QJEEM22Ma7f07: JEE Main 2022 Mathematics - Class 12 JEEM22Ma7f07 Solution: The length of the perpendicular from the point on the line…

The line $x + y - z = 0 = x - 2y + 3z - 5$ is parallel to the vector

$$\overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& -1\\ 1& -2& 3\end{array}\right|=(1,4,-3)$$

Equation of the line through $P(1,2,4)$ and parallel to $\overrightarrow b$

${{x - 1} \over 1} = {{y - 2} \over { - 4}} = {{z - 4} \over { - 3}}$

Let $N \equiv (\lambda + 1, - 4\lambda + 2, - 3\lambda + 4)$

$\overrightarrow {QN} = (\lambda , - 4\lambda + 4, - 3\lambda - 1)$

$\overrightarrow {QN}$ is perpendicular to $\overrightarrow {b}$

$\Rightarrow (\lambda , - 4\lambda + 4, - 3\lambda - 1)\,.\,(1,4, - 3) = 0$

$\Rightarrow \lambda = {1 \over 2}$

Hence $\overrightarrow {QN} = \left( {{1 \over 2},2,{{ - 5} \over 2}} \right)$ and $\left| {\overrightarrow {QN} } \right| = \sqrt {{{21} \over 2}}$