​ Find the vector equation of the plane which is at a distance of 6/√(29) from the origin with perpendicular vector 2i – 3j + 4k. ​

Question

Find the vector equation of the plane which is at a distance of \(\frac{6}{\sqrt{29}}\) from the origin with perpendicular vector 2i – 3j + 4k. Convert into Cartesian form. Also, find the foot of the perpendicular drawn from the origin to the Plane.

Answer 1

Perpendicular distance from origin = d = \(\frac{6}{\sqrt{29}}\)

The equation of the Plane is \(\bar r. \hat n\) = d

Cartesian equation is 2x – 3y + 4z = 6

The direction cosines perpendicular to the Plane is

\(\frac{2}{\sqrt{29}}\), \(-\frac{3}{29}\), \(\frac{4}{29}\).

Perpendicular distance to the Plane is as \(\frac{6}{\sqrt{29}}\)

Hence the foot of the perpendicular is