Cross Product and Lagrange's Identity

$\begin{align}|x \times y|^2&=\begin{vmatrix}x\cdot x&x\cdot y\\x\cdot y & y\cdot y \end{vmatrix}\\&=|x|^2|y|^2-(x\cdot y)^2\\
&=|x|^2|y|^2(1-\cos^2 \theta)\\
&=|x|^2|y|^2\sin^2 \theta\end{align}$

Proof of $|x \times y|^2=|x|^2|y|^2-(x\cdot y)^2$ (Lagrange's Identity):

$\large{\begin{align}RHS
&=\sum_{i=1}^n {x_i}^2 \sum_{i=1}^n {y_i}^2-(\sum_{i=1}^n {x_i y_i})^2\\
&=\sum_{i=1}^n {x_i}^2 \sum_{j=1}^n {y_j}^2-\sum_{i=1}^n {x_i y_i}\sum_{j=1}^n {x_j y_j}\\
&=\sum_{i,j=1}^n {x_i}^2 {y_j}^2-\sum_{i,j=1}^n {x_i y_i x_j y_j}\\
&=\sum_{i<j} {x_i}^2{y_j}^2+\sum_{i=1}^n {x_i}^2 {y_i}^2+\sum_{i>j} {x_i}^2{y_j}^2\\
& -\sum_{i<j} x_i y_i x_j y_j- \sum_{i=1}^n{x_i}^2{y_i}^2-\sum_{i>j} x_i y_i x_j y_j\\
&=\sum_{i<j} {x_i}^2 {y_j}^2+\sum_{i<j} {x_j}^2 {y_i}^2-2\sum_{i<j} x_i y_i x_j y_j\\
&=\sum_{i<j} ({x_i}^2{y_j}^2-2x_i y_i x_j y_j+{x_j}^2{y_i}^2)\\
&=\sum_{i<j} (x_i y_j-x_j y_i)^2\\
&=LHS \end{align}}$

Computations with cross products:

Area of triangle in space
Let $u,v,w$ be vertices of a triangle. Treat them as arrows from the origin. Then, two sides of the triangle can be given by $v-u$ and $w-u$. Thus, the area of the triangle is $\frac{1}{2}|(v-u)\times (w-u)|$.
By the bilinearity of cross products, $(\boldsymbol{v-u})\times (\boldsymbol{w-u})=v\times w-u\times w-v\times u+u\times u$.
By antisymmetry, $u\times u=0$.
Therefore, the area of the triangle is given by $\frac{1}{2}|v\times w+w\times u+u\times v|$. Note the cyclic pattern.

Remark: We can make use of this result to prove the three dimensional Pythagorean theorem. Consider the tetrahedron with vertices $A=(a,0,0)$, $B=(0,b,0)$, $C=(0,0,c)$ and $O=(0,0,0)$. For any three points $X,Y,Z$ in space, write $[X,Y,Z]$ for the area of the triangle with vertices $X,Y,Z$. Then, the three dimensional Pythagorean theorem states that $[ABC]^2=[OAB]^2+[OBC]^2+[OCA]^2$.

Equation of the plane spanned by two vectors (skipped)

Intersection of two planes
Say we have two non-parallel planes $ax+by+cz=0$ and $a'x+b'y+c'z=0$. They have normal vectors $n=(a,b,c)$ and $n’=(a',b',c')$ respectively. The vector $n \times n’$ is perpendicular to both $n$ and $n’$, so it lies on both planes. The line of intersection of the planes is therefore the line through $n \times n’$.

References:
http://www.owlnet.rice.edu/~fjones/chap7.pdf
https://www2.bc.edu/~reederma/Linalg13.pdf

Useful:
http://www.odeion.org/pythagoras/pythag3d.html
http://www.netcomuk.co.uk/~jenolive/homevec.html