Question

If the line drawn from the point (−2,−1,−3) meets a plane at right angle at the point (1,−3,3), find the equation of the plane.

Answer 1

Any line which is perpendicular to the plane is the normal

Equation of a plane is ax+by+cz+d=0ax+by+cz+d=0 where (a,b,c)(a,b,c) is normal

Let the given points be P(−2,−1,−3)P(−2,−1,−3) and (1,−3,3)(1,−3,3) and the line PQPQ be perpendicular to the plane.

Therefore PQ=(x2−x1),(y2−y1),(z2−z1)

=(1−(−2)),(−3−(−1)),(3−(−3))

PQ=(x2−x1),(y2−y1),(z2−z1)

=(1−(−2)),(−3−(−1)),(3−(−3))

=(3,−2,6)=(3,−2,6)

It is given that QQ is the foot of the perpendicular to PP on the plane.

This implies that point Q lies on the plane.

Equation of the plane is ax+by+cz+dax+by+cz+d,where (a,b,c)(a,b,c) is normal.

(ie)3x−2y+6z+d=0(ie)3x−2y+6z+d=0

Since Q lies on the plane,let us substitute for (x,y,z)(x,y,z)

(ie)(3×1)+(−2)×(−3)+(6×3)+d=0(ie)(3×1)+(−2)×(−3)+(6×3)+d=0

(ie)3+6+18+d=0(ie)3+6+18+d=0

=>d=−27=>d=−27

Therefore equation of the plane is

3x−2y+6z−27=03x−2y+6z−27=0