3D Geometry: A plane intersects the yz -plane at x=0 , 2y-3z=5 and it intersects… | JEE Main 2021

JEE Main 2021 (Online) 3rd August Evening ShiftQuestion 1

A plane intersects the $yz$-plane at $x=0$, $2y-3z=5$ and it intersects the $xy$-plane at $z=0$, $7x+4y=10$. Then the distance of the point $(1,-2,-1)$ from this plane is

3D Geometry

$\frac{1}{\sqrt{101}}$$\frac{3}{\sqrt{101}}$$\frac{5}{\sqrt{101}}$✓ Correct answer$\frac{7}{\sqrt{101}}$

Solution

Plane $7x+4y+cz=d$; at $x=0$: $4y+cz=d\propto2y-3z=5$ gives $c=-6$, $d=10$. Plane $7x+4y-6z=10$. Distance of $(1,-2,-1)$: $\frac{|7-8+6-10|}{\sqrt{101}}=\frac5{\sqrt{101}}$.

3D Geometry - JEE Mathematics question (JEE Main 2021 (Online) 3rd August Evening Shift)

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