Vector Algebra · Lesson 5 of 9

The Dot Product and Angles

“Measure directional alignment with a scalar product, and use it to calculate angles and test perpendicularity.”

Learning Objectives

• Explain the dot product and its dependence on the angle between vectors. • Derive the component formula from basis-vector products. • Apply commutative, distributive and scalar-multiplication rules. • Calculate angles and test perpendicularity while handling zero vectors correctly.

A product that measures alignment

Multiplying two vectors can produce either a scalar or another vector, depending on the operation. The dot product produces a scalar. It compares how strongly two nonzero vectors point along the same direction, taking both their lengths and their angle into account.

Definition

Angle between vectors

Translate two nonzero vectors to a common tail without rotating them. Their angle $θ$ is chosen in the interval $0 \leq θ \leq π$ .

Scalar or dot productLaTeX

a \cdot b = ∣ a ∣ ∣ b ∣ cos θ

θ

is the angle between the two nonzero vectors. The result is a real number, not a vector.

The angle between two vectors— Translate arrows to a common tail, then use the angle between their directions.

If either input is $0$ , an angle is not defined, but the dot product is defined to be zero. This separate convention makes the algebra consistent. Whenever an angle formula divides by vector magnitudes, both vectors must be nonzero.

What the sign tells us

Cosine is positive for an acute angle, zero for a right angle, and negative for an obtuse angle. The sign of the dot product therefore records a geometric relationship. At the extreme angles, the product equals either the positive or negative product of the lengths.

Relative direction	Dot product
Same direction	$θ = 0$ , so $a \cdot b = ∣ a ∣∣ b ∣$ .
Perpendicular nonzero vectors	$θ = π /2$ , so $a \cdot b = 0$ .
Opposite directions	$θ = π$ , so $a \cdot b = - ∣ a ∣∣ b ∣$ .
Acute or obtuse angle	The product is respectively positive or negative.

Taking a vector with itself gives angle zero and hence $a \cdot a = ∣ a ∣^{2}$ . A zero dot product between two nonzero vectors proves perpendicularity. A zero dot product alone does not prove that either input is zero.

Example — An angle from magnitudes and a product

Problem
If $∣ a ∣ = 1$ , $∣ b ∣ = 2$ and $a \cdot b = 1$ , find the angle between them.

1.Rearrange the definition: $cos θ = (a \cdot b) / (∣ a ∣∣ b ∣)$ .
2.Substitute to get $cos θ = 1/ (1 \times 2) = 1/2$ .
3.Because $0 \leq θ \leq π$ , the angle is $π /3$ , or 60 degrees.

Properties that make the product useful

Interchanging the two vectors leaves their lengths and angle unchanged, which explains commutativity. The dot product also distributes over addition, and a scalar factor can be taken outside it. These rules permit expansion, but the scalar nature of the result must be respected.

Dot-product lawsLaTeX

a \cdot b a \cdot (b + c) (λ a) \cdot b = b \cdot a = a \cdot b + a \cdot c = λ (a \cdot b) = a \cdot (λ b)

The distributive law also holds over subtraction. Each dot product is a scalar.

The perpendicular unit basis makes component calculations simple. A basis vector dotted with itself gives 1; two different basis vectors give 0. Expanding a product therefore removes every mixed-axis term.

Products of the unit basisLaTeX

\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1, \hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0

Use commutativity for the reversed mixed products.

For example, $a_{1} \hat{i} \cdot b_{1} \hat{i} = a_{1} b_{1}$ , whereas $a_{1} \hat{i} \cdot b_{2} \hat{j} = 0$ . The same argument in all nine terms leaves just the three products of matching components.

Dot product in componentsLaTeX

a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}

This applies to

a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}

and

b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}

Finding angles in coordinates

Calculate the dot product and the two lengths separately, then substitute into the cosine formula. A quick sign check is valuable: a negative product must lead to an obtuse angle. If the question asks for an angle at a vertex, form both side vectors starting at that vertex.

Angle from a dot productLaTeX

θ = cos^{- 1} (\frac{a \cdot b}{∣ a ∣ ∣ b ∣})

Both vectors must be nonzero. A value outside the interval

[- 1, 1]

for the cosine signals inconsistent data or a calculation error.

Example — An obtuse angle

Problem
Find the angle between $a = \hat{i} + \hat{j} - \hat{k}$ and $b = \hat{i} - \hat{j} + \hat{k}$ .

1.The dot product is $1 (1) + 1 (- 1) + (- 1) (1) = - 1$ .
2.Each magnitude is $3$ , so $cos θ = - 1/3$ .
3.Therefore $θ = cos^{- 1} (- 1/3)$ , an obtuse angle. Leaving the answer in inverse-cosine form is exact.

Example — Choosing the correct vertex vectors

Problem
Find $∠ A B C$ for $A (1, 2, 3)$ , $B (- 1, 0, 0)$ and $C (0, 1, 2)$ .

1.The angle is at $B$ , so use $B A = (2, 2, 3)$ and $B C = (1, 1, 2)$ .
2.Their dot product is $2 + 2 + 6 = 10$ , while their magnitudes are $17$ and $6$ .
3.Thus $∠ A B C = cos^{- 1} (10/ 102)$ . Using $A B$ in place of $B A$ would give the supplementary angle.

Perpendicularity and opposed directions

The dot product can establish a right angle without calculating an inverse cosine. It also distinguishes vectors that point in the same line but opposite directions. In both situations, state the nonzero conditions before interpreting the result as an angle.

Example — Perpendicular sum and difference

Problem
For $a = 5 \hat{i} - \hat{j} - 3 \hat{k}$ and $b = \hat{i} + 3 \hat{j} - 5 \hat{k}$ , show that $a + b$ is perpendicular to $a - b$ .

1.First find $a + b = 6 \hat{i} + 2 \hat{j} - 8 \hat{k}$ and $a - b = 4 \hat{i} - 4 \hat{j} + 2 \hat{k}$ .
2.Their dot product is $6 (4) + 2 (- 4) + (- 8) (2) = 24 - 8 - 16 = 0$ .
3.Both are nonzero, so they are perpendicular. Equivalently, expanding gives $(a + b) \cdot (a - b) = ∣ a ∣^{2} - ∣ b ∣^{2}$ ; the original vectors have equal squared magnitude 35.

Example — An angle between directed segments

Problem
Points $A, B, C, D$ have position vectors $(1, 1, 1)$ , $(2, 5, 0)$ , $(3, 2, - 3)$ and $(1, - 6, - 1)$ . Find the angle between $A B$ and $C D$ .

1.Subtract endpoint positions: $A B = (1, 4, - 1)$ and $C D = (- 2, - 8, 2)$ .
2.The dot product is $- 2 - 32 - 2 = - 36$ . The magnitudes are $18 = 32$ and $72 = 62$ .
3.Therefore $cos θ = - 36/36 = - 1$ and $θ = π$ . The vectors are collinear and oppositely directed, as also follows from $C D = - 2 A B$ .

Common mistake

A zero dot product does not imply a zero input vector. For example, $\hat{i} \cdot \hat{j} = 0$ , although neither vector is zero. Conversely, if an input is zero, do not try to calculate an angle by dividing by its magnitude.

Quick check

What is $(2 \hat{i} - \hat{j}) \cdot (\hat{i} + 3 \hat{j})$ ?

Quick check

For two nonzero vectors with a negative dot product, their angle is

Quick check

Which statement follows from $a \cdot a = 0$ ?

Quick check

Which vectors determine $∠ A B C$ ?

Quick check

If $∣ a ∣ = ∣ b ∣$ and $a + b, a - b$ are nonzero, then their dot product is

Practice Problems

Find the angle between vectors of magnitudes $3$ and $2$ whose dot product is $6$ .
Find the angle between $\hat{i} - 2 \hat{j} + 3 \hat{k}$ and $3 \hat{i} - 2 \hat{j} + \hat{k}$ .
Show that $(2 \hat{i} + 3 \hat{j} + 6 \hat{k}) /7$ , $(3 \hat{i} - 6 \hat{j} + 2 \hat{k}) /7$ and $(6 \hat{i} + 2 \hat{j} - 3 \hat{k}) /7$ are mutually perpendicular unit vectors.
If $a = \hat{i} + 2 \hat{j}$ and $b = λ \hat{i} - \hat{j}$ , find $λ$ when they are perpendicular.
Prove that $(a + b) \cdot (a - b) = ∣ a ∣^{2} - ∣ b ∣^{2}$ . State the qualification needed to interpret a zero result as a right angle.
Give two nonzero vectors whose dot product is zero, and two whose dot product is negative.
If two equal-magnitude vectors make an angle of 60 degrees and have dot product $1/2$ , find their common magnitude.

Key Takeaways

The dot product is a scalar that measures alignment through cosine. Matching components are multiplied and added to calculate it. Nonzero perpendicular vectors have zero dot product; the zero-input case needs separate interpretation. Angles at a vertex require two vectors directed away from that vertex. The sign of a dot product provides a quick check on the resulting angle.

Previous · Lesson 4

Vectors Between Points and the Section Formula

Next · Lesson 6

Projections and Dot Product Applications

A product that measures alignment

What the sign tells us

Properties that make the product useful

Finding angles in coordinates

Perpendicularity and opposed directions

Quiz

Practice Problems

Key Takeaways