Working with vectors in the Cartesian plane

This method handles a vector once it sits in the Cartesian plane, in component form $x\mathbf{i} + y\mathbf{j}$, or as a column vector $\begin{pmatrix} x \\ y \end{pmatrix}$. From that form you can read off everything the Vectors chapter asks for: the horizontal part $x$ along $\mathbf{i}$, the vertical part $y$ along $\mathbf{j}$, the magnitude (length) $\sqrt{x^{2}+y^{2}}$, and the unit vector that points the same way but has length $1$.

It answers questions such as 'express $\overrightarrow{AB}$ in terms of $\mathbf{i}$ and $\mathbf{j}$', 'find the magnitude of the vector', and 'find the unit vector in the direction of $\mathbf{r}$'. Because points and vectors share the same axes, it also turns two coordinates into a directed segment in one clean step.

Reach for the Cartesian approach whenever a vector is given in component form, as a column vector, or through the coordinates of points. Tell-tale phrases are 'in terms of $\mathbf{i}$ and $\mathbf{j}$', 'the magnitude of', 'the unit vector in the direction of', or a diagram drawn on labelled $x$- and $y$-axes.

If the question gives two points $A$ and $B$, first turn them into position vectors and use $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ to obtain a single $x\mathbf{i} + y\mathbf{j}$. Once a vector is in components, magnitude and direction become pure arithmetic, nothing needs to be measured off a diagram, which is exactly why the Cartesian form is the reliable one to fall back on under exam pressure.

It is also the form the next skills, such as parallel vectors and ratios, build directly upon.

1. 1

   Write the vector in component form

   Put the vector as $x\mathbf{i} + y\mathbf{j}$. From two points, use $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$, end minus start.

2. 2

   Read off the components

   Identify $x$ (the $\mathbf{i}$ part) and $y$ (the $\mathbf{j}$ part), keeping their signs.

3. 3

   Find the magnitude

   Apply $|\mathbf{r}| = \sqrt{x^{2} + y^{2}}$: square each component, add, then take the positive root.

4. 4

   Find the unit vector

   Divide the whole vector by its magnitude: $\hat{\mathbf{r}} = \frac{1}{|\mathbf{r}|}(x\mathbf{i} + y\mathbf{j})$.

5. 5

   Scale or combine if asked

   For $k\mathbf{r}$, multiply each component by $k$; to combine vectors, add or subtract matching components.

6. 6

   Check the result

   A unit vector must satisfy $\sqrt{x^{2}+y^{2}} = 1$; a magnitude is always positive.

• Forgetting the square root, leaving the answer as $x^{2} + y^{2}$ instead of $\sqrt{x^{2} + y^{2}}$.
• Reporting a negative magnitude; a length is always zero or positive, so $\sqrt{\dots}$ is the non-negative root.
• Squaring a negative component wrongly: $(-4)^{2} = 16$, not $-16$.
• Dividing only one component by the magnitude when forming the unit vector, instead of the whole vector.
• Reversing a directed segment by writing $\overrightarrow{OA} - \overrightarrow{OB}$ in place of $\overrightarrow{OB} - \overrightarrow{OA}$.

The step that quietly costs marks is the magnitude, students square the components correctly, then forget the root or let a stray minus survive under it. In one-to-one lessons our teachers make you write $|\mathbf{r}| = \sqrt{x^{2}+y^{2}}$ as a fixed template and read the value straight off, so $(-4)^{2}$ can never turn into $-16$ and the answer is always the positive root.

We build the unit vector as 'the whole vector divided by its length', then check together that it squares back to $1$. Because SPM Add Math uses analytic marking, a correctly set-up $\sqrt{6^{2}+8^{2}}$ already earns method marks even if the arithmetic slips.

Lessons are taught in English, though the SPM paper is set bilingually. Every teacher on spmaddmath.com.my is experienced, and lessons run online at your own pace.