Question

Find the angles between the planes 2x – 3y – 6z = 6 and 6x + 3y – 2z = 18

Answer 1

Given planes are 2x – 3y – 6z = 6 and

6x + 3y – 2z = 18

The normal vector to the given planes are

$\vec{n_1}=2\vec{i}-3\vec{j}-6\vec{k}$

$\vec{n_2}=6\vec{i}+3\vec{j}-2\vec{k}$

The angle between the given two planes is

$\boxed{\bf\:sin\theta=\frac{\vec{n_1}\;.\;\vec{n_2}}{|\vec{n_1}|\;|\vec{n_2}|}}$

$sin\theta=\frac{2(6)-3(3)-6(-2)}{\sqrt{4+9+36}\:\sqrt{36+9+4}}$

$sin\theta=\frac{12-9+12}{\sqrt{49}\:\sqrt{49}}$

$sin\theta=\frac{15}{49}$

$sin\theta=\frac{15}{49}$

$\implies\bf\theta=sin^{-1}(\frac{15}{49})$