Problem: Let A be an arbitrary vector, and let e be a unit vector in some fixed direction. Show that, A=e(A\(\bullet\)e) + e x (Axe)

Work:
A=e(A\(\bullet\)e)+ex(Axe)
A=|A|cos(θ)e + e x (|A|sin(θ)n)
A=|A|cos(θ)e + Asin(θ)sin(90°)
A=|A|cos(θ)e + Asin(θ)

This is as far as I got. I'm not sure if I'm going in the right direction.

 

Aren't the two angles the same?

How did the vector A pop back into your second term in your third line?

You might try writing things in terms of vector components and not even introduce trig functions at all.

 

Aren't the two angles the same?

How did the vector A pop back into your second term in your third line?

You might try writing things in terms of vector components and not even introduce trig functions at all.

You're right the angles are the same. I realized that vector e x |A|sin(θ)n wouldn't give me A but a vector perpendicular to n. I didn't think about solving it in vector components, but trying to solve it with undefined variables made the work messy really quick. I chose arbitrary values that fit that problem and it worked. Thanks!

 

So now go back and use generic vectors:

\( \vec A \, = \, A_x \hat x \, + \, A_y \hat y \, + \, A_z \hat z\)
\( \vec e \, = \, e_x \hat x \, + \, e_y \hat y \, + \, e_z \hat z\)
\( e_x^2 \, + \, e_y^2 \, + \, e_z^2 \, = \, 1\)