Vectors in Three Dimensions

You already know how to add and scale vectors in the plane: work component by component. In space nothing changes except that a vector has three components, $\mathbf{v} = \langle v_1, v_2, v_3 \rangle$. Think of it as a displacement: $v_1$ units in the $x$-direction, $v_2$ in the $y$-direction, $v_3$ in the $z$-direction. Where you start does not matter; only the size and direction of the move do.

Idea clave

From $P$ to $Q$: $\overrightarrow{PQ} = Q - P$ ("end minus start"). Vectors add and scale component by component, and $$|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}, \qquad \text{unit vector in the direction of } \mathbf{v}: \ \frac{\mathbf{v}}{|\mathbf{v}|}.$$ With $\mathbf{i} = \langle 1, 0, 0 \rangle$, $\mathbf{j} = \langle 0, 1, 0 \rangle$, $\mathbf{k} = \langle 0, 0, 1 \rangle$, every vector is $v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k}$.

Por qué funciona

Adding two displacements means doing one move and then the other. The $x$-parts of the two moves simply add, and so do the $y$-parts and the $z$-parts, because the three directions are independent. The length formula is the 3-D distance formula applied to the move from the tail $(0, 0, 0)$ to the tip $(v_1, v_2, v_3)$. Multiplying by $c$ multiplies every component, so it multiplies the length by $|c|$; dividing by $|\mathbf{v}|$ therefore makes the length exactly ${}1$ without changing the direction.

Tip to tail: $\mathbf{u} + \mathbf{v}$ goes from the tail of $\mathbf{u}$ to the tip of $\mathbf{v}$. Change the components in the boxes and turn the picture to see it in 3-D.

Worked examples

Ejemplo resuelto: From two points

Find $\overrightarrow{PQ}$, its length, and the unit vector in its direction for $P(2, -1, 4)$ and $Q(5, 3, 4)$.

$$\overrightarrow{PQ} = \langle 5 - 2, 3 - (-1), 4 - 4 \rangle = \langle 3, 4, 0 \rangle.$$
$|\overrightarrow{PQ}| = \sqrt{9 + 16 + 0} = 5$.
Unit vector: $\frac{1}{5}\langle 3, 4, 0 \rangle = \left\langle \frac{3}{5}, \frac{4}{5}, 0 \right\rangle$. Its length is $\sqrt{\frac{9}{25} + \frac{16}{25}} = 1$.

Respuesta: $\langle 3, 4, 0 \rangle$, length ${}5$, unit vector $\left\langle \frac{3}{5}, \frac{4}{5}, 0 \right\rangle$

Ejemplo resuelto: A combination

For $\mathbf{u} = \langle 1, -2, 2 \rangle$ and $\mathbf{v} = \langle 3, 0, -1 \rangle$, find ${}2\mathbf{u} - 3\mathbf{v}$ and its length.

${}2\mathbf{u} = \langle 2, -4, 4 \rangle$ and ${}3\mathbf{v} = \langle 9, 0, -3 \rangle$.
Subtract matching components: $$\langle 2 - 9, -4 - 0, 4 - (-3) \rangle = \langle -7, -4, 7 \rangle.$$
Length: $\sqrt{49 + 16 + 49} = \sqrt{114}$.

Respuesta: $\langle -7, -4, 7 \rangle$, length $\sqrt{114}$

Ejemplo resuelto: A prescribed length

Find the vector of length ${}10$ pointing the same way as $\mathbf{w} = \langle 2, -1, 2 \rangle$.

$|\mathbf{w}| = \sqrt{4 + 1 + 4} = 3$, so the unit vector is $\frac{1}{3}\langle 2, -1, 2 \rangle$.
Stretch it to length 10: $\frac{10}{3}\langle 2, -1, 2 \rangle = \left\langle \frac{20}{3}, -\frac{10}{3}, \frac{20}{3} \right\rangle$.

Respuesta: $\left\langle \frac{20}{3}, -\frac{10}{3}, \frac{20}{3} \right\rangle$

Two vectors are parallel when one is a scalar multiple of the other: $\langle 2, -4, 6 \rangle = 2\langle 1, -2, 3 \rangle$. A negative multiple points the opposite way but is still parallel.

Error común

Start minus end. $P - Q$ is the vector from $Q$ to $P$, the opposite direction. Say "end minus start" out loud. A half-made unit vector. Dividing only some components by the length, or dividing by the sum of the components, gives a vector that is not length 1. Check by computing the length of your answer.

Worked examples

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