Prove that for $x,y\in \mathbb{R}^3,\lVert x \times y \rVert^2 = \lVert x \rVert^2 \lVert y \rVert^2 - \langle x,y \rangle^2$ Prove  that for $x,y\in \mathbb{R}^3,\lVert x \times y \rVert^2 = \lVert x \rVert^2 \lVert y \rVert^2 - \langle x,y \rangle^2$
Use this identity to show that for $x,y \neq 0$, with the angle $\alpha$ between x and y it follows that:
$\lVert x \times y \rVert = \lVert x \rVert  \lVert y \rVert  \sin(\alpha)$
I don't have any idea for a starting point, can someone help me prove these please?
 A: As a point of interest, though almost certainly not the way you're intended to solve the problem, this is a trivial question when attacked with clifford algebra:
Define the "geometric product" of vectors as an associative product:  $x(yz) = (xy)z$ for instance.  Define $xx = x \cdot x = x^2$ and, for $a, b$ orthogonal, $ab = -ba$.  Define the grade of a product as the least number of orthogonal vectors that produce the same overall product.  For instance, if $a, b, c$ are all mutually orthogonal, then the product $abc$ is a grade-3 "multivector".
Result:  for any $x, y$, the geometric product $xy$ can be broken down as follows:
$$xy = x \cdot y + x \wedge y, \quad x \cdot y = \langle xy \rangle_0, \quad x \wedge y = \langle x y \rangle_2$$
where $\langle A \rangle_n$ denotes the grade-$n$ part of a multivector.

Given all these definitions, consider $xyyx$:
$$xyyx = x (yy)x = x (y^2) x = x^2 y^2$$
But, the product is associative, so you can group the vectors differently:
$$xyyx = (xy)(yx) = (x \cdot y + x \wedge y) (x \cdot y + y \wedge x)$$
It follows from the properties of the product that $x \wedge y = - y \wedge x$, so the result is
$$xyyx = (x \cdot y)^2 - (x \wedge y)^2 = x^2 y^2$$
On its face, this seems to run counter to the instructed result.  That's not the case:  it turns out that any 2-vector of the form $x \wedge y$ will always square to a negative number under the product.  (Consider:  for $a, b$ orthogonal, $abab = -abba = -a^2 b^2 < 0$.)
So, written in terms of magnitudes, we get
$$(x \cdot y)^2 + |x \wedge y|^2 = |x|^2 |y|^2$$
The magnitude of the "wedge product" is exactly that of the cross product, which is a whole topic in unto itself.
