Vectors

Writing a vector

A vector has both size and direction. Two notations appear on this syllabus and they mean exactly the same thing:

$$\mathbf{a} = 3\mathbf{i} + 4\mathbf{j} \qquad\text{and}\qquad \mathbf{a} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$$

Here $\mathbf{i}$ is one unit to the right and $\mathbf{j}$ is one unit up. Use whichever form the question uses.

Magnitude

The magnitude — the length — comes from Pythagoras on the two components:

Adding, subtracting and scaling

Work component by component. Add the $\mathbf{i}$ parts, add the $\mathbf{j}$ parts, and keep them apart:

$$(2\mathbf{i} + \mathbf{j}) + (3\mathbf{i} - 4\mathbf{j}) = 5\mathbf{i} - 3\mathbf{j}$$

Multiplying by a scalar multiplies both components: $3(2\mathbf{i} - \mathbf{j}) = 6\mathbf{i} - 3\mathbf{j}$.

Position vectors and the vector joining two points

The position vector of a point $A$ is $\overrightarrow{OA}$, the vector from the origin to $A$. So the point $A(1,2)$ has position vector $\mathbf{i} + 2\mathbf{j}$.

Destination minus start. Going from $A(1,2)$ to $B(5,5)$ gives $\overrightarrow{AB} = (5-1)\mathbf{i} + (5-2)\mathbf{j} = 4\mathbf{i} + 3\mathbf{j}$.

Unit vectors

A unit vector has magnitude 1. To find the unit vector in the direction of $\mathbf{a}$, divide $\mathbf{a}$ by its own magnitude:

For $6\mathbf{i} - 8\mathbf{j}$ the magnitude is $\sqrt{36+64} = 10$, so the unit vector is $0.6\mathbf{i} - 0.8\mathbf{j}$. Check it: $\sqrt{0.36 + 0.64} = 1$.

Parallel vectors and collinear points

So $2\mathbf{i} - 3\mathbf{j}$ and $-4\mathbf{i} + 6\mathbf{j}$ are parallel, because the second is $-2$ times the first. Both components must scale by the same factor — doubling one and leaving the other is not a scalar multiple.

Three points are collinear if they lie on one straight line. To prove it, show that two vectors between them are parallel and that they share a point:

For $A(1,2)$, $B(4,8)$, $C(6,12)$: $\overrightarrow{AB} = 3\mathbf{i}+6\mathbf{j}$ and $\overrightarrow{BC} = 2\mathbf{i}+4\mathbf{j} = \tfrac23 \overrightarrow{AB}$. Parallel, and both pass through $B$, so $A$, $B$ and $C$ are collinear.

The scalar product

Multiply matching components and add. The result is a number, not a vector — which is why it is called the scalar product.

Its most-used consequence:

So $2\mathbf{i}+3\mathbf{j}$ and $6\mathbf{i}-4\mathbf{j}$ are perpendicular, since $12 - 12 = 0$.