Determine whether the given planes are parallel or perpendicular, and in case they are neither, then find the angles between them. $(a)\; 7x + 5y + 6z + 30 = 0 \: and \: 3x -y -10z + 4 = 0 $

Question

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

Answer 1

Toolbox:

• The direction ratios of normal to the plane $l_1=a_1x+b_1y+c_1 x=0$ are $a_1,b_1$ and $c_1$ and $l_2=a_2x+b_2y+c_2z=0$ are $a_2,b_2,c_2$
• If $L_1\parallel L_2$ then $\large\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
• If $L_1\perp L_2$,$a_1a_2+b_1b_2+c_1c_2=0$
• $\cos\theta=\begin{vmatrix}\large\frac{\overrightarrow n_1.\overrightarrow n_2}{\mid\overrightarrow n_1\mid\mid\overrightarrow n_2\mid}\end{vmatrix}$
• $\Rightarrow \begin{vmatrix}\large\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\end{vmatrix}$

Step 1:

The equations of the given planes are :

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

The direction cosines of $L_1$ are (7,5,6).

The direction cosines of $L_2$ are (3,-1,-10).

Step 2:

Therefore $\cos\theta=\begin{vmatrix}\large\frac{7\times 3+5\times -1+6\times -10}{\sqrt{7^2+5^2+6^2}\sqrt{3^2+(-1)^2+(-10)^2}}\end{vmatrix}$

$\Rightarrow \begin{vmatrix}\large\frac{21-5-60}{\sqrt{49+25+36}\sqrt{9+1+100}}\end{vmatrix}$

$\Rightarrow \large\frac{44}{\sqrt{49+25+36}\sqrt{9+1+100}}$

$\Rightarrow \large\frac{44}{\sqrt{110}\sqrt{110}}$

$\Rightarrow \large\frac{44}{110}$

$\Rightarrow \large\frac{2}{5}$

$\theta=\cos^{-1}\big(\large\frac{2}{5}\big)$