Distance in space is Pythagoras, twice.

A third axis joins the other two, and every rule gains a third term. The length of a 3D vector comes from two right-angled triangles, one flat on the floor and one standing up.

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$P$ is $(3, 4, 2)$: go $3$ along $x$, $4$ along $y$ and $2$ up. Its position vector is $\overrightarrow{OP} = 3\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}$. How far is $P$ from $O$?

First the point $Q = (3, 4, 0)$ on the floor, straight below $P$. The triangle with sides $3$ and $4$ lies flat, with its right angle on the $x$-axis, so

$P$ is $2$ straight above $Q$, at right angles to the floor, so the triangle $OQP$ stands upright:

The $5^2$ was $3^2 + 4^2$, so in one step, $|x\mathbf{i} + y\mathbf{j} + z\mathbf{k}| = \sqrt{x^2 + y^2 + z^2}$.

The component along an axis is the length times the cosine of the angle with that axis. For $\overrightarrow{OP}$, the angle with the $z$-axis has $\cos\theta_z = \tfrac{2}{\sqrt{29}}$, so $\theta_z = 68.2^\circ$.

Three components, one square root.

Unit vectors, adding and scaling in 3D, parallelograms and collinear points, and twelve graded exam questions with full worked solutions are waiting.

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