1. ## 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?

2. 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.

3. 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.