3D Geometry: The distance of the point (7,-3,-4) from the plane containing the… | JEE Main 2023 | Yukti Vault

Yukti Vault

JEE Main 2023 (Online) 24th January Morning ShiftQuestion 1

The distance of the point $(7,-3,-4)$ from the plane containing the points $(2,-3,1)$, $(-1,1,-2)$ and $(3,-4,2)$ is equal to:

3D Geometry

$4$$5$$5\sqrt2$✓ Correct answer$4\sqrt2$

Solution

Normal $=(-3,4,-3)\times(1,-1,1)=(1,0,-1)$; plane $x-z=1$. Distance of $(7,-3,-4)$: $\frac{|7+4-1|}{\sqrt2}=5\sqrt2$.

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