Question

Find the angles between the planes 2x – 3y – 6z = 6 and 6x + 3y – 2z = 18​

Answer 1

Answer:

Let A be the angle between the planes

Cos A=[2. 6+(-3)3+(-6)(-2)]/v(4+9+36)(v(36+9+4)

=[12-9+12]/7*7

=15/49

A=cos inverse (15/49)

Answer 2

Given planes

2x - 3y - 6z = 6,

6x + 3y - 2z = 18

Angle between the planes is the angle between their normals.

For a plane lx + my + nz = p, l, m, n are the direction ratios of the normal to the plane.

D.r's of the normal to the plane 2x - 3y - 6z = 6 are 2, - 3, - 6

D.r's of the normal to the plane 6x + 3y - 2z = 18 are 6, 3, - 2.

D.c's of the normal to the plane 2x - 3y - 6z = 6 are 2/7 , - 3/7 , - 6/7

D.c's of the normal to the plane 6x + 3y - 2z = 18 are 6/7, 3/7, - 2/7

Two lines having D.c's a, b, c and p, q, r are at angle

$\cos( \theta) = |ap + bq + cr|$

So Angle between the normals of the given planes is

$\cos( \theta) = | \frac{2}{7} \times \frac{6}{7} + \frac{ - 3}{7} \times \frac{3}{7} + \frac{ - 6}{7} \times \frac{ - 2}{7} | \\ \\ \cos( \theta) = \frac{12 - 9 + 12}{49} \\ \\ \theta = {cos}^{ - 1} ( \frac{15}{49})$

Therefore, Angle between the given planes is cos^-1 ( 15/49).