Scalar product of vectors
Master the scalar product of vectors for Year 11 Specialist Mathematics in Queensland (QCAA). The scalar (dot) product multiplies two vectors to give a single number, and it is the key to measuring the angle between two vectors.
You will learn to compute the scalar product in Cartesian form, use it to find angles and to test whether vectors are perpendicular or parallel, and read the sign to classify an angle as acute, right or obtuse — core skills for the algebra of vectors.
Theory
The scalar (dot) product combines two vectors to give a single number. In Year 11 Specialist Mathematics (QCAA, Queensland) it is the tool for finding the angle between two vectors and for testing whether they are perpendicular or parallel. This page shows both formulas and how to apply them to vectors in Cartesian form.
The scalar product (also called the dot product) of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is written \(\mathbf{a}\cdot\mathbf{b}\). The result is a scalar — an ordinary number, not a vector.
There are two equivalent formulas. In component form, multiply matching components and add: \(\mathbf{a}\cdot\mathbf{b}=a_1 b_1 + a_2 b_2\). In geometric form, \(\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta\), where \(\theta\) is the angle between the two vectors.
Setting the two forms equal makes the scalar product a way to find an angle: \(\cos\theta=\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\). A useful special case is \(\mathbf{a}\cdot\mathbf{a}=|\mathbf{a}|^2\).
Two non-zero vectors are perpendicular exactly when \(\mathbf{a}\cdot\mathbf{b}=0\) (because \(\cos 90^\circ=0\)). The sign of the scalar product tells you the angle at a glance: positive means acute, zero means a right angle, and negative means obtuse.
For \(\mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}\) and \(\mathbf{b}=b_1\mathbf{i}+b_2\mathbf{j}\), the scalar product in component form is:
The geometric form links the scalar product to the angle \(\theta\) between the vectors:
Rearranging gives the angle, and the self-product gives the squared magnitude:
How to use the scalar product
- Read the components of each vector, keeping the signs, as \((a_1,a_2)\) and \((b_1,b_2)\).
- Multiply and add: compute \(\mathbf{a}\cdot\mathbf{b}=a_1 b_1 + a_2 b_2\) — a single number.
- To test perpendicularity, check whether \(\mathbf{a}\cdot\mathbf{b}=0\); if an unknown is involved, set the scalar product to \(0\) and solve.
- To find an angle, also compute \(|\mathbf{a}|\) and \(|\mathbf{b}|\), then evaluate \(\theta=\cos^{-1}\!\left(\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\right)\).
Multiply matching components, then add:
| \(\mathbf{a}\cdot\mathbf{b}\) | \(=\) | \(a_1 b_1 + a_2 b_2\) |
| \(=\) | \((2)(3) + (5)(4)\) | |
| \(=\) | \(6 + 20\) | |
| \(=\) | \(26\) |
\(\mathbf{a}\cdot\mathbf{b}=26\).
Find the scalar product and both magnitudes:
| \(\mathbf{a}\cdot\mathbf{b}\) | \(=\) | \((3)(2) + (1)(4)\) |
| \(=\) | \(6 + 4 = 10\) | |
| \(|\mathbf{a}|\) | \(=\) | \(\sqrt{3^2+1^2} = \sqrt{10}\) |
| \(|\mathbf{b}|\) | \(=\) | \(\sqrt{2^2+4^2} = \sqrt{20}\) |
Substitute into the cosine formula and simplify:
| \(\cos\theta\) | \(=\) | \(\dfrac{10}{\sqrt{10}\times\sqrt{20}}\) |
| \(=\) | \(\dfrac{10}{\sqrt{200}}\) | |
| \(=\) | \(\dfrac{10}{10\sqrt{2}} = \dfrac{1}{\sqrt{2}}\) | |
| \(\theta\) | \(=\) | \(45^\circ\) |
The angle is \(45^\circ\).
Perpendicular means the scalar product is zero, so set it to \(0\):
| \(\mathbf{a}\cdot\mathbf{b}\) | \(=\) | \((2)(6) + (-3)(m)\) |
| \(=\) | \(12 - 3m\) | |
| \(12 - 3m\) | \(=\) | \(0\) |
| \(3m\) | \(=\) | \(12\) |
| \(m\) | \(=\) | \(4\) |
\(m=4\).
Compute the scalar product and both magnitudes:
| \(\mathbf{a}\cdot\mathbf{b}\) | \(=\) | \((5)(1) + (1)(2)\) |
| \(=\) | \(5 + 2 = 7\) | |
| \(|\mathbf{a}|\) | \(=\) | \(\sqrt{5^2+1^2} = \sqrt{26}\) |
| \(|\mathbf{b}|\) | \(=\) | \(\sqrt{1^2+2^2} = \sqrt{5}\) |
Apply the cosine formula, then take the inverse cosine:
| \(\cos\theta\) | \(=\) | \(\dfrac{7}{\sqrt{26}\times\sqrt{5}}\) |
| \(=\) | \(\dfrac{7}{\sqrt{130}}\) | |
| \(\theta\) | \(=\) | \(\cos^{-1}\!\left(\dfrac{7}{\sqrt{130}}\right)\) |
| \(\approx\) | \(52^\circ\) |
The angle is about \(52^\circ\).
Common pitfalls
Frequently asked questions
What is the scalar product of two vectors?
It is the number \(\mathbf{a}\cdot\mathbf{b}=a_1 b_1 + a_2 b_2\), found by multiplying matching components and adding. It equals \(|\mathbf{a}||\mathbf{b}|\cos\theta\).
How do you find the angle between two vectors?
Use \(\cos\theta=\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\): work out the scalar product and both magnitudes, then take the inverse cosine.
How can you tell if two vectors are perpendicular?
Two non-zero vectors are perpendicular exactly when their scalar product is zero, \(\mathbf{a}\cdot\mathbf{b}=0\), because \(\cos 90^\circ=0\).
What does a negative scalar product mean?
A negative scalar product means \(\cos\theta<0\), so the angle between the vectors is obtuse (between \(90^\circ\) and \(180^\circ\)).
What is \(\mathbf{a}\cdot\mathbf{a}\)?
The scalar product of a vector with itself equals its magnitude squared: \(\mathbf{a}\cdot\mathbf{a}=|\mathbf{a}|^2\).
Is the scalar product a vector or a number?
It is a scalar — an ordinary number. That is why it is called the scalar (dot) product.