} \\ \Rightarrow|\vec{b}|=\sqrt{(3)^{2}+(1)^{2}+(2)^{2}}=\sqrt{9+1+4}=\sqrt{14} \$

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

$=\hat{i}(-4-4)-\hat{j}(4-12)+\hat{k}(2+6) \\$

$=-8 \hat{i}+8 \hat{j}+8 \hat{k} \\$

$|\vec{a} \times \vec{b}|=\sqrt{(-8)^{2}+(8)^{2}+(8)^{2}}=\sqrt{64+64+64}=\sqrt{192} \\ \qquad$

$\because \sin \theta=\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}=\frac{\sqrt{192}}{\sqrt{24} \sqrt{14}}=\frac{\sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3}}{\sqrt{2 \times 2 \times 2 \times 3} \sqrt{2 \times 7}}=\frac{\sqrt{4}}{\sqrt{7}}=\frac{2}{\sqrt{7}}$

$\qquad \sin \theta=\frac{2}{\sqrt{7}}$