Question

Find the angles between the planes whose vector equation are $r.(2\hat{i}+2\hat{j}-3\hat{k})=5andr.(3\hat{i}-3\hat{j}+5\hat{k})=3$

Answer 1

Given planes are $r.(2\hat{i}+2\hat{j}-3\hat{k})=5andr.(3\hat{i}-3\hat{j}+5\hat{k})=3$

It is known that if $n_{1}and{n}_2$ are normal to the plane, $r.n_1=d_1$ and $r.n_2=d_2$, then the angle between them is given by

$cos\theta =\left|\frac{n_1.n_2}{|n_1||n_2|}\right|$

Here, $n_1=2\hat{i}+2\hat{j}-3\hat{k}and{n}_2=3\hat{i}-3\hat{j}+5\hat{k}{n}_1.n_2=(2\hat{i}+2\hat{j}-3\hat{k}).(3\hat{i}-3\hat{j}+5\hat{k})2.3+2.(-3)+(-3).5=6-6-15=-15|n_1|=\sqrt{(2{)}^2+(2{)}^2+(-3{)}^2}=\sqrt{4+4+9}=\sqrt{17}|n_2|=\sqrt{(3{)}^2+(-3{)}^2+(5{)}^2}=\sqrt{9+9+25}=\sqrt{43}$

Substituting the value of $n_1.n_2,|n_1|and|n_2|$ in Eq. (i), we obtain

$cos\theta =\left|\frac{-15}{\sqrt{17}\sqrt{43}}\right|\Rightarrow cos\theta =\frac{15}{\sqrt{731}}\Rightarrow \theta =co{s}^{-1}\left(\frac{15}{\sqrt{731}}\right)$