SL 3.1—Three-dimensional geometry

Use distance and midpoint in three dimensions.

Treat a 3D point as coordinates (x,y,z). Distance extends Pythagoras and midpoint averages corresponding coordinates.

Worked example

Between A(1,−2,3) and B(5,4,−1), distance is √68 and midpoint is (3,1,1).

The z-coordinate contributes just like x and y.

A midpoint is a point; divide each coordinate by 2 after adding.

Solid and spatial checks: $V_{pyramid}=Bh/3$, $V_{cone}=\pi r^2h/3$, $V_{sphere}=4\pi r^3/3$; total sphere area is $4\pi r^2$, while a solid hemisphere including its base has area $3\pi r^2$. In a $3\times4\times12$ cuboid, the base diagonal is $5$ and the space diagonal is $\sqrt{5^2+12^2}=13$. Its angle $\alpha$ with the base satisfies $\tan\alpha=12/5$, so $\alpha\approx67.38^\circ$.