#### Need solution for RD sharma maths class 12 chapter 24 vector or cross product Exercise Multiple choice questions question 19 maths textbook solution

Answer: $\frac{1}{2}\sqrt{229}$

Given: The vectors from origin O to the points A and B are $\overrightarrow{a}=2\hat{i}-3\hat{j}+2\hat{k};\overrightarrow{b}=2\hat{i}+3\hat{j}+\hat{k}$respectively.

Hint: Area of

Explanation: Here $\overrightarrow{a}=2\hat{i}-3\hat{j}+2\hat{k};\overrightarrow{b}=2\hat{i}+3\hat{j}+\hat{k}$

$\overrightarrow{O A}=( P. V of A- P.V of O)$

$=(2 \hat{i}-3 \hat{j}+2 \hat{k})-(0 \hat{i}+0 \hat{j}+0 \hat{k})$

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

$\overrightarrow{A B}=(P . V \text { of } B-P . V \text { of } A) \\$

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

$=0 \hat{i}+6 \hat{j}-\hat{k}$

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

\begin{aligned} &\text { Now, }(\overrightarrow{O A} \times \overrightarrow{A B})=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -2 \\ 0 & 6 & -1 \end{array}\right| \\ \end{aligned}

$= \hat{i}(3-12)-\hat{j}(-2-0)+\hat{k}(12+0) \\$

$=-9 \hat{i}+2 \hat{j}+12 \hat{k} \\$

$\Rightarrow|\overrightarrow{O A} \times \overrightarrow{A B}|=\sqrt{(-9)^{2}+(2)^{2}+(12)^{2}}=\sqrt{81+4+144}=\sqrt{229}$

$Area of \triangle O A B=\frac{1}{2}\sqrt{229}$