Question

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

7x +5y +6z+ 30 =0 and 3x -y- 10z+ 4=0

2x+y+3z-2=0 and x-2y+5=0

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

2x-y+3z-1=0 and 2x-y+3z+3=0

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

Answer 1

Given planes are

$7x+5y+6z+30=0and3x-y-10z+4=0$

Here $a_1=7,b_1=5,c_1=6and{a}_2=3,b_2=-1,c_2=-10$

$\therefore ,a_1{a}_2+b_1{b}_2+c_1{c}_2=7\times 3+5\times (-1)+6\times (-10)$

$21-5-60=-44\ne 0$

Therefore, the given planes are not perpendicular.

Here, $\frac{a_1}{a_2}=\frac{7}{3},\frac{b_1}{b_2}=\frac{5}{-1}=-5,\frac{c_1}{c_2}=\frac{6}{-10}=\frac{-3}{5}$

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.

Let $\theta$ be the acute angle between the given planes, then

$cos\theta =\left|\frac{7\times 3+5\times (-1)+6\times (-10)}{\sqrt{7^2+5^2+6^2}\sqrt{3^2+(-1{)}^2+(-{10}^2)}}\right|=\left|\frac{21-5-60}{\sqrt{110}\sqrt{110}}\right|\Rightarrow cos\theta =\frac{44}{110}=\frac{2}{5}\Rightarrow \theta =co{s}^{-1}\left(\frac{2}{5}\right)$

Hence, the angles between the given planes is $co{s}^{-1}\left(2/5\right)$

Given planes are 2x +y +3z -2=0 and x-2y+0z+5=0

Here, $a_1=2,b_1=1,c_1=3and{a}_2=1,b_2=-2,c_2=0$

$\therefore a_1{a}_2+b_1{b}_2+c_1{c}_2=2\times 1+1\times (-2)+3\times 0=0$

Thus, the given planes are perpendicular to each other.

Given planes are

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

Here, $a_1=2,b_1=-2,c_1=4and{a}_2=3,b_2=-3,c_2=6$

$\therefore a_1{a}_2+b_1{b}_2+c_1{c}_2=2\times 3+(-2)\times (-3)+4\times 6$

$6+6+24=36\ne 0$

Therefore, the given planes are not perpendicular.

Here $\frac{a_1}{a_2}=\frac{2}{3},\frac{b_1}{b_2}=\frac{-2}{-3},\frac{c_1}{c_2}=\frac{4}{6}=\frac{2}{3}$

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

Thus , the given planes are parallel to each other.

Given planes are

2x-y+3z-1=0 and 2x-y+3z+3=0

Here, $a_1=2,b_1=-1,c_1=3and{a}_2=2,b_2=-1,c_2=3$

$\therefore a_1{a}_2+b_1{b}_2+c_1{c}_2=2\times 2+(-1)\times (-1)+3\times 3$

$=4+1+9=14\ne 0$

Here, $\frac{a_1}{a_2}=\frac{2}{2}=1,\frac{b_1}{b_2}=\frac{-1}{-1},\frac{c_1}{c_2}=\frac{3}{3}=1$

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

Thus, the given lines are parallel to each other.

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.

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