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 =\(\bigg(\frac{15}{\sqrt{731}}\bigg)\)