In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

Solution:

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 $a_{2}, b_{2}, c_{2}$.

$L_{1} \| L_{2}$, 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,

$Q=\cos ^{-1}\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}}\right|$

(a) The equations of the planes are $7 x+5 y+6 z+30=0$ and

$3 x-y-10 z+4=0$

Here, $a_{1}=7, b_{1}=5, c_{1}=6$

$a_{2}=3, b_{2}=-1, c_{2}=-10$

$a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=7 \times 3+5 \times(-1)+6 \times(-10)=-44 \neq 0$

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, $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

Therefore, the given planes are not parallel.

The angle between them is given by,

$Q=\cos ^{-1}\left|\frac{7 \times 3+5 \times(-1)+6 \times(-10)}{\sqrt{(7)^{2}+(5)^{2}+(6)^{2}} \times \sqrt{(3)^{2}+(-1)^{2}+(-10)^{2}}}\right|$

$=\cos ^{-1}\left|\frac{21-5-60}{\sqrt{110} \times \sqrt{110}}\right|$

$=\cos ^{-1} \frac{44}{110}$

$=\cos ^{-1} \frac{2}{5}$

(b) The equations of the planes are $2 x+y+3 z-2=0$ and $x-2 y+5=0$

Here, $a_{1}=2, b_{1}=1, c_{1}=3$ and $a_{2}=1, b_{2}=-2, c_{2}=0$

$\therefore a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=2 \times 1+1 \times(-2)+3 \times 0=0$

Thus, the given planes are perpendicular to each other.

(c) The equations of the given planes are $2 x-2 y+4 z+5=0$ and $3 x-3 y+6 z-1=0$

Here, $a_{1}=2, b_{1}-2, c_{1}=4$ and $a_{2}=3, b_{2}=-3, c_{2}=6 \quad a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=2 \times 3+(-2)(-3)+4 \times 6=6+6+24=36 \neq 0$

Thus, the given planes are not perpendicular to each other.

$\frac{a_{1}}{a_{2}}=\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{-2}{-3}=\frac{2}{3}$ 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.

(d) The equations of the planes are $2 x-y+3 z-1=0$ and $2 x-y+3 z+3=0$

Here, $a_{1}=2, b_{1}=-1, c_{1}=3$ and $a_{2}=2, b_{2}=-1, c_{2}=3$

$\frac{a_{1}}{a_{2}}=\frac{2}{2}=1, \frac{b_{1}}{b_{2}}=\frac{-1}{-1}=1$ and $\frac{c_{1}}{c_{2}}=\frac{3}{3}=1$

$\therefore \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

Thus, the given lines are parallel to each other.

(e) The equations of the given planes are $4 x+8 y+z-8=0$ and $y+z-4=0$

Here, $a_{1}=4, b_{1}=8, c_{1}=1$ and $a_{2}=0, b_{2}=1, c_{2}=1$

$a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=4 \times 0+8 \times 1+1=9 \neq 0$

Therefore, the given lines are not perpendicular to each other.

$\frac{a_{1}}{a_{2}}=\frac{4}{0}, \frac{b_{1}}{b_{2}}=\frac{8}{1}=8, \frac{c_{1}}{c_{2}}=\frac{1}{1}=1$

$\therefore \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq, \frac{c_{1}}{c_{2}}$

Therefore, the given lines are not parallel to each other.

The angle between the planes is given by,

$Q=\cos ^{-1}\left|\frac{4 \times 0+8 \times 1+1 \times 1}{\sqrt{4^{2}+8^{2}+1^{2}} \times \sqrt{0^{2}+1^{2}+1^{2}}}\right|=\cos ^{-1}\left|\frac{9}{9 \times \sqrt{2}}\right|=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ}$