Getting comfortable with vectors in Add Math

Vectors tend to feel strange the first time you meet them, and there is a fair reason for that. Up to this point in Add Math, every quantity you have handled has been a single number, a length, an area, a price.

A vector is the first quantity that insists on carrying two pieces of information at once: how big it is, and which way it points.

That second piece, direction, is the whole idea. A vector has both a magnitude (a size) and a direction.

A quantity with size but no direction is called a scalar. "Walk 3 km" is a scalar; "walk 3 km due north" is a vector, because now you also know where you would end up.

• Scalars, distance, speed, mass, temperature, time: size only.
• Vectors, displacement, velocity, force: a size and a direction bound together.

Because a vector holds two things, we write it in particular ways. The vector that runs from point $A$ to point $B$ is written $\vec{AB}$.

A vector named with a single letter is usually printed in bold, like $\mathbf{a}$; when you write the same vector by hand you underline it or draw a small arrow or wavy line beneath the letter. Different clothing, same object, do not let the notation put you off.

The move that turns vectors from pictures into something you can calculate with is breaking each one into components along two fixed directions. We use $\mathbf{i}$ for one unit step in the $x$-direction and $\mathbf{j}$ for one unit step in the $y$-direction.

Every vector in the plane can then be written as so many $\mathbf{i}$'s plus so many $\mathbf{j}$'s.

So $\mathbf{r} = 3\mathbf{i} + 4\mathbf{j}$ simply means: go 3 across and 4 up. The very same vector can be stacked as a column, which is often tidier when you calculate:

$$\mathbf{r} = x\mathbf{i} + y\mathbf{j} = \begin{pmatrix} x \\ y \end{pmatrix}$$

The magnitude, the length of the vector, comes straight from Pythagoras, because the two components sit at right angles to each other:

$$|\mathbf{r}| = \sqrt{x^{2} + y^{2}}$$

For $\mathbf{r} = 3\mathbf{i} + 4\mathbf{j}$, the magnitude is $\sqrt{3^{2}+4^{2}} = \sqrt{25} = 5$. A unit vector is a vector of length exactly $1$ that points the same way; you build it by dividing a vector by its own magnitude:

$$\hat{r} = \frac{1}{|\mathbf{r}|}\,\mathbf{r}$$

So the unit vector along $3\mathbf{i}+4\mathbf{j}$ is $\frac{1}{5}(3\mathbf{i}+4\mathbf{j}) = \frac{3}{5}\mathbf{i} + \frac{4}{5}\mathbf{j}$. Check it if you like: $\sqrt{(3/5)^{2} + (4/5)^{2}} = \sqrt{9/25 + 16/25} = 1$, exactly as a unit vector should be.

To add two vectors you place them nose to tail: draw the first, start the second where the first ends, and the resultant runs from the very start to the very end. That is the triangle law, and it is worth picturing, but in component form the whole thing becomes almost effortless, because you just add the matching components:

$$(x_{1}\mathbf{i} + y_{1}\mathbf{j}) + (x_{2}\mathbf{i} + y_{2}\mathbf{j}) = (x_{1}+x_{2})\mathbf{i} + (y_{1}+y_{2})\mathbf{j}$$

Subtraction works the same way, component by component. The vector $-\mathbf{a}$ has the same length as $\mathbf{a}$ but points the opposite way, so subtracting a vector is really adding its reverse.

Multiplying a vector by a number $k$, a scalar, stretches or shrinks it without turning it, unless $k$ is negative, in which case it also flips. This gives one of the most useful facts in the whole chapter: two non-zero vectors are parallel exactly when one is a scalar multiple of the other, $\mathbf{a} = k\mathbf{b}$.

A position vector is simply the vector from the origin $O$ to a point. The position vector of $A$ is $\vec{OA}$, which we often shorten to $\mathbf{a}$.

Position vectors are how a diagram of points gets turned into vectors you can compute with.

The single most-used result in the chapter connects two points to the vector between them:

$$\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a}$$

Read it aloud as "end minus start". If $A$ has position vector $2\mathbf{i} + \mathbf{j}$ and $B$ has $5\mathbf{i} + 6\mathbf{j}$, then $\vec{AB} = (5-2)\mathbf{i} + (6-1)\mathbf{j} = 3\mathbf{i} + 5\mathbf{j}$.

From there the distance $AB$ is just $|\vec{AB}| = \sqrt{3^{2}+5^{2}} = \sqrt{34}$.

Getting the order right, end minus start, never start minus end, is where a surprising number of quiet marks are won or lost. Write the formula down first, in letters, then substitute the numbers underneath it.

That habit keeps the direction the correct way round.

Most lost marks in this chapter come from a handful of slips that have nothing to do with difficulty and everything to do with treating a vector like an ordinary number. Knowing them in advance is worth more than another hour of practice on the parts you already do well.

• Adding magnitudes when you should add vectors: $|\mathbf{a}| + |\mathbf{b}|$ is almost never the same as $|\mathbf{a} + \mathbf{b}|$, because direction changes how vectors combine.
• Dropping the $\mathbf{i}$ and $\mathbf{j}$ partway through and leaving a bare number where a vector belongs.
• Forgetting to divide by the magnitude when the question asks for a unit vector, a unit vector must have length $1$.

Vectors are a chapter that clicks the instant the picture behind the algebra clicks, and stays cloudy for as long as the symbols are pushed around without a mental image to check them against. Working one-to-one, a teacher can watch how you read a diagram, catch the exact point where the direction gets reversed or a component gets dropped, and rebuild the picture so the formulas feel obvious rather than fiddly.

Our teachers are experienced; lessons are online and taught in English, while the SPM papers are set bilingually in Bahasa Melayu and English.

If vectors are a sticking point, you are welcome to begin with a one-hour paid trial class at the teacher's own rate (from RM50 per hour, depending on experience). We will not promise a grade, but we can help you turn vectors from a source of confusion into one of the more dependable topics on your paper.