Compute the scalar product of two vectors in component or geometric form and use it to find the angle between vectors and test orthogonality

What this dot point is asking

NESA wants you to compute the scalar (dot) product of two vectors using either the component formula or the geometric formula, understand it as a measure of how much two vectors point the same way, and use it for the two jobs it does in this course: finding the angle between vectors, and testing whether they are perpendicular. The dot product is also the engine of the next dot point, projection, and shows up constantly in geometric proofs, so it is worth knowing cold.

The answer

For two-dimensional vectors $\mathbf{a} = (a_1, a_2)$ and $\mathbf{b} = (b_1, b_2)$ there are two formulas for the same number.

Component form (the one you compute with):

Geometric form (the one that gives it meaning):

where $\theta$ is the angle between the two arrows when their tails are placed together.

Why the two formulas agree

This is worth seeing once. Apply the cosine rule to the triangle whose sides are $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{a} - \mathbf{b}$ (the side opposite the angle $\theta$):

Now expand the left side in components: $|\mathbf{a} - \mathbf{b}|^2 = (a_1 - b_1)^2 + (a_2 - b_2)^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2(a_1 b_1 + a_2 b_2)$. Comparing the two right-hand sides, the $|\mathbf{a}|^2 + |\mathbf{b}|^2$ terms cancel and we are left with $a_1 b_1 + a_2 b_2 = |\mathbf{a}|\,|\mathbf{b}|\cos\theta$. The component formula and the geometric formula are the same number, which is exactly why one can be used to compute the other.

The sign of the dot product tells you the angle

Because $|\mathbf{a}|$ and $|\mathbf{b}|$ are positive, the sign of $\mathbf{a} \cdot \mathbf{b}$ matches the sign of $\cos\theta$. So without computing the angle at all you can read off whether it is acute, right or obtuse.

Acute angle: the dot product is positive. The vectors broadly agree in direction, $\cos\theta > 0$, so $\mathbf{a} \cdot \mathbf{b} > 0$.

Right angle: the dot product is zero. Perpendicular vectors have $\cos 90^\circ = 0$, so $\mathbf{a} \cdot \mathbf{b} = 0$. This is the orthogonality test.

Obtuse angle: the dot product is negative. The vectors broadly oppose, $\cos\theta < 0$, so $\mathbf{a} \cdot \mathbf{b} < 0$.

Finding the angle between two vectors

Rearranging the geometric formula gives the angle directly:

This works for any two non-zero vectors. Because $0 \le \theta \le \pi$, $\arccos$ returns exactly the right range (no quadrant ambiguity to resolve), which is one reason the dot product is the standard tool for angles between vectors.

Properties (bilinearity)

The scalar product behaves like ordinary multiplication, with one caveat: the output is a scalar. It is commutative, $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$, and bilinear (linear in each slot):

These rules let you expand brackets exactly as in algebra. In particular, since $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$,

which is the vector analogue of $(a + b)^2 = a^2 + 2ab + b^2$ and a workhorse in proofs.

How exam questions ask about the scalar product

• "Show that $\mathbf{a}$ and $\mathbf{b}$ are perpendicular" or "... orthogonal": compute $\mathbf{a} \cdot \mathbf{b}$ and show it is $0$.
• "Find the value of $k$ so that $\mathbf{a}$ and $\mathbf{b}$ are perpendicular": set $\mathbf{a} \cdot \mathbf{b} = 0$ and solve for $k$.
• "Find the angle between ...": use $\theta = \arccos\!\big(\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\big)$, then round if asked.
• "Find a vector perpendicular to $\mathbf{a}$": in 2D, swap the components and negate one, $(a_1, a_2) \to (-a_2, a_1)$; check with a dot product.
• "Hence find $|\mathbf{a} + \mathbf{b}|$ given $|\mathbf{a}|$, $|\mathbf{b}|$ and the angle": expand $|\mathbf{a} + \mathbf{b}|^2 = |\mathbf{a}|^2 + 2\mathbf{a}\cdot\mathbf{b} + |\mathbf{b}|^2$ with $\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$.
• "$\mathbf{a} \cdot \mathbf{b} = \ldots$ and $\mathbf{a} \cdot \mathbf{c} = \ldots$; find $\mathbf{a} \cdot (p\mathbf{b} + q\mathbf{c})$": a bilinearity drill, distribute the dot product.