#### Need solution for RD sharma maths class 12 chapter 24 vector or cross product Exercise Multiple choice questions question 21 maths textbook solution

Answer: $3\sqrt{5}$

Given: \begin{aligned} &\overrightarrow{O A}=\hat{i}+2 \hat{j}+3 \hat{k} ; \overrightarrow{O B}=-3 \hat{i}-2 \hat{j}+\hat{k} \\ \end{gathered}

Hint:Area of $OAB=\frac{1}{2}\left | \overrightarrow{AO}\times \overrightarrow{AB} \right |$

Explanation:

\begin{aligned} &\overrightarrow{O A}=\hat{i}+2 \hat{j}+3 \hat{k} ; \overrightarrow{O B}=-3 \hat{i}-2 \hat{j}+\hat{k} \\ \end{gathered}

$\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \\$

$=-3 \hat{i}-2 \hat{j}+\hat{k}-(\hat{i}+2 \hat{j}+3 \hat{k}) \\$

$=-4 \hat{i}-4 \hat{j}-2 \hat{k}$

$Area of \triangle O A B=\frac{1}{2}|\overrightarrow{O A} \times \overrightarrow{A B}|$

$\overrightarrow{O A} \times \overrightarrow{A B}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -4 & -4 & -2 \end{array}\right| \\$

$=\hat{i}(-4+12)-\hat{j}(-2+12)+\hat{k}(-4+8) \\$

$=8 \hat{i}-10 \hat{j}+4 \hat{k} \\$

$|\overrightarrow{O A} \times \overrightarrow{A B}|=\sqrt{(8)^{2}+(-10)^{2}+(4)^{2}}=\sqrt{64+100+16}=\sqrt{180}$

$\text { Area of } \triangle \mathrm{OAB}=\frac{1}{2} \sqrt{180}=\frac{2 \sqrt{45}}{2}=\sqrt{45}=\sqrt{9 \times 5}=3 \sqrt{5} \text { units }$