QJEEM19Ma259b: JEE Main 2019 Mathematics - Class 12 JEEM19Ma259b Solution: The length of the perpendicular from the point (2, –1, 4) on…

Let ${{x + 3} \over {10}}$ = ${{y - 2} \over {-7}}$ = ${{z} \over {1}}$ = $\lambda$

$\therefore$ General point Q(10 $\lambda$ - 3, -7 $\lambda$ + 2, $\lambda$ ).

$\overrightarrow {PQ}$ = (10 $\lambda$ - 5) $\widehat i$ + (3 - 7 $\lambda$ ) $\widehat j$ + ( $\lambda$ - 4) $\widehat k$

The vector parrallel to the given line is

= 10 $\widehat i$ - 7 $\widehat j$ + $\widehat k$

As $\overrightarrow {PQ}$ and (10 $\widehat i$ - 7 $\widehat j$ + $\widehat k$ ) are perpendicular to the each other.

$\overrightarrow {PQ}$ . (10 $\widehat i$ - 7 $\widehat j$ + $\widehat k$ ) = 0

$\Rightarrow$ (10 $\lambda$ - 5)10 + (3 - 7 $\lambda$ )(-7) + ( $\lambda$ - 4)(1) = 0

$\Rightarrow$ 100 $\lambda$ – 50 + 49 $\lambda$ – 21 + $\lambda$ – 4 = 0

$\Rightarrow$ 150 $\lambda$ = 75

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

$\therefore$ The point Q = (2, $- {3 \over 2}$ , ${1 \over 2}$ )

$\therefore$ Length of perpendicular (PQ) = $\sqrt {0 + {1 \over 4} + {{49} \over 4}}$

= $\sqrt {{{50} \over 4}} = {5 \over {\sqrt 2 }}$ = 3.53