Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 14

If either vector is zero, both sides vanish. Otherwise let $\theta$ be the angle between them. The magnitudes of the cross product and dot product are described bySquaring and using ${\sin }^2\theta =1-{\cos }^2\theta$ proves Lagrange's identity:

Assume the two line-defining points are distinct. Put $\mathbf{u}={\mathbf{r}}_1-{\mathbf{r}}_0$ and $\mathbf{v}={\mathbf{r}}_2-{\mathbf{r}}_0$. The triangle formed by the three points has area $|\mathbf{u}\times \mathbf{v}|/2$. The same area is half the base length $|{\mathbf{r}}_2-{\mathbf{r}}_1|$ times the perpendicular height $D$ from the third point to the line. Hence the distance from a point to a line in three dimensions isThis remains valid when the nearest point lies outside the segment, since the base line is infinite. If all three points are collinear, the numerator and distance are both zero.

Use $(0,1,1)$ as an anchor. Two directions in the required plane areTheir cross product is $(-2,8,2)$, so a normal vector is $\mathbf{n}=(-1,4,1)$. The Cartesian equation of a plane isDirect substitution verifies all three given points. The independent direction vectors ensure the plane is uniquely determined.

The axis intercepts of the plane areUse two edge vectors from $P$:The triangle area is half this cross product magnitude: