Vector prob.: Find the angle which a = 3i - 6j + 2k makes...

[Guest]

Find the angles which the vector a = 3i - 6j + 2k makes with the coordinate axes.

If the angles are alpha, beta, and theta show that for any 3-dimensional vector cos^2 alpha + cos ^2 beta + cos^2 theta = 1.

Thank you in advance

 

[soroban]

Re: Vector problem

Hello, americo74!

Find the angles which the vecotr \(\displaystyle \vec{v} \:= \:3i\,-\,6j\,+\,2k\) makes with the coordinate axes.
If the angles are \(\displaystyle \alpha,\:\beta,\:\theta\), show that for any 3-dimensional vector:
. . . \(\displaystyle \cos^2\alpha\,+\,cos^2\beta\,+\,\cos^2\theta \:=\: 1\)

You're expected to be familiar with the formulas for this problem.

Given vector \(\displaystyle \vec{v}\:=\;ai\,+\,bj\,+\,ck\), the angle cosines are:
. . \(\displaystyle \:\cos\alpha\:=\:\frac{a}{\sqrt{a^2\,+\,b^2\,+\,c^2}}\;\;\cos\beta \:=\:\frac{b}{\sqrt{a^2\,+\,b^2\,+\,c^2}}\;\;\cos\theta\:=\:\frac{c}{\sqrt{a^2\,+\,b^2\,+\,c^2}}\)