JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Three Dimensional Geometry

The direction ratios of the line are given by the cross product of the normals to the two planes: $\vec{d}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& -1\\ 1& -2& 3\end{array}\right|=\hat{i}-4\hat{j}-3\hat{k}$. The line through $(1,2,4)$ with this direction is $\vec{r}=\hat{i}+2\hat{j}+4\hat{k}+\lambda (\hat{i}-4\hat{j}-3\hat{k})$. A general point on the line is $P(1+\lambda ,2-4\lambda ,4-3\lambda )$. For the perpendicular from $(1,-2,5)$ to the line, $\overrightarrow{AP}\cdot (\hat{i}-4\hat{j}-3\hat{k})=0$ where $A=(1,2,4)$. We have $\overrightarrow{AP}=(\lambda ,-4\lambda ,-3\lambda )$ and $(1,-2,5)-(1,2,4)=(0,-4,1)$. Actually, let $Q=(1,-2,5)$ and the foot of perpendicular be $P(1+\lambda ,2-4\lambda ,4-3\lambda )$. Then $\overrightarrow{QP}\cdot (1,-4,-3)=0$. This gives $(\lambda ,4-4\lambda ,-1-3\lambda )\cdot (1,-4,-3)=0\Rightarrow \lambda -16+16\lambda +3+9\lambda =0\Rightarrow 26\lambda =13\Rightarrow \lambda =\frac{1}{2}$. So $P=(\frac{3}{2},0,\frac{5}{2})$ and $|QP|=\sqrt{(\frac{1}{2}{)}^2+4+(\frac{5}{2}{)}^2}=\sqrt{\frac{1}{4}+4+\frac{25}{4}}=\sqrt{\frac{42}{4}}=\sqrt{\frac{21}{2}}$.