I need to prove that, for two linearly independent vectors $\mathbf{a},\mathbf{b}\in\mathbb{R}^3$,
$$(\mathbf{a} · \mathbf{a}) (\mathbf{b} · \mathbf{b}) - (\mathbf{a} · \mathbf{b})(\mathbf{a} · \mathbf{b}) = (\mathbf{a} \times \mathbf{b})·(\mathbf{a} \times \mathbf{b})$$
Could someone give me a demonstration of this identity? Or a hint to prove it?
On
The left hand side is the determinant of the Gramian matrix of $\mathbf a$ and $\mathbf b$: $$ \det\begin{pmatrix}\mathbf a\cdot\mathbf a & \mathbf a\cdot\mathbf b \\ \mathbf b\cdot\mathbf a & \mathbf b\cdot\mathbf b\end{pmatrix}. $$ In general, the determinant of a Gramian matrix is the square of the $n$-dimensional volume of a parallelotope, in this case the square of the area of the parallelogram spanned by $\mathbf a$ and $\mathbf b$.
The right hand side is $\|\mathbf a\times\mathbf b\|^2$, where $\|\mathbf a\times\mathbf b\|$ also is the area of the parallelogram spanned by $\mathbf a$ and $\mathbf b$, hence both sides are equal.
On
For a unit vector $u$, $|u\cdot v|$ is the length of the projection of $v$ onto $u$ (which is also the distance of $v$ from the plane perpendicular to $u$) and $|u\times v|$ is the length of the projection of $v$ onto the plane perpendicular to $u$ ($u\times v$ is the projection of $v$ onto the plane rotated by $\frac\pi2=90^{\large\circ}$ counter-clockwise).
The Pythagorean Theorem says $$ |u\cdot v|^2+|u\times v|^2=|v|^2\tag1 $$ Due to the linearity of "$\cdot$" and "$\times$", we can remove the restriction on $u$ by making the equation homogeneous; here, substitute $u\mapsto\frac{u}{|u|}$ (which changes nothing if $|u|=1$) and multiply by $|u|^2$: $$ |u\cdot v|^2+|u\times v|^2=|u|^2|v|^2\tag2 $$
Observe that $(A\times B)\cdot (A\times B) = |A\times B|^2 = |A|^2|B|^2\sin^2\theta$, where $\theta$ is the angle between them. Now we have $\sin^2\theta = 1-\cos^2\theta$, and I will remind you that $A\cdot B = |A||B|\cos\theta$.