Dot product and projection | Year 12

01 / LEARN

Understand the idea.

Measure alignment

The dot product is a scalar measuring alignment: multiply matching components and add. For nonzero vectors, $a \cdot b = ∣ a ∣ ∣ b ∣ cos θ$ . Positive values give $0 \leq θ < \frac{π}{2}$ , zero gives $θ = \frac{π}{2}$ , and negative values give $\frac{π}{2} < θ \leq π$ .

For an angle, divide the dot product by both magnitudes before applying inverse cosine. Use the requested angle unit. A zero dot product indicates perpendicular nonzero vectors; the zero vector itself has no defined angle.

Resolve along a direction

A projection keeps only the part of $a$ parallel to a nonzero direction $b$ . Its vector form is ${proj}_{b} a = \frac{a \cdot b}{b \cdot b} b$ . The squared magnitude in the denominator accounts for the length of $b$ .

Subtracting the projection from $a$ leaves a perpendicular remainder. Check it by dotting with $b$ . A negative projection coefficient means the projected vector points opposite to $b$ ; it is not an arithmetic error.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

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

Worked solution

1
Calculate the dot product.
$(1, 1, 0) \cdot (1, 0, 1) = 1$
Multiply matching components and add their products.
2
Calculate both magnitudes.
$∣ a ∣ = \sqrt{1^{2} + 1^{2}} = \sqrt{2}, ∣ b ∣ = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$
Each vector has two unit components along perpendicular axes.
3
Form the cosine ratio.
$cos θ = \frac{1}{\sqrt{2} \sqrt{2}} = \frac{1}{2}$
The dot-product identity isolates the angle between the nonzero vectors.
4
Select the vector angle.
$θ = {cos}^{- 1} \frac{1}{2} = 60^{\circ}$
Inverse cosine gives the unique angle in $[0^{\circ}, 180^{\circ}]$ .

Answer

$60^{\circ}$

EXAMPLE 02

Worked example 2

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

Worked solution

1
Calculate the projection numerator.
$a \cdot b = (2, 1, 3) \cdot (1, 2, 2) = 2 + 2 + 6 = 10$
The dot product extracts alignment with the target direction.
2
Calculate the squared target magnitude.
$b \cdot b = 1^{2} + 2^{2} + 2^{2} = 9$
The projection coefficient divides by squared magnitude, not magnitude.
3
Multiply the target direction by the coefficient.
${proj}_{b} a = \frac{10}{9} (1, 2, 2) = (\frac{10}{9}, \frac{20}{9}, \frac{20}{9})$
The result must be a scalar multiple of the direction onto which we project.
4
Check the perpendicular remainder.
$(\frac{8}{9}, - \frac{11}{9}, \frac{7}{9}) \cdot (1, 2, 2) = \frac{8 - 22 + 14}{9} = 0$
Subtracting the projection from $(2, 1, 3)$ leaves no component along the target.

Answer

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

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$k = - 1$

Perpendicularity requires $(k, 1, 2) \cdot (2, - 4, 3) = 2 k - 4 + 6 = 0$ . Hence $2 k + 2 = 0$ and $k = - 1$ . Both vectors are nonzero.

Keep practising

6 worksheet questions · 12 marks

Open worksheet 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 ${cos}^{- 1} (\frac{1}{\sqrt{3}})$ .
${54.7}^{\circ}$
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.
$k = - \frac{1}{2}$
Force $(6, 2, - 1)$ N acts through displacement $(3, - 1, 2)$ m. Find the work.
2 marks
Work equals the dot product.
$18 - 2 - 2 = 14$ .
$14 J$

Checking saved progress…

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 ↗

In this lesson

Understand the idea.

Measure alignment

Resolve along a direction

See the method in action.

Worked example 1

Calculate the dot product.

Calculate both magnitudes.

Form the cosine ratio.

Select the vector angle.

Worked example 2

Calculate the projection numerator.

Calculate the squared target magnitude.

Multiply the target direction by the coefficient.

Check the perpendicular remainder.

Try it yourself.

Check your understanding

Keep practising