Chapter 11.4, Problem 37E

To determine

Tofind: whether the planes are parallel, perpendicular, or neither. If they are neither parallel nor perpendicular, find the angle of intersection.

Answer to Problem 37E

Given planes are neither perpendicular nor parallel and the angle between the planes are ${\cos }^{-1}\left(\frac{4}{\sqrt{78}}\right)$

Explanation of Solution

Given planes: $3x+2z=4$ and $6x-2y-3z=8$

Calculation:

The direction ratios of normal to given planes are $\langle 3,0,2\rangle$ and $\langle 6,-2,-3\rangle$ .

We know that if the direction ratios of normal to the two planes are $\langle a_1,b_1,c_1\rangle$ and $\langle a_2,b_2,c_2\rangle$ , then the planes are perpendicular if

  $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

And the planes are parallel if

  $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Here,

  $a_1{a}_2+b_1{b}_2+c_1{c}_2=3\left(6\right)+\left(0\right)\left(-2\right)+2\left(-3\right)=12\ne 0$

That is, the given planes are not perpendicular.

Also,

  $\frac{3}{6}\ne \frac{0}{-3}\ne \frac{2}{-3}$

That is, the planes are also not parallel.

Now, the angle between the planes , let it be $\theta$ , is evaluated as follows.

  $\begin{array}{l}\cos \theta =\frac{\left|a_1{a}_2+b_1{b}_2+c_1{c}_2\right|}{\sqrt{a_1{}^2+b_1{}^2+c_1{}^2}\sqrt{a_2{}^2+b_2{}^2+c_2{}^2}}\\ \cos \theta =\frac{\left|12\right|}{\sqrt{3^2+0^2+2^2}\sqrt{6^2+{\left(-3\right)}^2+{\left(-3\right)}^2}}\\ \cos \theta =\frac{12}{\sqrt{13}\sqrt{54}}\\ \cos \theta =\frac{12}{3\sqrt{78}}\\ \theta ={\cos }^{-1}\left(\frac{4}{\sqrt{78}}\right)\end{array}$