#### Provide solution for RD Sharma maths class 12 chapter Vector or Cross Product exercise 24.1 question 38

Answer: $\frac{\sqrt{24}}{7}$

Hint: To solve this we use $\sin ^{2} \theta+\cos ^{2} \theta=1$ formula

Given:

\begin{aligned} &\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k} \\\\ &\vec{b}=3 \hat{i}-2 \hat{j}+\hat{k} \end{aligned}

Solution:

\begin{aligned} &\cos \theta=\vec{u} \cdot \vec{v}=|\vec{u}||\vec{v}| \cos \theta \\\\ &\cos \theta=\frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|} \end{aligned}

\begin{aligned} &\cos \theta=\frac{\hat{i}-2 \hat{j}+3 \hat{k} \cdot(3 \hat{i}-2 \hat{j}+\hat{k})}{\sqrt{1^{2}+(-2)^{2}+(3)^{2}} \sqrt{3^{2}+(-2)^{2}+(1)^{2}}} \\\\ &\cos \theta=\frac{3+4+3}{\sqrt{14} \sqrt{14}}=\frac{10}{14} \end{aligned}

\begin{aligned} &\cos \theta=\frac{5}{7} \\\\ &\sin ^{2} \theta+\cos ^{2} \theta=1 \\\\ &\sin ^{2} \theta=1-\cos ^{2} \theta \\\\ &\sin ^{2} \theta=1-\left(\frac{5}{7}\right)^{2} \end{aligned}

\begin{aligned} &\sin ^{2} \theta=\frac{49-25}{49} \\\\ &\sin ^{2} \theta=\frac{24}{49} \\\\ &\sin \theta=\sqrt{\frac{24}{49}}=\frac{\sqrt{24}}{7} \end{aligned}