Distances and Angles in Three Dimensions

For \(A(x_1,y_1,z_1)\) and \(B(x_2,y_2,z_2)\), use the same coordinate order throughout.

\(AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\)

\(M=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2}\right)\)

In this schematic cuboid, \(H\) is the perpendicular foot of \(P\) on the base. The angle between \(AP\) and the base is \(\angle PAH\), not the angle with a convenient edge. First find \(AH\) in the base, then use right triangle \(AHP\).

\(AP^2=AH^2+PH^2,\qquad \tan\angle PAH=\dfrac{PH}{AH}\)

For two intersecting lines, isolate the triangle containing both lines. For a line and a plane, use the line and its perpendicular projection onto the plane.

Example

Find the midpoint and distance between \(A(-1,2,3)\) and \(B(3,-4,15)\).

\(M=\left(\dfrac{-1+3}{2},\dfrac{2-4}{2},\dfrac{3+15}{2}\right)=(1,-1,9)\)

\(\begin{aligned}AB&=\sqrt{(3-(-1))^2+(-4-2)^2+(15-3)^2}\\&=\sqrt{16+36+144}=14.\end{aligned}\)