Let $P$ be the plane, passing through the point $(1,-1,-5)$ and perpendicular to the line joining the points $(4,1,-3)$ and $(2,4,3)$. Then the distance of $P$ from the point $(3,-2,2)$ is :
Solution
Equation of Plane :
<br/><br/>$2(x-1)-3(y+1)-6(z+5)=0$
<br/><br/>or $2 x-3 y-6 z=35$
<br/><br/>$\Rightarrow$ Required distance $=$
$\frac{|2(3)-3(-2)-6(2)-35|}{\sqrt{4+9+36}}$ = 5
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
This question is part of PrepWiser's free JEE Main question bank. 182 more solved questions on Three Dimensional Geometry are available — start with the harder ones if your accuracy is >70%.