Vector Help: Prove $B^2 = (U \cdot B)^2 + (U \times B)^2$ $U$ is a fixed unit vector and $B$ is any vector. 
My attempt so far. Am I even starting this right?
$$B^2 = (U_xB_x + U_yB_y)(U_xB_x + U_yB_y) + (U_xB_y - B_xU_y)(U_xB_y - B_xU_y)$$
$$B^2 = (U_x^2B_y^2 + U_xB_xU_yB_y + U_xB_xU_yB_y + U_y^2B_y^2) + (U_x^2B_y^2 - U_x^2B_y^2 - U_x^2B_y^2 +U_x^2B_y^2)$$
The cross product portion cancels to $0$.
I'm left with $B^2 = U_x^2B_x^2 + U_y^2B_y^2 + 2U_xB_xU_yB_y$
I'm confused.
 A: Perhaps instead of working with the coordinates of $U$ and $B$, you should consider the geometry of the dot and cross products. Recall:
$$U\cdot B = |U||B|\cos\theta, ~ |U\times B| = |U||B|\sin\theta$$
where $\theta$ is the angle between $U$ and $B$. Since $|U| = 1$ the desired result becomes obvious:
$$(U\cdot B)^2 + |U\times B|^2 = (|U||B|\cos\theta)^2 + (|U||B|\sin\theta)^2 = |B|^2(\sin^2\theta + \cos^2\theta) = |B|^2,$$
where I have used the trigonometric identity $\sin^2\theta+\cos^2\theta = 1$.
This proof has the advantage of working in both 2 and 3 dimensions. It is also independent of the choice of basis for $\mathbb{R}^3$, since it makes no reference to coordinates, hence one can consider this an "intrinsic" proof.
A: As noted in my comment above, a flaw in your work above is that you've not accurately expanded the squares in going to the second line. Once you've fixed that, your derivation should go through if you recall that $U$ is a unit vector.
However, one issue is that you've assumed that $B$ and $U$ are both vectors in the $xy$-plane. The identity actually holds more generally: $B$ and $U$ can be any two vectors in $\mathbb{R}^3$ with $U$ a fixed unit vector. Both follow as special cases of the more general vector quadrupole product identity for any four vectors in $\mathbb{R}^3$, which states
$$(A\times B)\cdot(C\times D)=(A\cdot C)(B\cdot D)-(A\cdot D)(B\cdot C).$$
The identity in the OP follows by taking $D=B$ and $A=C=U$ with $U\cdot U=1$. For a proof of the quadrupole product identity, see this answer. (For the reader familiar with Levi-Civita symbols, it follows automatically from the identity $\epsilon_{ijk}\epsilon_{klm}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$.)
