Reduce the equation 3x – 4y + 12z = 5 or r (3i – 4j + 12k) = 5 to normal form and hence find the length

Question

Reduce the equation 3x – 4y + 12z = 5 or \(\vec{r}\).(3i – 4j + 12k) = 5 to normal form and hence find the length of perpendicular from the origin to the plane. Also find direction cosines of the normal to the plane.

Answer 1

First method : Given

II method : On dividing 3x – 4y + 12z – 5 by its absolute value 5.

Let the lengths of perpendicular drawn from origin in p and dc's of normal are l, m, n.