Question

Find the angle between two planes A and B represented by A = 2i + 3j + 2k + 1 and B = -2i - 2j + 3k - 2

Answer 1

Correct Answer - Option 3 : cos^-1 \(\rm 4\over17\)

Concept:

The angle between the planes is the same as the angle between their respective normals.

The angle between any two planes ai + bj + ck = 0 and pi + qj + rk = 0 is:

cos θ = \(\rm \left|ap+bq+cr\over \sqrt{a^2+b^2+c^2}\times\sqrt{p^2+q^2+r^2}\right|\)

Calculation:

Plane A = 2i + 3j + 2k + 1 

Normal to plane A n_A = 2i + 3j + 2k

Plane B = -2i - 2j + 3k - 2 

Normal to plane B nB = -2i - 2j + 3k

Now let the angle between the planes be θ 

∴ cos θ = \(\rm \left|2\times(-2)+3\times(-2)+2\times3\over \sqrt{2^2+3^2+2^2}\times\sqrt{(-2)^2+(-2)^2+3^2}\right|\)

cos θ = \(\rm \left|-4\over \sqrt{17}\times\sqrt{17}\right|\)

cos θ = \(\rm 4\over17\)

θ = cos^-1\(\boldsymbol{\rm \left(4\over 17\right)}\)