# The Scalar Product and Projections

> Multiply two vectors to get a number. Find angles, test for right angles, project one vector onto another and prove geometry results.

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/scalar-product-and-projections/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: Pulling a sled
Kai pulls a sled with a rope angled up at $30^\circ$. He pulls hard, but only part of his effort moves the sled forward. The rest just lifts the front.

How much of his pull actually works along the ground? That "useful part" of one vector in the direction of another is found with the **scalar product**, also called the **dot product**. Physicists use it to calculate work, and engineers use it to check whether beams meet at right angles.

**In short:** The scalar (dot) product multiplies two vectors to give a number: $\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 = |\mathbf{a}||\mathbf{b}|\cos\theta$. It finds the angle between two vectors, and a dot product of zero means they are perpendicular.

## Starting point: what you already know

- Vectors in component form, magnitude $|\mathbf{a}| = \sqrt{x^2 + y^2}$, and unit vectors (previous chapter).
- The **cosine rule** $c^2 = a^2 + b^2 - 2ab\cos C$.
- $\cos 90^\circ = 0$, and $\cos\theta < 0$ for obtuse angles.
- Perpendicular lines have gradients that multiply to $-1$.

## Two ways to calculate it

### Key idea: The scalar product
For $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j}$ and $\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j}$ with angle $\theta$ between them:
$$\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 = |\mathbf{a}||\mathbf{b}|\cos\theta$$

The answer is a **number** (a scalar), not a vector. The two formulas agree; the cosine rule shows why.

### Worked example: Find the angle
$\mathbf{a} = 3\mathbf{i} + 4\mathbf{j}$ and $\mathbf{b} = 5\mathbf{i} - 12\mathbf{j}$.

$\mathbf{a} \cdot \mathbf{b} = 15 - 48 = -33$, $|\mathbf{a}| = 5$, $|\mathbf{b}| = 13$.
$$\cos\theta = \frac{-33}{65} \quad\Rightarrow\quad \theta \approx 120.5^\circ$$
A negative dot product means the angle is obtuse.

### Calculator tip: Angle between vectors in one line
Work out $\mathbf{a} \cdot \mathbf{b}$, $|\mathbf{a}|$ and $|\mathbf{b}|$, then type `cos⁻¹( -20 ÷ ( √(40) × √(20) ) )` with your numbers. Brackets around the whole bottom are essential. Use degree mode unless radians are asked for.

## Perpendicular vectors

### Key idea: The right angle test
Non-zero vectors $\mathbf{a}$ and $\mathbf{b}$ are perpendicular exactly when $\mathbf{a} \cdot \mathbf{b} = 0$.

### Worked example: Find k
For which $k$ is $2\mathbf{i} + k\mathbf{j}$ perpendicular to $3\mathbf{i} - 6\mathbf{j}$?

$6 - 6k = 0$, so $k = 1$.

## Useful properties

- $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$
- $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$
- $\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}$
- $(k\mathbf{a}) \cdot \mathbf{b} = k(\mathbf{a} \cdot \mathbf{b})$
- $\mathbf{i} \cdot \mathbf{i} = \mathbf{j} \cdot \mathbf{j} = 1$ and $\mathbf{i} \cdot \mathbf{j} = 0$

## Projections

The **projection** of $\mathbf{a}$ onto $\mathbf{b}$ is the "shadow" of $\mathbf{a}$ on the line of $\mathbf{b}$, when light shines at right angles to $\mathbf{b}$.

%%FIG:projection%%

### Key idea: Projection of a onto b
$$\operatorname{proj}_{\mathbf{b}}\mathbf{a} = \left(\frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}}\right)\mathbf{b}$$
Its length is the **scalar projection** $\dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|} = |\mathbf{a}|\cos\theta$.

### Worked example: Kai's sled
Kai pulls with force $\mathbf{F} = 50\cos 30^\circ\,\mathbf{i} + 50\sin 30^\circ\,\mathbf{j}$ newtons. The ground direction is $\mathbf{i}$. The useful part is
$$\frac{\mathbf{F} \cdot \mathbf{i}}{|\mathbf{i}|} = 50\cos 30^\circ \approx 43.3 \text{ N}$$

### Worked example: Projection in components
Project $\mathbf{a} = 4\mathbf{i} + 2\mathbf{j}$ onto $\mathbf{b} = 3\mathbf{i} + \mathbf{j}$.

$\mathbf{a} \cdot \mathbf{b} = 14$ and $\mathbf{b} \cdot \mathbf{b} = 10$, so the projection is $\dfrac{14}{10}(3\mathbf{i} + \mathbf{j}) = 4.2\mathbf{i} + 1.4\mathbf{j}$.

**Mind map**

```text
Scalar product
  Formulas
    a₁b₁ + a₂b₂
    |a||b| cos θ
  Results
    Number, not a vector
    Negative means obtuse
    Zero means perpendicular
  Properties
    a · a = |a|²
    Order does not matter
  Projection
    (a · b / b · b) b
    Length |a| cos θ
  Proofs
    Diagonals of a rhombus
    Angle in a semicircle
```

### Year 11: Gradients and vectors
A line with gradient $m$ points in the direction of $\mathbf{i} + m\mathbf{j}$. Two lines with gradients $m_1$ and $m_2$ are perpendicular when $(\mathbf{i} + m_1\mathbf{j}) \cdot (\mathbf{i} + m_2\mathbf{j}) = 1 + m_1m_2 = 0$, which is the Year 10 rule $m_1m_2 = -1$.

### Year 12: Proving geometry with vectors
**The diagonals of a rhombus are perpendicular.** Let the sides from one corner be $\mathbf{a}$ and $\mathbf{b}$ with $|\mathbf{a}| = |\mathbf{b}|$. The diagonals are $\mathbf{a} + \mathbf{b}$ and $\mathbf{a} - \mathbf{b}$.
$$(\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = |\mathbf{a}|^2 - |\mathbf{b}|^2 = 0$$
So they are perpendicular.

**The angle in a semicircle is a right angle.** Let the centre be $O$, the diameter run from $-\mathbf{a}$ to $\mathbf{a}$, and $P$ have position vector $\mathbf{p}$ with $|\mathbf{p}| = |\mathbf{a}|$. Then $(\mathbf{p} - \mathbf{a}) \cdot (\mathbf{p} + \mathbf{a}) = |\mathbf{p}|^2 - |\mathbf{a}|^2 = 0$.

**Length of a sum.** Expand a dot product: $|\mathbf{u} + \mathbf{v}|^2 = (\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = |\mathbf{u}|^2 + 2\mathbf{u} \cdot \mathbf{v} + |\mathbf{v}|^2$. With $|\mathbf{u}| = 3$, $|\mathbf{v}| = 5$ and $60^\circ$ between them, $\mathbf{u} \cdot \mathbf{v} = 7.5$, so $|\mathbf{u} + \mathbf{v}| = \sqrt{9 + 15 + 25} = 7$. Adding $|\mathbf{a} + \mathbf{b}|^2$ and $|\mathbf{a} - \mathbf{b}|^2$ gives $2|\mathbf{a}|^2 + 2|\mathbf{b}|^2$ (the parallelogram law). For unit vectors at angle $2\theta$, $|\mathbf{a} + \mathbf{b}|^2 = 2 + 2\cos 2\theta = 4\cos^2\theta$.

**Work done** by a constant force $\mathbf{F}$ over a displacement $\mathbf{s}$ is $\mathbf{F} \cdot \mathbf{s}$. A force $2\mathbf{i} + 5\mathbf{j}$ N moving an object by $3\mathbf{i} + \mathbf{j}$ m does $6 + 5 = 11$ J of work.

**A perpendicular vector.** Swap the components and change one sign: $5\mathbf{i} - 12\mathbf{j}$ is perpendicular to $12\mathbf{i} + 5\mathbf{j}$. Scale it to the length you need.

**Triangle area.** Find the angle between $\overrightarrow{OA}$ and $\overrightarrow{OB}$ from the dot product, then use $\frac{1}{2}|\mathbf{a}||\mathbf{b}|\sin\theta$.

**Reflection in a line.** The reflection of $\mathbf{b}$ in the line through $O$ with direction $\mathbf{a}$ is $2\operatorname{proj}_{\mathbf{a}}\mathbf{b} - \mathbf{b}$.

**Distance from a point to a line.** If the line through $O$ has direction $\mathbf{d}$, the scalar projection of $\mathbf{q}$ onto it is $p = \dfrac{\mathbf{q} \cdot \mathbf{d}}{|\mathbf{d}|}$, and the distance is $\sqrt{|\mathbf{q}|^2 - p^2}$. The closest point on the line is $\operatorname{proj}_{\mathbf{d}}\mathbf{q}$.

## Quick summary

- $\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 = |\mathbf{a}||\mathbf{b}|\cos\theta$, a number.
- Use it to find the angle between vectors.
- $\mathbf{a} \cdot \mathbf{b} = 0$ means perpendicular. Negative means obtuse.
- $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$.
- Projection of $\mathbf{a}$ onto $\mathbf{b}$: $\left(\dfrac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}}\right)\mathbf{b}$.

## Quiz

Try these questions from the chapter quiz.

1. Find $(2\mathbf{i} + 3\mathbf{j}) \cdot (4\mathbf{i} - \mathbf{j})$.
   - A. 11
   - B. 5
   - C. $8\mathbf{i} - 3\mathbf{j}$
   - D. 14
   - Answer: B. 5. $8 - 3 = 5$.

2. The scalar product of two vectors is:
   - A. a number
   - B. a vector
   - C. always positive
   - D. always zero
   - Answer: A. a number. That is why it is called the scalar product.

3. If $\mathbf{a} \cdot \mathbf{b} = 0$ for non-zero vectors, then they are:
   - A. parallel
   - B. perpendicular
   - C. equal
   - D. opposite
   - Answer: B. perpendicular. $\cos\theta = 0$, so $\theta = 90^\circ$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/scalar-product-and-projections/#quiz