Class 12

Math

3D Geometry

Three Dimensional Geometry

Show that the distance of the point of intersection of the line $3x−2 =4y+1 =12z−12 $ and the plane $x−y+z=5$ from the point $(−1,−5,−10)$ is $13$ units.

Given line is $3x−2 =4y+1 =12z−2 =λ$(say)

A general point on this line is $P(3λ+2,4λ−1,12λ+2)$.

If this point lies on the plane $x−y+z=5$, then

$(3λ+2)−(4λ−1)+(12λ+2)=5⇒11λ=0⇒λ=0$.

$∴$ point P is $P(2,−1,2)$ and the other point is $Q(−1,−5,−10)$

$PQ=(−1−2)_{2}+(−5+1)_{2}+(−10−2)_{2} =9+16+144 =169 $

$PQ=13$ units.