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

Two planes

$L_{1}:a_{1}x+b_{1}y+c_{1}z = 0$ whose direction ratios are $a_{1},b_{1},c_{1}$ and $L_{2}:a_{2}x+b_{2}y+c_{2}z = 0$ whose direction ratios are $a_{2},b_{2},c_{2}$,

are said to Parallel:

If, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

and Perpendicular:

If, $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0$

And the angle between $L_{1}\ and\ L_{2}$ is given by the relation,

$A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |$
So, given two planes  $7x + 5y + 6z + 30 = 0\ and\ 3x -y - 10z + 4 = 0$

Here,

$a_{1} = 7,b_{1} = 5, c_{1} = 6$   and   $a_{2} = 3,b_{2} = -1, c_{2} = -10$

So, applying each condition to check:

Parallel check:   $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{a_{1}}{a_{2}} =\frac{7}{3}, \frac{b_{1}}{b_{2}}=\frac{5}{-1},\frac{c_{1}}{c_{2}} = \frac{6}{-10}$

Clearly, the given planes are NOT parallel.$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

Perpendicular check: $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0$

$\Rightarrow 7(3)+5(-1)+6(-10) = 21-5-60 = -44 \neq 0$.

Clearly, the given planes are NOT perpendicular.

Then find the angle between them,

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

$= \cos^{-1}\left | \frac{-44}{\sqrt{7^2+5^2+6^2}.\sqrt{3^2+(-1)^2+(-10)^2}} \right |$

$= \cos^{-1}\left | \frac{-44}{\sqrt{110}.\sqrt{110}} \right |$

$= \cos^{-1}\left ( \frac{44}{110} \right )$

$= \cos^{-1}\left ( \frac{2}{5} \right )$

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