Scalar/vector projections | Year 11

01 / LEARN

Understand the idea.

The part along a direction

A scalar projection is the signed amount of one vector along a chosen direction. Dotting $u$ with the unit vector $v / ∣ v ∣$ gives this amount. Its sign tells whether that component follows or opposes the chosen direction.

The vector projection includes the direction as well as the amount. Multiply the signed scalar projection by the same unit vector, giving a denominator $∣ v ∣^{2} = v \cdot v$ . Only the direction vector must be non-zero.

\begin{aligned} {comp}_{v} u & = \frac{u \cdot v}{∣ v ∣} \\ {proj}_{v} u & = \frac{u \cdot v}{v \cdot v} v \end{aligned}

Equal-scale coordinate axes with u=(4,2), v=(1,1) and the labelled vector projection of u onto v ending at (3,3). All arrows start at the origin; tick intervals are one unit.

The leftover is perpendicular

The projected vector is the parallel part of $u$ . Subtract it from $u$ to obtain the perpendicular remainder. Check the decomposition in two ways: the parts must add back to $u$ , and the remainder must have dot product zero with $v$ .

Use the vector being projected onto in the denominator. Reversing that direction reverses the signed scalar projection but leaves the vector projection unchanged. The projected vector's magnitude is the absolute value of the scalar projection.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Projection on a diagonal direction

Find the scalar and vector projections of u=(4,2) onto v=(1,1).

Worked solution

1
Calculate the shared dot product and direction length.
$u \cdot v = 4 (1) + 2 (1) = 6, ∣ v ∣ = \sqrt{2}$
The chosen direction is the vector being projected onto.
2
Calculate the signed scalar projection.
${comp}_{v} u = \frac{6}{\sqrt{2}} = 3 \sqrt{2}$
Dividing by the direction length gives the component along a unit direction.
3
Calculate the vector projection.
${proj}_{v} u = \frac{6}{1^{2} + 1^{2}} (1, 1) = (3, 3)$
The vector projection multiplies the direction vector by the dot product divided by its squared length.
4
Check the perpendicular remainder.
$\begin{aligned} u - (3, 3) & = (1, - 1), \\ (1, - 1) \cdot (1, 1) & = 1 - 1 = 0. \end{aligned}$
Removing the parallel component must leave a vector perpendicular to the chosen direction.

Answer

Scalar $3 \sqrt{2}$ ; vector $(3, 3)$ .

EXAMPLE 02

A negative component

Find the projection of a=(-3,4) onto b=(1,0), then its perpendicular remainder.

Worked solution

1
Find the signed component along the given unit direction.
${comp}_{b} a = \frac{(- 3) (1) + 4 (0)}{1} = - 3$
The direction vector is already a unit vector, and the horizontal component opposes it.
2
Restore the direction to obtain the vector projection.
${proj}_{b} a = - 3 (1, 0) = (- 3, 0)$
A negative scalar sends the projected vector opposite to the positive horizontal direction.
3
Subtract the parallel part.
$a - {proj}_{b} a = (- 3, 4) - (- 3, 0) = (0, 4)$
The remainder contains the part not accounted for by the parallel projection.
4
Verify perpendicularity.
$(0, 4) \cdot (1, 0) = 0$
The remaining vector is perpendicular to the direction of projection.

Answer

Parallel component $(- 3, 0)$ ; perpendicular component $(0, 4)$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Project $(5, 2)$ onto the y-axis direction $(0, 1)$ .

Show answer & reasoning

$(0, 2)$ .

The direction $(0, 1)$ is a unit vector and $(5, 2) \cdot (0, 1) = 2$ . The vector projection is therefore $2 (0, 1) = (0, 2)$ ; the discarded horizontal part is perpendicular to the axis.

Keep practising

6 worksheet questions · 16 marks

Open worksheet Worked solutions

Read the questions online

Find the scalar projection of $(3, 4)$ onto $(1, 0)$ .
2 marks
Find the vector projection of $(2, 5)$ onto $(0, 2)$ .
2 marks
Project $(5, 1)$ onto $(1, 1)$ .
2 marks
Resolve $(4, 2)$ into components parallel and perpendicular to $(1, 1)$ .
3 marks
Find the scalar projection of $(- 2, 3)$ onto $(2, 0)$ and the length of the projected vector.
3 marks
Show that replacing $v$ by $2 v$ does not change ${proj}_{v} u$ .
4 marks

Read the worked solutions

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

Find the scalar projection of $(3, 4)$ onto $(1, 0)$ .
2 marks
The direction has length one and the dot product is three.
$3$ .
Find the vector projection of $(2, 5)$ onto $(0, 2)$ .
2 marks
$Coefficient = (\frac{10}{4}) = \frac{5}{2}$ .
Multiply by $(0, 2)$ .
$(0, 5)$ .
Project $(5, 1)$ onto $(1, 1)$ .
2 marks
The dot product is six and squared direction length is two.
$(3, 3)$ .
Resolve $(4, 2)$ into components parallel and perpendicular to $(1, 1)$ .
3 marks
Calculate the projection, then subtract it.
Check $(1, - 1) \cdot (1, 1) = 0$ .
Parallel $(3, 3)$ ; perpendicular $(1, - 1)$ .
Find the scalar projection of $(- 2, 3)$ onto $(2, 0)$ and the length of the projected vector.
3 marks
The dot product is $- 4$ , divided by direction length $2$ .
Take the absolute value only for the length.
Scalar $- 2$ ; length $2$ .
Show that replacing $v$ by $2 v$ does not change ${proj}_{v} u$ .
4 marks
The numerator doubles, the denominator quadruples and the final direction vector doubles.
The factors cancel.
Both give $\frac{u \cdot v}{v \cdot v} v$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: vectors in the plane. Covers magnitude, components, geometry, scalar products, projections and proofs; follows the approved Specialist preparation structure.

Official curriculum source ↗

In this lesson

Understand the idea.

The part along a direction

The leftover is perpendicular

See the method in action.

Projection on a diagonal direction

Calculate the shared dot product and direction length.

Calculate the signed scalar projection.

Calculate the vector projection.

Check the perpendicular remainder.

A negative component

Find the signed component along the given unit direction.

Restore the direction to obtain the vector projection.

Subtract the parallel part.

Verify perpendicularity.

Try it yourself.

Check your understanding

Keep practising