Find the angle between two planes \vec{r}.(2\hat{i}-\hat{j}+\hat{k})=3 and \vec{r}.(3\hat{i}+2\hat{j}-3\hat{k})=5.

(a) cos^{-1}\frac{1}{\sqrt{22}}

(b) cos^{-1}\frac{1}{\sqrt{6}}

(c) cos^{-1}\frac{1}{\sqrt{132}}

(d) cos^{-1}\frac{1}{\sqrt{13}}

This question was posed to me by my school principal while I was bunking the class.

This intriguing question comes from Three Dimensional Geometry topic in chapter Three Dimensional Geometry of Mathematics – Class 12

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The correct answer is (c) cos^{-1}\frac{1}{\sqrt{132}}

The explanation: Given that, the normal to the planes are \vec{n_1}=2\hat{i}-\hat{j}+\hat{k} and \vec{n_2}=3\hat{i}+2\hat{j}-3\hat{k}

cosθ=\left |\frac{\vec{n_1}.\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}\right |

|\vec{n_1}|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{6}

|\vec{n_2}|=\sqrt{3^2+2^2+(-3)^2}=\sqrt{22}

\vec{n_1}.\vec{n_2}=2(3)-1(2)+1(-3)=6-2-3=1

cosθ=\frac{1}{\sqrt{22}.\sqrt{6}}=\frac{1}{\sqrt{132}}

∴θ=cos^{-1}\frac{1}{\sqrt{132}}.