Question

Determine whether the given planes are parallel or perpendicular and in case they are neither, find the angle between them.

4x+8y+z-8=0 and y+z-4=0

Answer 1

Given planes are

4x +8y+z-8=0 and 0x+1y+1z-4=0

Here $a_1=4,b_1=8,c_1=1and{a}_2=0,b_2=1,c_2=1$

$\therefore a_1{a}_2+b_1{b}_2+c_1{c}_{@}=4\times 0+8\times 1+1\times 1$

$=0+8+1=9\ne 0$

Therefore, the given planes are not perpendicular.

Here, $\frac{a_1}{a_2}=\frac{4}{0},\frac{b_1}{b_2}=\frac{8}{1},\frac{c_1}{c_2}=\frac{1}{1}=1$

It can be seen that
$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}\ne \frac{c_1}{c_2}.$

Therefore, the given planes are not parallel.

Let $\theta$ be the acute angle between the given planes.

$\therefore cos\theta =\left|\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\right|=\left|\frac{4\times +8\times 1+1\times 1}{\sqrt{4^2+8^2+1^2}\sqrt{0^2+1^2+1^2}}\right|=\left|\frac{9}{9\times \sqrt{2}}\right|cos\theta =\frac{1}{\sqrt{2}}\Rightarrow \theta =co{s}^{-1}\left(\frac{1}{\sqrt{2}}\right)={45}^{\circ }$