Vectors in two dimensionsEdexcel A-Level Maths: Revision notes

A vector has both magnitude (size) and direction; a scalar has size only. In two dimensions a vector can be written as a column vector $\begin{pmatrix}x \\ y\end{pmatrix}$ or in terms of the unit vectors $\mathbf{i}$ (one unit in the positive $x$ direction) and $\mathbf{j}$ (one unit in the positive $y$ direction): $\begin{pmatrix}3\\-4\end{pmatrix}=3\mathbf{i}-4\mathbf{j}$. The numbers $3$ and $-4$ are the components of the vector. Two vectors are equal only if both components are equal, so $x_1\mathbf{i}+y_1\mathbf{j}=x_2\mathbf{i}+y_2\mathbf{j}$ gives $x_1=x_2$ and $y_1=y_2$. In handwriting a vector is underlined, $\underline{a}$; in print it is bold, $\mathbf{a}$. A vector from $A$ to $B$ is written $\overrightarrow{AB}$.

The magnitude of $\mathbf{a}=x\mathbf{i}+y\mathbf{j}$ is $|\mathbf{a}|=\sqrt{x^2+y^2}$, by Pythagoras. The direction is the angle $\theta$ the vector makes with the positive $x$-axis (the $\mathbf{i}$ direction), measured anticlockwise, with $\tan\theta=\frac{y}{x}$ taking the quadrant into account. Component form to magnitude and direction: $\mathbf{p}=-4\mathbf{i}+4\sqrt3\,\mathbf{j}$ has $|\mathbf{p}|=\sqrt{16+48}=8$. The reference angle is $\tan^{-1}\sqrt3=60^\circ$ and the vector is in the second quadrant, so $\theta=120^\circ$. Magnitude and direction to component form: a vector of magnitude $r$ at angle $\theta$ is $r\cos\theta\,\mathbf{i}+r\sin\theta\,\mathbf{j}$. For $r=10$, $\theta=30^\circ$: $5\sqrt3\,\mathbf{i}+5\mathbf{j}$.

A unit vector has magnitude $1$. The unit vector in the direction of $\mathbf{a}$ is written $\hat{\mathbf{a}}$ and found by dividing by the magnitude: $\hat{\mathbf{a}}=\frac{\mathbf{a}}{|\mathbf{a}|}.$ For $\mathbf{a}=5\mathbf{i}-12\mathbf{j}$, $|\mathbf{a}|=13$ and $\hat{\mathbf{a}}=\frac{1}{13}(5\mathbf{i}-12\mathbf{j})$. To find a vector of a given magnitude $m$ in the direction of $\mathbf{a}$, take $m\hat{\mathbf{a}}$. A vector of magnitude $26$ in the direction of $\mathbf{a}$ is $2\mathbf{a}$.

To add vectors diagrammatically, draw the second vector starting from the end of the first (the triangle law): $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$. In a parallelogram, two vectors drawn from the same point form adjacent sides and their sum is the diagonal from that point (the parallelogram law). The sum is also called the resultant. Algebraically, add the components: $(2\mathbf{i}+3\mathbf{j})+(-5\mathbf{i}+\mathbf{j})=-3\mathbf{i}+4\mathbf{j}$. Subtracting a vector means adding its negative: $\mathbf{a}-\mathbf{b}=\mathbf{a}+(-\mathbf{b})$, and $\overrightarrow{BA}=-\overrightarrow{AB}$. Example: $\overrightarrow{AB}=6\mathbf{i}+8\mathbf{j}$ and $\overrightarrow{BC}=-2\mathbf{i}+5\mathbf{j}$ give $\overrightarrow{AC}=4\mathbf{i}+13\mathbf{j}$, and $AC=\sqrt{185}$.

Multiplying a vector by a scalar $\lambda$ multiplies each component by $\lambda$ and the magnitude by $|\lambda|$. If $\lambda>0$ the direction is unchanged; if $\lambda<0$ the direction is reversed. Two non-zero vectors are parallel exactly when one is a scalar multiple of the other, $\mathbf{b}=\lambda\mathbf{a}$. To test, compare ratios of components: $4\mathbf{i}-6\mathbf{j}$ is parallel to $2\mathbf{i}-3\mathbf{j}$ since $\lambda=2$. Worked example: $\mathbf{a}=2\mathbf{i}+3\mathbf{j}$ and $\mathbf{b}=(k+1)\mathbf{i}+(k-2)\mathbf{j}$ are parallel. Then $\frac{k+1}{2}=\frac{k-2}{3}$, so $3k+3=2k-4$ and $k=-7$.