C. Geometry and Algebra of Vectors · Dot and Cross Products

Scalar and Vector Projections

C2.4 C2.8About 70 minutes

Learning goals

I can find the scalar projection and vector projection of one vector on another in two-space and three-space.
I can explain what a negative scalar projection means.
I can split a vector into a part parallel to a given direction and a part perpendicular to it.
I can use projections to resolve a force along a slope.

Words to know

scalar projection
The signed length of the shadow of u on the line of v: (u · v)/|v| = |u| cos θ.

vector projection
proj_v u = ((u · v)/|v|²)v, the shadow of u on the line of v written as a vector.

unit vector
A vector of length 1. The unit vector in the direction of v is v/|v|.

resolve a vector
Write a vector as a sum of perpendicular parts, for example along a ramp and perpendicular to it.

How much of a vector points along a line?

A box sits on a ramp. Gravity pulls it straight down, but the box can only slide along the ramp. How much of the pull acts along the ramp? A plane flies northeast through a wind from the west. How much of the wind pushes the plane forward? Questions like these ask for the part of one vector that lies along another. That part is called a projection.

In lesson 1 you met the "shadow" |u| cos θ. Now you will give it a name, find it from components, and turn it into a vector.

Scalar projection

The scalar projection of u on v is the signed length of the shadow of u on the line of v:

scalar projection of u on v = |u| cos θ = (u · v) / |v|

The second form comes from u · v = |u||v| cos θ: divide both sides by |v|. It is a number. It is positive when θ is acute, 0 when u ⊥ v, and negative when θ is obtuse (the shadow points backwards).

Vector projection

The vector projection of u on v, written proj_v u, is the shadow as a vector. Take the scalar projection and multiply it by the unit vector v/|v|:

proj_v u = ((u · v) / |v|) (v / |v|) = ((u · v) / |v|²) v = ((u · v) / (v · v)) v

The number in front is just a scalar, so the vector projection is always parallel to v (or 0).

The vector projection of u on v (thick arrow) lies along the line of v. The dashed arrow from its tip to the tip of u is the part of u perpendicular to v. Together they add up to u.

Which vector goes on the bottom?

Projecting u on v: v is the direction you project onto, so |v| goes in the denominator. Swapping the roles gives a different answer. Also, u minus its projection is always perpendicular to v. Use that as a built-in check.

Worked example

Example 1: Two-space

Find the scalar and vector projections of u = [5, 2] on v = [3, 4]. Then write u as the sum of a vector parallel to v and a vector perpendicular to v.

u · v = 15 + 8 = 23 and |v| = 5.
Scalar projection: 23/5 = 4.6.
Vector projection: (23/25)[3, 4] = [69/25, 92/25] = [2.76, 3.68].
Perpendicular part: u − proj = [5 − 2.76, 2 − 3.68] = [2.24, −1.68].
Check: [2.24, −1.68] · [3, 4] = 6.72 − 6.72 = 0 ✓.

Scalar projection 4.6; vector projection [2.76, 3.68]; u = [2.76, 3.68] + [2.24, −1.68].

Worked example

Example 2: Three-space

Find the scalar and vector projections of a = [2, −1, 3] on b = [1, 2, 2].

a · b = 2 − 2 + 6 = 6 and |b| = √(1 + 4 + 4) = 3.
Scalar projection: 6/3 = 2.
Vector projection: (6/9)[1, 2, 2] = (2/3)[1, 2, 2] = [2/3, 4/3, 4/3].
Check the length: |[2/3, 4/3, 4/3]| = (2/3)(3) = 2, matching the scalar projection ✓.

Scalar projection 2; vector projection [2/3, 4/3, 4/3].

When the angle is obtuse, the projection points opposite to v and the scalar projection is negative. Here u · v = −8, so the scalar projection of u on v is −4.

Worked example

Example 3: A negative projection

Find the scalar and vector projections of u = [−4, 1] on v = [2, 0].

u · v = −8 + 0 = −8 and |v| = 2.
Scalar projection: −8/2 = −4.
Vector projection: (−8/4)[2, 0] = −2[2, 0] = [−4, 0].

Scalar projection −4; vector projection [−4, 0], which points opposite to v.

Worked example

Example 4: Order matters

Using a = [2, −1, 3] and b = [1, 2, 2] from Example 2, find the projections of b on a.

a · b = 6 as before, but now we divide by |a| = √14.
Scalar projection of b on a: 6/√14 ≈ 1.60.
Vector projection: (6/14)[2, −1, 3] = (3/7)[2, −1, 3] = [6/7, −3/7, 9/7].

Scalar projection 6/√14 ≈ 1.60, vector projection [6/7, −3/7, 9/7]: not the same as in Example 2.

Components are projections

The scalar projection of v = [v₁, v₂, v₃] on i is v · i / |i| = v₁. So each component of a vector is its scalar projection on an axis, and v = (projection on i) + (projection on j) + (projection on k). Projections let you use any direction as an "axis": along a ramp, along a road, along a runway.

The weight of a 10 kg box on a ramp that rises 3 m for every 4 m across splits into 58.8 N pulling down the slope and 78.4 N pressing into the ramp. Both parts come from projections. (Arrow lengths are drawn to scale.)

Worked example

Example 5: A box on a ramp

A 10 kg box rests on a ramp that rises 3 m for every 4 m horizontally. Its weight is w = [0, −98] N (using g = 9.8 N/kg). Find the part of the weight that acts along the ramp and the part that presses into the ramp.

Direction up the ramp: d = [4, 3], with |d| = 5.
Scalar projection of w on d: (0 − 294)/5 = −58.8. Negative: it acts down the ramp.
Vector projection: (−294/25)[4, 3] = [−47.04, −35.28].
Perpendicular part: w − proj = [0 + 47.04, −98 + 35.28] = [47.04, −62.72], with length √(47.04² + 62.72²) = 78.4.
Check with trigonometry: sin θ = 3/5 and cos θ = 4/5, so 98 sin θ = 58.8 and 98 cos θ = 78.4 ✓.

58.8 N down the slope and 78.4 N into the ramp. Friction must supply 58.8 N up the slope to keep the box still.

Just for fun

A vector looked at its projection and said, "You're just a shadow of my former self." The projection replied, "Maybe, but at least I know which way I'm going."

Hands-on activity

A kitchen scale on a slope

You need: a digital kitchen scale, a heavy unopened can (about 400 g), a long cutting board or a sturdy shelf board, a few books, a ruler and a calculator. Do this on the floor so nothing can fall far.

Weigh the can on the flat scale. Call the reading m.
Put the scale on the board, and raise one end of the board on a stack of books. Measure the board length L between its supports and the rise h. The tilt angle θ satisfies sin θ = h/L.
Press the scale's zero (tare) button while it is empty on the slope, then put the can on it. A scale measures only the push into its surface, which is the perpendicular part of the weight. Predict the reading: m cos θ.
Try two or three different heights. Keep the slope gentle (h no more than half of L) so the can does not slide.

What to look for: with m = 400 g, L = 50 cm and h = 20 cm, sin θ = 0.4, cos θ ≈ 0.917, so the reading should be about 367 g. The missing part of the weight is the component along the slope, the part that would make the can slide.

Your turn

Find the scalar and vector projections of [6, 8] on [1, 0].

Find the scalar and vector projections of [3, 1] on [1, 1].

Find the scalar and vector projections of [1, 2, 3] on [2, −2, 1].

Find the scalar and vector projections of [2, 5] on [−3, 1].

Write [7, 1] as the sum of a vector parallel to [3, 4] and a vector perpendicular to [3, 4].

Find the scalar projection of [4, −2, 5] on k and on j.

More practice: warm-up

u · v = 12 and |v| = 4. Find the scalar projection of u on v.

Same vectors: write the vector projection of u on v as a multiple of v.

u ⊥ v. What is the vector projection of u on v?

Find the vector projection of [5, −3] on i.

When is a scalar projection negative?

More practice: on your own

|u| = 10 and the angle between u and v is 120°. Find the scalar projection of u on v.

Find the vector projection of [2, −1, 2] on [4, 4, −2]. Explain the result.

Write a = [3, −1, 4] as a sum of a vector parallel to b = [1, 1, 1] and a vector perpendicular to b.

A 20 kg toboggan (weight 196 N) sits on a snowy hill that rises 5 m for every 12 m horizontally. Find the parts of its weight along the hill and into the hill.

Show that the vector projection of u on 2v is the same as the vector projection of u on v.

Challenge

Find the distance from the point P(5, 2) to the line through the origin with direction [3, 4].

When is the scalar projection of u on v equal to the scalar projection of v on u (both non-zero vectors)?

Prove that projecting twice changes nothing: proj_v(proj_v u) = proj_v u.

Check your understanding

Quiz

What is the scalar projection of [3, 4] on [1, 0]?

What is the vector projection of [2, 3] on [1, 1]?

The scalar projection of u on v is negative. What does that tell you?

Which expression is the scalar projection of u on v?

What is the scalar projection of a = [1, 2, 2] on b = [0, 0, 3]?

u = [4, 2] and v = [1, 0]. What is the part of u perpendicular to v?

For parents and teachers

Big idea (C2.4, C2.8): the scalar projection (u · v)/|v| is the signed length of the "shadow" of u on the line of v, and the vector projection ((u · v)/|v|²)v is that shadow as a vector. Subtracting the projection from u leaves a part perpendicular to v. This is how physics splits a force into parts along and across a surface.

How to help: The kitchen-scale activity is a memorable way to see a perpendicular component. Ask your teen to explain why the scale reads less on a slope.

Common difficulties:

Dividing by the wrong length: projecting on v means dividing by |v| (or |v|²).
Mixing up the scalar projection (a number) with the vector projection (a vector).
Forgetting the square in |v|² in the vector projection.
Dropping the negative sign when the angle is obtuse.