Question

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

Answer 1

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 \(\vec{n_2}=3\hat{i}+2\hat{j}-3\hat{k}\)

cos⁡θ=\(\left |\frac{\vec{n_1}.\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}\right |\)

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

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

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

cos⁡θ=\(\frac{1}{\sqrt{22}.\sqrt{6}}=\frac{1}{\sqrt{132}}\)

∴θ=\(cos^{-1}⁡\frac{1}{\sqrt{132}}\).