find the vector a {1, 1, 1} length of the projection in the vector b

Question

I need to find the vector a {1, 1, 1} length of the projection in the vector b

Given: vector b with the OX axis forms a 30 degree angle

vector b with the OZ axis forms a 60 degree angle

What formulas do I need to use?

Answer 1

Hint:

$\vec{a} \cdot \vec{b} = ||a|| \,||b|| \,\cos \theta$

Say $\vec {b} = (b_1, b_2, b_3)$

Unit vector in the direction of $x-$axis $= (1,0,0)$

So $(b_1, b_2, b_3) \cdot (1,0,0) = \sqrt{b_1^2 + b_2^2 + b_3^2} \, \cos 30^0$

Or the direction angle of a vector with $x$ and $z$ axis is simply given by

$\cos \alpha = \frac{b_1}{\sqrt{b_1^2 + b_2^2 + b_3^2}}, \cos \gamma = \frac{b_3}{\sqrt{b_1^2 + b_2^2 + b_3^2}} \,$ respectively.

That gives you two equations.

Solve for $b_1, b_2, b_3$ in terms of one of them and divide by magnitude to get the unit vector in the direction of $\vec{b}$.

Now dot product of $(1,1,1)$ with unit vector in the direction of $\vec{b}$ will give the projection of $(1,1,1)$ in vector $\vec{b}$.