Question

Find the angle between the planes whose vector equations are

$\overrightarrow r.(2\hat i+2\hat j-3\hat k)=5$ and $\overrightarrow r.(3\hat i-3\hat j+5\hat k)=3$

Answer 1

The equations of the given planes are $\overrightarrow r.(2\hat i+2\hat j-3\hat k)=5$ and $\overrightarrow r.(3\hat i-3\hat j+5\hat k)=3$

It is known that if $\overrightarrow n_1$ and $\overrightarrow n_2$ are normal to the planes, $\overrightarrow r.\overrightarrow n_1=\overrightarrow d_1$ and $\overrightarrow r.\overrightarrow n_2=\overrightarrow d_2$
then the angle between them, Q, is given by,

cos Q= $\begin{vmatrix}\frac{\overrightarrow n_1.\overrightarrow n_2}{\|\overrightarrow n_1\| \overrightarrow n_2 \|}\end{vmatrix}$...(1)

Here, $\overrightarrow n_1=2\hat i+2\hat j-3\hat k$ and $\overrightarrow n_2=3\hat i-3\hat j+5\hat k$

∴ $\overrightarrow n_1.\overrightarrow n_2$= (2\hat i+2\hat j-3\hat k)$$(3\hat i-3\hat j+5\hat k)

=2.3+2.(-3)+(-3).5 =-15

|$\overrightarrow n_1$|=$\sqrt{(2)^2+(2)^2+(-3)^2}=\sqrt{17}$

|$\overrightarrow n_2$|=$\sqrt{(3)^2+(-3)^2+(5)^2}=\sqrt{43}$
Substituting the value of $\overrightarrow n_1.\overrightarrow n_2$,|$\overrightarrow n_1$|and|$\overrightarrow n_2$| in equation(1), we obtain

cos Q=|$\frac{-15}{\sqrt{17}.\sqrt{43}}$|

$\Rightarrow$ cos Q=$\frac{15}{\sqrt{731}}$

$\Rightarrow$ cos Q-1 =$\bigg(\frac{15}{\sqrt{731}}\bigg)$