Dot product and projection | Year 12

01 / LEARN

The idea, step by step

Measure alignment

The dot product $a \cdot b = a_{x} b_{x} + a_{y} b_{y} + a_{z} b_{z}$ is a scalar. For nonzero vectors it also equals $∣ a ∣ ∣ b ∣ cos θ$ , with $0 \leq θ \leq π$ . A negative result indicates an obtuse vector angle.

To find an angle, divide by both magnitudes before taking inverse cosine. Check degree/radian mode. A zero dot product proves perpendicularity only when both input vectors are nonzero.

Resolve along a direction

The vector projection of a along nonzero b is ${proj}_{b} a = (a \cdot b) / (b \cdot b) b$ . Its signed scalar component along the unit direction is $a \cdot \hat{b}$ .

The perpendicular remainder is a minus its projection. Dot that remainder with b to check it is perpendicular. A negative projection coefficient is meaningful: the component points opposite to the chosen direction.

A common trap

A dot product is a vector perpendicular to both inputs.

Instead: It is a scalar. The perpendicular-vector operation is the cross product.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example 1

Find the angle between (1,1,0) and (1,0,1).

The dot product is 1; both magnitudes are $\sqrt{2}$ .
Thus $cos θ = 1 / 2$ .

60^{\circ}

Example 2

Worked example 2

Project (2,1,3) along (1,2,2).

The dot product is 10 and the squared target magnitude is 9.
Multiply the target vector by $10 / 9$ .

(10 / 9, 20 / 9, 20 / 9)

Check your understanding

Find k so (k,1,2) is perpendicular to (2,-4,3).

Show answer

$k = - 1$

$2 k - 4 + 6 = 0$ , so $2 k + 2 = 0$ .

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Calculate $(2, - 1, 3) \cdot (4, 2, - 2)$ .
2 marks
Find the angle between (1,0,0) and (1,1,1), to one decimal degree.
2 marks
Project (3,4,0) along (1,0,0).
2 marks
Find the component of (2,-1,2) perpendicular to (1,0,0).
2 marks
Find k for perpendicular vectors (1,k,2) and (3,2,-1).
2 marks
Force (6,2,-1) N acts through displacement (3,-1,2) m. Find the work.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Calculate $(2, - 1, 3) \cdot (4, 2, - 2)$ .
2 marks
Multiply matching components.
$8 - 2 - 6 = 0$ .
$0$
Find the angle between (1,0,0) and (1,1,1), to one decimal degree.
2 marks
The dot product is 1 and the magnitudes are 1 and $\sqrt{3}$ .
Take ${cos}^{- 1} (1 / \sqrt{3})$ .
${54.7}^{\circ}$
Project (3,4,0) along (1,0,0).
2 marks
The target is a unit direction.
Keep the x component and set the perpendicular components to zero.
$(3, 0, 0)$
Find the component of (2,-1,2) perpendicular to (1,0,0).
2 marks
Its parallel component is $(2, 0, 0)$ .
Subtract that component.
$(0, - 1, 2)$
Find k for perpendicular vectors (1,k,2) and (3,2,-1).
2 marks
Their dot product is $3 + 2 k - 2$ .
Set it to zero and solve.
$k = - 1 / 2$
Force (6,2,-1) N acts through displacement (3,-1,2) m. Find the work.
2 marks
Work equals the dot product.
$18 - 2 - 2 = 14$ .
$14 J$

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics vector algebra, vector/Cartesian equations and linear systems. Original explanations and tasks.

Official curriculum source ↗

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