Question

Find the angle between the two planes 2x+2y+z=2 and x-y+z=1?

(a) \(cos^{-1}\frac{⁡1}{3}\)

(b) \(cos^{-1}⁡\sqrt{3}\)

(c) \(cos^{-1}⁡\frac{1}{3}\)

(d) \(cos^{-1}⁡\frac{1}{3\sqrt{3}}\)

I had been asked this question in an online interview.

Enquiry is from Three Dimensional Geometry in portion Three Dimensional Geometry of Mathematics – Class 12

Answer 1

The correct answer is (d) \(cos^{-1}⁡\frac{1}{3\sqrt{3}}\)

Best explanation: The angle between two planes of the form \(A_1 x+B_1 y+C_1 z+D_1\)=0 and \(A_2 x+B_2 y+C_2 z+D_2\)=0 is given by

cos⁡θ=\(\left |\frac{A_1 A_2+B_1 B_2+C_1 C_2}{\sqrt{A_1^2+B_1^2+C_1^2} \sqrt{A_2^2+B_2^2+C_2^2}}\right |\)

According to the given question, \(A_1=2,B_1=2,C_1=1 \,and \,A_2=1,B_2=-1,C_2=1\)

cos⁡θ=\(\left |\frac{2(1)+2(-1)+1(1)}{|\sqrt{2^2+2^2+1^2} \sqrt{1^2+(-1)^2+1^2}|}\right |\)

cos⁡θ=\(\left |\frac{1}{\sqrt{9}.\sqrt{3}}\right |\)

θ=\(cos^{-1}⁡\frac{1}{3\sqrt{3}}\).