Three Dimensional Geometry Question 30
Question 30 - 31 January - Shift 2
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
(1) 6
(2) 4
(3) 5
(4) 7
Show Answer
Answer: (3)
Solution:
Formula: A Plane and A Point (7.1)
Equation of Plane :
$2(x-1)-3(y+1)-6(z+5)=0$
Or $2 x-3 y-6 z=35$
$\Rightarrow$ Required distan ce $=$
$\frac{|2(3)-3(-2)-6(2)-35|}{\sqrt{4+9+36}}$
$\sqrt{4+9+36}$
$=5$
