JEE 2023 — Mathematics PYQ

Given that $\overrightarrow{OP}=-\hat{i}-2\hat{j}+3\hat{k},\overrightarrow{OA}=-2\hat{i}+\hat{j}-3\hat{k}$
$\overrightarrow{OB}=2\hat{i}+4\hat{j}-2\hat{k}$, and $\overrightarrow{OC}=-4\hat{i}+2\hat{j}-\hat{k}$
Now $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=4\hat{i}+3\hat{j}+\hat{k}$
and $\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=-2\hat{i}+\hat{j}+2\hat{k}$

and $\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& amp;\hat{j}& amp;k\\ 4& amp;3& amp;1\\ -2& amp;1& amp;2\end{array}\right|=5\hat{i}-10\hat{j}+10\hat{k}=\vec{a}$ (say)
Since $\overrightarrow{AB}\times \overrightarrow{AC}=\vec{a}$, which is perpendicular to both $\overrightarrow{AB}$ and $\overrightarrow{AC}$ therefore, projection of $\overrightarrow{OP}$ on $\vec{a}$ is given by $\frac{\overrightarrow{OP}\cdot \vec{a}}{|\vec{a}|}$
$\frac{\overrightarrow{OP}\cdot \vec{a}}{|\vec{a}|}=\frac{(-\hat{i}-2\hat{j}+3\hat{k})\cdot (5\hat{i}-10\hat{j}+10\hat{k})}{\sqrt{(5{)}^2+(-10{)}^2+(10{)}^2}}$
$=\frac{-5+20+30}{\sqrt{25+100+100}}=\frac{45}{15}=3$