Question

If α,β and γ are the angles made by the vector 3i - 6j + 2k with the positive direction of the coordinate axes then find cos α, cos β, cos γ.

Answer 1

The Answer is:  cos α = 3/7   ,  cos β = -6/7  ,   cos γ = 2/7

The given vector is

                                     "  3i - 6j + 2k  "

$MagnitudeOfVector=\sqrt{(3)^{2} +(-6)^{2}+(2)^{2}}=7 \\DirectionCosinesOfVector=[\frac{3}{7},\frac{-6}{7},\frac{2}{7}]$

So

cos α = 3/7   ,    cos β = -6/7    ,     cos γ = 2/7

ANGLES:

$With\ x-axis = Cos^{-1}\frac{3}{7} =64.62degrees \\With\ y-axis = Cos^{-1}\frac{-6}{7} = 149degrees \\With\ z-axis = Cos^{-1}\frac{2}{7} =73.39degrees$

Answer 2

Answer:

I hope this may help you..