Dot Product and Angles (AA HL)
The dot product turns two vectors into a single number, and that number is the key to unlocking the angle between them - useful for checking perpendicularity, finding a projection, or answering an angle question directly. This page covers the scalar product formula and its use in angle problems, with worked examples and the mistakes examiners see most. It's part of the broader Vectors, Lines & Planes topic.
11 questions on this sub-topic.
The scalar product and the angle formula
Covered under IB syllabus reference AHL3.13: the scalar (dot) product of two vectors and its properties, the angle between two vectors, and perpendicular and parallel vectors. Both formulas below are in the formula booklet.
Scalar (dot) product
\(\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3\)
Multiply matching components and add - the result is always a single number. If \(\mathbf a\cdot\mathbf b=0\) and neither vector is zero, the vectors are perpendicular.
Scalar product and angle
\(\cos\theta=\dfrac{\mathbf a\cdot\mathbf b}{|\mathbf a||\mathbf b|}\)
Rearrange the dot product formula to isolate \(\theta\), then take the inverse cosine to get the angle between \(\mathbf a\) and \(\mathbf b\).
Need the cross product, projections, or the full syllabus wording? See Vectors, Lines & Planes.
Worked examples
Given \(\mathbf a=(2,-1,3)\) and \(\mathbf b=(4,0,-2)\), find \(\mathbf a\cdot\mathbf b.\)
Worked solution
\(\mathbf a\cdot\mathbf b = (2)(4)+(-1)(0)+(3)(-2)\) M1
\(= 8+0-6 = 2.\) A1
Given \(\mathbf a=(1,2,2)\) and \(\mathbf b=(2,-1,2)\), find the angle between \(\mathbf a\) and \(\mathbf b\), correct to 1 decimal place.
Worked solution
\(\mathbf a\cdot\mathbf b = (1)(2)+(2)(-1)+(2)(2) = 4.\) M1
\(|\mathbf a| = \sqrt{1^2+2^2+2^2}=3, |\mathbf b| = \sqrt{2^2+(-1)^2+2^2}=3.\) M1
\(\cos\theta = \dfrac{\mathbf a\cdot\mathbf b}{|\mathbf a||\mathbf b|} = \dfrac{4}{3\times3} = \dfrac{4}{9}.\) M1
\(\theta = \cos^{-1}\!\left(\dfrac{4}{9}\right) \approx 63.6^\circ.\) A1
On the GDC: use the inverse cosine function to evaluate \(\cos^{-1}(4/9)\) directly once you have the ratio.
Common mistakes
- Forgetting to take the inverse cosine. The dot product formula gives \(\cos\theta\) - the final answer needs \(\theta=\cos^{-1}(\ldots)\), not the cosine value itself.
- Multiplying the magnitudes instead of adding the products. The dot product is \(a_1b_1+a_2b_2+a_3b_3\), a sum of three products - it's easy to slip into multiplying whole vectors together component-by-component and forgetting to add.
- Rounding the dot product or magnitudes too early. Truncating \(\mathbf a\cdot\mathbf b\) or \(|\mathbf a||\mathbf b|\) to a couple of decimal places before dividing can shift the final angle by a degree or more - keep exact values or full calculator precision until the last step.
- Ignoring what the sign of the dot product means. A positive dot product means the angle between the vectors is acute, negative means obtuse, and exactly zero means they're perpendicular - a quick way to sanity-check an answer before finishing the calculation.
Ready to practise properly?
11 dot-product questions, marked instantly like the real exam.
Quick answers
What is the formula for the dot product of two vectors?
\(\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3\), summing the products of matching components. It's a single number (a scalar), not a vector.
How do I find the angle between two vectors?
Rearrange the dot product formula: \(\cos\theta = \dfrac{\mathbf a\cdot\mathbf b}{|\mathbf a||\mathbf b|}.\) Compute the dot product and both magnitudes, divide, then take the inverse cosine to get \(\theta.\)