Question

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

Answer 1

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.