Question

Find the angle between the planes 6x-3y+7z=8 and 2x+3y-2z=5?

(a) \(cos^{-1}\frac{⁡11}{\sqrt{98}}\)

(b) \(cos^{-1}⁡\frac{⁡11}{\sqrt{1598}}\)

(c) \(cos^{-1}\frac{⁡⁡13}{\sqrt{198}}\)

(d) \(cos^{-1}\frac{⁡⁡11}{1598}\)

This question was addressed to me in quiz.

I want to ask this question from Three Dimensional Geometry topic in chapter Three Dimensional Geometry of Mathematics – Class 12

Answer 1

Right option is (b) \(cos^{-1}⁡\frac{⁡11}{\sqrt{1598}}\)

Easy explanation: We know that, 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=\(\left |\frac{A_1 A_2+B_1 B_2+C_1 C_2}{\sqrt{A_1^2+B_1^2+C_1^2} \sqrt{A_2^2+B_2^2+C_2^2}}\right |\)

Given that, \(A_1=6,B_1=-3,C_1=7\) and \(A_2=2,B_2=3,C_2=-2\)

cos⁡θ=\(\left |\frac{6(2)-3(3)+7(-2)}{|\sqrt{6^2+(-3)^2+7^2} \sqrt{2^2+3^2+(-2)^2}|}\right |\)

cos⁡θ=\(|\frac{-11}{\sqrt{94}.\sqrt{17}}|\)

θ=\(cos^{-1}⁡\frac{⁡11}{\sqrt{1598}}\).