What is the formula to find the angle between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0?

(a) cos θ=\frac {a1a2+b1b2+c1c2}{\sqrt {a1^2+b1^2+c1^2} \sqrt {a2^2+b2^2+c^2 }}

(b) sec θ=\frac {a1a2+b1b2+c1c2}{\sqrt {a1^2+b1^2+c1^2} \sqrt{a2^2+b2^2+c2^2 }}

(c) cos θ=\frac {a1a2.b1b2.c1c2}{\sqrt {a1^2+b1^2+c1^2} \sqrt{a2^2+b2^2+c2^2 }}

(d) cot θ=\frac {a1a2+b1b2+c1c2}{\sqrt {a1^2+b1^2+c1^2} \sqrt{a2^2+b2^2+c2^2 }}

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

Origin of the question is Three Dimensional Geometry in section Three Dimensional Geometry of Mathematics – Class 12

NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options

Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience

The correct option is (a) cos θ=\frac {a1a2+b1b2+c1c2}{\sqrt {a1^2+b1^2+c1^2} \sqrt {a2^2+b2^2+c^2 }}

The explanation is: The formula to find the angle between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is cos θ=\frac {a1a2+b1b2+c1c2}{\sqrt {a1^2+b1^2+c1^2} \sqrt {a2^2+b2^2+c2^2 }}. θ is the angle between the normal of two planes.