Vectors

3.7 Vectors

Notation

A vector has size and direction. The same vector can be written in several ways:

$\overrightarrow{AB} = \mathbf{a} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}$

The vectors $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ have length 1 and point along the $x$, $y$ and $z$ axes.

Operations

Add, subtract and scale component by component.

• $\mathbf{a} + \mathbf{b}$ is one vector followed by the other.
• $k\mathbf{a}$ is parallel to $\mathbf{a}$, and $k$ times as long. A negative $k$ reverses the direction.
• Two vectors are parallel if one is a multiple of the other.

The position vector of $A$ is $\overrightarrow{OA}$, from the origin. Then:

$\overrightarrow{AB} = 2\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}$, and $\overrightarrow{OC} = \overrightarrow{OA} + \tfrac{1}{3}\overrightarrow{AB}$:

$\overrightarrow{OC} = \frac{5}{3}\mathbf{i} + \mathbf{j} - \frac{1}{3}\mathbf{k}$

Magnitude and unit vectors

The magnitude is $\sqrt{4 + 1 + 4} = 3$, so the unit vector is $\tfrac{1}{3}(2\mathbf{i} - \mathbf{j} + 2\mathbf{k})$.

The scalar product

Rearranged, this gives the angle between two vectors. If $\mathbf{a}\cdot\mathbf{b} = 0$ for non-zero vectors, they are perpendicular.

$\cos\theta = \frac{1 - 1 - 2}{\sqrt{6}\sqrt{3}} = \frac{-2}{\sqrt{18}} \quad\Rightarrow\quad \theta = 118.1^\circ$

A negative scalar product means an obtuse angle. The acute angle between the lines is $180^\circ - 118.1^\circ = 61.9^\circ$.

The equation of a line

Each value of $t$ gives a different point on the line.

The direction is $\overrightarrow{AB} = \mathbf{i} - \mathbf{j} + 2\mathbf{k}$, so

$\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} + s(\mathbf{i} - \mathbf{j} + 2\mathbf{k})$

How two lines can meet

In three dimensions, two lines are in one of three positions:

Equate components:

$1 + s = 1 + t \qquad 2 - s = t \qquad 3 + 2s = 6 - t$

The first two give $s = t = 1$. Check the third: $5 = 5$. The lines meet at $(2, 1, 5)$.

The foot of a perpendicular

To find the point on a line closest to a point $P$, write a general point $F$ on the line and make $\overrightarrow{PF}$ perpendicular to the direction.

A general point on $l_1$ is $F(1 + s, 2 - s, 3 + 2s)$, so $\overrightarrow{PF} = (s - 3)\mathbf{i} + (-1 - s)\mathbf{j} + (2s - 2)\mathbf{k}$.

Using the scalar product with the direction:

$(s - 3) - (-1 - s) + 2(2s - 2) = 0 \quad\Rightarrow\quad 6s - 6 = 0 \quad\Rightarrow\quad s = 1$

The foot is $F(2, 1, 5)$. Then $\overrightarrow{PF} = -2\mathbf{i} - 2\mathbf{j}$, and the shortest distance is $\sqrt{8} = 2\sqrt{2}$.