Question

Find the angle between the planes \(\vec{r}.(4\hat{i}+\hat{j}-2\hat{k})\)=6 and \(\vec{r}.(5\hat{i}-6\hat{j}+\hat{k})\)=7?

(a) \(cos^{-1}⁡\frac{12}{\sqrt{1302}}\)

(b) \(cos^{-1}⁡\frac{1}{\sqrt{1392}}\)

(c) \(cos^{-1}\frac{⁡23}{\sqrt{102}}\)

(d) \(cos^{-1}⁡\frac{15}{\sqrt{134}}\)

The question was posed to me in a job interview.

I need to ask this question from Three Dimensional Geometry in division Three Dimensional Geometry of Mathematics – Class 12

Answer 1

Right answer is (a) \(cos^{-1}⁡\frac{12}{\sqrt{1302}}\)

To explain: Given that, the normal to the planes are \(\vec{n_1}=4\hat{i}+\hat{j}-2\hat{k} \,and \,\vec{n_2}=5\hat{i}-6\hat{j}+\hat{k}\)

The angle between two planes of the form \(\vec{r}.\vec{n_1}=d_1 \,and \,\vec{r}.\vec{n_2}=d_2\) is given by

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

\(|\vec{n_1}|=\sqrt{4^2+1^2+(-2)^2}=\sqrt{21}\)

\(|\vec{n_2}|=\sqrt{5^2+(-6)^2+1^2}=\sqrt{62}\)

\(\vec{n_1}.\vec{n_2}\)=4(5)+1(-6)-2(1)=20-6-2=12

cos⁡θ=\(\frac{12}{\sqrt{21}.\sqrt{62}}=\frac{12}{\sqrt{1302}}\)

∴θ=\(cos^{-1}⁡\frac{12}{\sqrt{1302}}\).