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.
$2x-2y+4z+5=0$ and $3x-3y+6z-1=0$.

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 planse are $2x-2y+4z+5=0$ and $3x-3y+6z-1=0$.
Here $a_1=2,b_1=-2,c_1=4$ and
$a_2=3,b_2=-3,c_2=6$
$a_1{a}_2+b_1{b}_2+c_1{c}_2=2\times 3+(-2)(-3)+4\times 6=6+6+24=36\ne 0$
Thus, the given planes are nor perpendicular to each other.
Now $\frac{a_1}{a_2}=\frac{2}{3},\frac{b_1}{b_2}=\frac{-2}{-3}=\frac{2}{3}$ and $\frac{c_1}{c_2}=\frac{4}{6}=\frac{2}{3}$
$\therefore$ $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
Thus, the given planes are parallel to each other