# Vector Arithmetic Problem

• Aug 30th 2010, 06:41 PM
davesface
Vector Arithmetic Problem
Let $\vec{u}$ be an arbitrary fixed unit vector and show that any vector $\vec{b}$ satisfies $b^2=(\vec{u}\cdot\vec{b})^2+ (\vec{u}\times\vec{b})^2$

Explain this result in words, with the help of a picture.

So I got to the point where I can see the end in sight: $b_1(u_1^2+u_2^2+u_3^2)+b_2(u_1^2+u_2^2+u_3^2)+b_3( u_1^2+u_2^2+u_3^2)-2(u_2 u_3 b_2 b_3+u_1 u_3 b_1 b_3 +u_1 u_2 b_1 b_2)$

Obviously the $(u_1^2+u_2^2+u_3^2)$'s will all go away since u is a unit vector. My only problem is that I have some leftover terms from the dotting of the cross products which I don't see how to get rid of since u and b are so generalized. Since the first part seems to be right, I have to assume that I either did something wrong in the cross product arithmetic or else I'm missing some property of dot products that will make that last term go away. Any thoughts?
• Aug 31st 2010, 12:55 AM
Opalg
Quote:

Originally Posted by davesface
Let $\vec{u}$ be an arbitrary fixed unit vector and show that any vector $\vec{b}$ satisfies $b^2=(\vec{u}\cdot\vec{b})^2+ (\vec{u}\times\vec{b})^2$

Explain this result in words, with the help of a picture.

Here's a very brief explanation in words, without the help of a picture. The quantity $\vec{u}\cdot\vec{b}$ represents the magnitude of the component of $\vec{b}$ in the direction of $\vec{u}$. The quantity $\vec{u}\times\vec{b}$ represents the component of $\vec{b}$ perpendicular to $\vec{u}$. The result then follows from Pythagoras' theorem.
• Aug 31st 2010, 09:37 AM
davesface
The picture isn't so much the issue, since the geometric interpretations of dot and cross products is pretty simple. I just included that last sentence as a part of the full problem statement. The real issue I'm having is with that -2... term that's leftover.

I'll add that, upon starting the problem from scratch again, I see that it is much easier to do using the trigonometric definitions of dot and cross products. I'm convinced that the equation holds true from that, but out of curiosity I still want to know what I did wrong the first time.