Question

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.
$7x+5y+6z+30=0$ and $3x-y-10z+4=0$.

Answer 1

The direction ratios of normal to the plane, $L_1:a_{1}x+b_{1}y+c_{1}z=0$ are $a_1,b_1,c_1$ and $L_2:a_{1}x+b_{2}y+c_{2}z=0$ are $a_2,b_2,c_2$
$L_1\parallel L_2$, if $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$L_1\perp L_2$, if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$
The angle between $L_1$ and $L_2$ is given by,
$\theta ={\cos }^{-1}\left|\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2.\sqrt{a_2^2+b_2^2+c_1^2}}}\right|$
The equations of the planes are $7x+5y+6z+30=0$ and $3x-y-10z+4=0$.
Here, $a_1=7,b_1=5,c_1=6$
$a_2,b_2=-1,c_2=-10$
$a_1{a}_2+b_1{b}_2+c_1{c}_2=7\times 3+5\times (-1)+6\times (-10)=-44\ne 0$
Therefore, the given planes are not perpendicular
$\frac{a_1}{a_2}=\frac{7}{3},\frac{b_1}{b_2}=\frac{5}{-1}=-5,\frac{c_1}{c_2}=\frac{6}{-10}=\frac{-3}{5}$
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.
The angle between them is given by,
$\theta ={\cos }^{-1}\left|\frac{7\times 3+5(-1)+6\times (-10)}{\sqrt{{\left(7\right)}^2+{\left(5\right)}^2+{\left(6\right)}^2}\times \sqrt{{\left(3\right)}^2+{(-1)}^2+{(-10)}^2}}\right|$
$={\cos }^{-1}\left|\frac{21-5-60}{\sqrt{110}\times \sqrt{110}}\right|$
$={\cos }^{-1}\left|\frac{44}{110}\right|={\cos }^{-1}\frac{2}{5}$