# Vector proof

Discussion in 'Homework Help' started by JasonL, Oct 5, 2013.

1. ### JasonL Thread Starter Active Member

Jul 1, 2011
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0
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.

Last edited: Oct 5, 2013
2. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
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.

JasonL likes this.
3. ### JasonL Thread Starter Active Member

Jul 1, 2011
47
0
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!

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Last edited: Oct 5, 2013
4. ### WBahn Moderator

Mar 31, 2012
18,087
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So now go back and use generic vectors:

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