Vector Projections

A projection answers the question “how much of $\vec{a}$ points in the direction of $\vec{b}$?” Think of the shadow $\vec{a}$ casts on the line of $\vec{b}$ when a light shines straight down onto that line. Projections are how you find the part of a force that pulls along a ramp, the part of the wind that pushes a boat forward, or the part of a pull that does work, and they come straight from the dot product.

Key ideas

The picture

Place $\vec{a}$ and $\vec{b}$ tail to tail. Drop a perpendicular from the tip of $\vec{a}$ to the line containing $\vec{b}$. The arrow from the common tail to the foot of that perpendicular is the projection of $\vec{a}$ onto $\vec{b}$.

The projection of $\vec{a}$ onto $\vec{b}$ (green). It points along $\vec{b}$ when $\theta$ is acute, and the opposite way when $\theta$ is obtuse.

Scalar projection

The scalar projection of $\vec{a}$ onto $\vec{b}$ is the signed length of that shadow. From the right triangle, it’s $|\vec{a}|\cos\theta$. Multiply and divide by $|\vec{b}|$ to write it with the dot product:

$$\text{scalar projection of } \vec{a} \text{ onto } \vec{b} = |\vec{a}|\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$

It’s positive when $\theta$ is acute (the shadow points along $\vec{b}$), negative when $\theta$ is obtuse (it points the opposite way), and $0$ when the vectors are perpendicular.

Vector projection

The vector projection is the shadow as an actual vector. It has length equal to the scalar projection, in the direction of the unit vector $\dfrac{\vec{b}}{|\vec{b}|}$:

$$\operatorname{proj}_{\vec{b}}\vec{a} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\right)\frac{\vec{b}}{|\vec{b}|} = \left(\frac{\vec{a} \cdot \vec{b}}{\vec{b} \cdot \vec{b}}\right)\vec{b}$$

The second form is the easiest to use: the bracket is just a number, so the answer is a scalar multiple of $\vec{b}$. (Recall $\vec{b} \cdot \vec{b} = |\vec{b}|^2$.)

Order matters: ”$\vec{a}$ onto $\vec{b}$” divides by $\vec{b}$‘s length. The projection of $\vec{b}$ onto $\vec{a}$ is a different vector, along $\vec{a}$.

Splitting a vector into two parts

Any vector $\vec{a}$ can be split into a part parallel to $\vec{b}$ and a part perpendicular to $\vec{b}$:

$$\vec{a} = \underbrace{\operatorname{proj}_{\vec{b}}\vec{a}}_{\text{parallel to } \vec{b}} + \underbrace{\left(\vec{a} - \operatorname{proj}_{\vec{b}}\vec{a}\right)}_{\text{perpendicular to } \vec{b}}$$

You can check your work: the perpendicular part should have a dot product of $0$ with $\vec{b}$. This split is exactly what you do with a force on a ramp: one part along the slope, one part pressing into it.

Direction angles and direction cosines

The direction angles of a vector $\vec{v} = [a, b, c]$ are the angles $\alpha$, $\beta$, $\gamma$ it makes with the positive $x$-, $y$-, and $z$-axes. Dot $\vec{v}$ with $\vec{i} = [1, 0, 0]$ and you get $a = |\vec{v}|\cos\alpha$, and likewise for the others, so the direction cosines are

$$\cos\alpha = \frac{a}{|\vec{v}|}, \qquad \cos\beta = \frac{b}{|\vec{v}|}, \qquad \cos\gamma = \frac{c}{|\vec{v}|}$$

They’re the components of the unit vector in the direction of $\vec{v}$, so $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$. Each component $a$, $b$, $c$ is just the scalar projection of $\vec{v}$ onto an axis.

Applications

• Work. $W = \vec{F} \cdot \vec{d}$ is the scalar projection of $\vec{F}$ onto $\vec{d}$ (the useful part of the force) times the distance $|\vec{d}|$.
• Forces on a slope. On a ramp inclined at angle $\theta$, gravity $\vec{F}_g$ (pointing straight down) splits into a part along the ramp, of size $|\vec{F}_g|\sin\theta$, and a part into the ramp, of size $|\vec{F}_g|\cos\theta$. Example 4 does this with projections.

Worked examples

Example 1: Projections in 2-D

Let $\vec{a} = [5, 1]$ and $\vec{b} = [3, 4]$. Find the scalar and vector projections of $\vec{a}$ onto $\vec{b}$, and split $\vec{a}$ into parts parallel and perpendicular to $\vec{b}$.

Solution. $\vec{a} \cdot \vec{b} = 15 + 4 = 19$ and $|\vec{b}| = \sqrt{9 + 16} = 5$.

$$\text{scalar projection} = \frac{19}{5} = 3.8$$ $$\operatorname{proj}_{\vec{b}}\vec{a} = \frac{19}{25}[3, 4] = \left[\frac{57}{25}, \frac{76}{25}\right] = [2.28, 3.04]$$

The perpendicular part is what’s left over:

$$\vec{a} - \operatorname{proj}_{\vec{b}}\vec{a} = [5 - 2.28,\ 1 - 3.04] = [2.72, -2.04]$$

Check: $[2.72, -2.04] \cdot [3, 4] = 8.16 - 8.16 = 0$, so it really is perpendicular to $\vec{b}$. ✓ And the two parts add back to $[5, 1]$. ✓

Example 2: A negative projection in 3-D

Let $\vec{a} = [2, -1, 4]$ and $\vec{b} = [1, 2, -2]$. Find the scalar and vector projections of $\vec{a}$ onto $\vec{b}$, and explain the sign.

Solution. $\vec{a} \cdot \vec{b} = 2 - 2 - 8 = -8$ and $|\vec{b}| = \sqrt{1 + 4 + 4} = 3$.

$$\text{scalar projection} = \frac{-8}{3} \approx -2.67$$ $$\operatorname{proj}_{\vec{b}}\vec{a} = \frac{-8}{9}[1, 2, -2] = \left[-\frac{8}{9}, -\frac{16}{9}, \frac{16}{9}\right]$$

The dot product is negative, so the angle between $\vec{a}$ and $\vec{b}$ is obtuse. The shadow of $\vec{a}$ falls behind the tail, and the projection points in the direction opposite to $\vec{b}$ (it’s a negative multiple of $\vec{b}$), like the right-hand diagram above.

Example 3: Direction angles

Find the direction cosines and direction angles of $\vec{v} = [2, -3, 6]$, to one decimal place.

Solution. $|\vec{v}| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.

$$\begin{aligned} \cos\alpha &= \frac{2}{7} &&\Rightarrow\quad \alpha \approx 73.4^\circ \\ \cos\beta &= -\frac{3}{7} &&\Rightarrow\quad \beta \approx 115.4^\circ \\ \cos\gamma &= \frac{6}{7} &&\Rightarrow\quad \gamma \approx 31.0^\circ \end{aligned}$$

Check: $\left(\dfrac{2}{7}\right)^2 + \left(-\dfrac{3}{7}\right)^2 + \left(\dfrac{6}{7}\right)^2 = \dfrac{4 + 9 + 36}{49} = 1$. ✓ The angle with the $y$-axis is obtuse because the $y$-component is negative.

Example 4: A box on a ramp

A $40$ kg box rests on a ramp inclined at $20^\circ$. Gravity pulls down on it with a force of $40 \times 9.8 = 392$ N. Find the component of gravity along the ramp and the component pressing into the ramp, to the nearest newton.

Gravity (blue) splits into a part along the ramp (green) and a part into the ramp (orange).

Solution. Use axes with $x$ horizontal and $y$ vertical. Gravity is $\vec{F} = [0, -392]$. A unit vector pointing up the ramp is $\vec{u} = [\cos 20^\circ, \sin 20^\circ]$, and a unit vector pointing out of the ramp (perpendicular to it) is $\vec{n} = [-\sin 20^\circ, \cos 20^\circ]$. Since both are unit vectors, the scalar projections are just dot products:

$$\begin{aligned} \vec{F} \cdot \vec{u} &= 0 - 392\sin 20^\circ \approx -134 \text{ N} \\ \vec{F} \cdot \vec{n} &= 0 - 392\cos 20^\circ \approx -368 \text{ N} \end{aligned}$$

So about $134$ N of gravity acts down the ramp (negative means opposite to “up the ramp”), and about $368$ N presses into the ramp. The ramp pushes back with $368$ N, and friction or a person has to supply $134$ N up the slope to stop the box sliding.

Check: $\sqrt{134.07^2 + 368.36^2} \approx 392$ ✓. These match $392\sin 20^\circ$ and $392\cos 20^\circ$ from the diagram.

Common mistakes

Dividing by the wrong magnitude. The projection of $\vec{a}$ onto $\vec{b}$ divides by $|\vec{b}|$ (or $\vec{b} \cdot \vec{b}$), the vector you’re projecting onto. Dividing by $|\vec{a}|$ gives the projection of $\vec{b}$ onto $\vec{a}$, which is a different answer.

Forgetting to square in the vector projection. $\operatorname{proj}_{\vec{b}}\vec{a} = \dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\vec{b}$, with $|\vec{b}|$ squared. One factor of $|\vec{b}|$ gives the scalar projection; the second one turns $\vec{b}$ into a unit vector.

Mixing up the scalar and the vector projection. The scalar projection is a number (possibly negative). The vector projection is a vector along the line of $\vec{b}$. If a question asks for “the projection” as a vector, give components.

Dropping the negative sign. A negative scalar projection isn’t an error. It tells you the angle is obtuse and the projection points opposite to $\vec{b}$. Don’t take the absolute value unless you’re asked for a length.

Swapping sine and cosine on a ramp. The part of gravity along a slope of angle $\theta$ is $|\vec{F}_g|\sin\theta$, not $\cos\theta$. Sense check: on a nearly flat ramp ($\theta$ close to $0^\circ$), almost nothing pulls the box along it, and $\sin\theta$ is close to $0$.

Practice

1. (Warm-up) Find the scalar and vector projections of $\vec{a} = [4, -2]$ onto $\vec{b} = [3, 4]$.

Solution

$\vec{a} \cdot \vec{b} = 12 - 8 = 4$ and $|\vec{b}| = 5$.

Scalar projection: $\dfrac{4}{5} = 0.8$.

Vector projection: $\dfrac{4}{25}[3, 4] = \left[\dfrac{12}{25}, \dfrac{16}{25}\right] = [0.48, 0.64]$.

2. (Warm-up) $|\vec{a}| = 10$, $|\vec{b}| = 4$, and the angle between them is $150^\circ$. Find the scalar projection of $\vec{a}$ onto $\vec{b}$. Does the vector projection point the same way as $\vec{b}$ or the opposite way?

Solution

$$|\vec{a}|\cos\theta = 10\cos 150^\circ = 10\left(-\frac{\sqrt{3}}{2}\right) = -5\sqrt{3} \approx -8.66$$

It’s negative (the angle is obtuse), so the vector projection points opposite to $\vec{b}$.

3. (Core) Let $\vec{a} = [1, 3, -2]$ and $\vec{b} = [2, -1, 2]$.

• (a) Find the scalar and vector projections of $\vec{a}$ onto $\vec{b}$.
• (b) Find the scalar and vector projections of $\vec{b}$ onto $\vec{a}$.

Solution

$\vec{a} \cdot \vec{b} = 2 - 3 - 4 = -5$, $|\vec{a}| = \sqrt{1 + 9 + 4} = \sqrt{14}$, $|\vec{b}| = \sqrt{4 + 1 + 4} = 3$.

(a) Scalar: $\dfrac{-5}{3} \approx -1.67$. Vector: $\dfrac{-5}{9}[2, -1, 2] = \left[-\dfrac{10}{9}, \dfrac{5}{9}, -\dfrac{10}{9}\right]$.

(b) Scalar: $\dfrac{-5}{\sqrt{14}} \approx -1.34$. Vector: $\dfrac{-5}{14}[1, 3, -2] = \left[-\dfrac{5}{14}, -\dfrac{15}{14}, \dfrac{5}{7}\right]$.

The two answers are different: projection is not symmetric.

4. (Core) Write $\vec{F} = [10, 6]$ as the sum of a vector parallel to $\vec{d} = [4, -3]$ and a vector perpendicular to $\vec{d}$.

Solution

$\vec{F} \cdot \vec{d} = 40 - 18 = 22$ and $\vec{d} \cdot \vec{d} = 16 + 9 = 25$.

$$\text{parallel part} = \frac{22}{25}[4, -3] = [3.52, -2.64]$$$$\text{perpendicular part} = [10, 6] - [3.52, -2.64] = [6.48, 8.64]$$

Check: $[6.48, 8.64] \cdot [4, -3] = 25.92 - 25.92 = 0$. ✓ So $\vec{F} = [3.52, -2.64] + [6.48, 8.64]$.

5. (Core)

• (a) Find the direction angles of $[1, -1, 1]$, to one decimal place.
• (b) A vector makes an angle of $60^\circ$ with the $x$-axis and $45^\circ$ with the $y$-axis, and an acute angle with the $z$-axis. Find that angle.

Solution

(a) The magnitude is $\sqrt{3}$, so $\cos\alpha = \dfrac{1}{\sqrt{3}}$, $\cos\beta = -\dfrac{1}{\sqrt{3}}$, $\cos\gamma = \dfrac{1}{\sqrt{3}}$. That gives $\alpha \approx 54.7^\circ$, $\beta \approx 125.3^\circ$, $\gamma \approx 54.7^\circ$.

(b) Use $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$:

$$\cos^2\gamma = 1 - \left(\frac{1}{2}\right)^2 - \left(\frac{\sqrt{2}}{2}\right)^2 = 1 - \frac{1}{4} - \frac{1}{2} = \frac{1}{4}$$

$\gamma$ is acute, so $\cos\gamma = \dfrac{1}{2}$ and $\gamma = 60^\circ$.

6. (Core) A sailboat is heading northeast. The wind pushes on its sail with a force of $\vec{F} = [300, 400]$ N, where the components are east and north. Taking the boat’s heading as the direction $\vec{d} = [1, 1]$, find the component of the wind’s force along the heading, to the nearest newton.

Solution

This is the scalar projection of $\vec{F}$ onto $\vec{d}$:

$$\frac{\vec{F} \cdot \vec{d}}{|\vec{d}|} = \frac{300 + 400}{\sqrt{2}} = \frac{700}{\sqrt{2}} \approx 495 \text{ N}$$

7. (Core) A $50$ N force pulls a cart $12$ m along a straight track. The force acts at $40^\circ$ to the track. Find the scalar projection of the force onto the direction of motion, and use it to find the work done, each to one decimal place.

Solution

Scalar projection: $50\cos 40^\circ \approx 38.3$ N. That’s the part of the force pulling along the track.

Work: $W = 50\cos 40^\circ \times 12 \approx 459.6$ J. (The same as $|\vec{F}||\vec{d}|\cos\theta$.)

8. (Challenge) Let $A(1, 0, 2)$, $B(4, 4, 2)$, and $C(1, 6, 10)$.

• (a) Find the scalar projection of $\overrightarrow{AB}$ onto $\overrightarrow{AC}$.
• (b) Find the point $D$ on the line $AC$ that is closest to $B$ (the foot of the perpendicular from $B$).
• (c) Find the distance from $B$ to the line $AC$, to two decimal places.

Solution

$\overrightarrow{AB} = [3, 4, 0]$ and $\overrightarrow{AC} = [0, 6, 8]$, so $\overrightarrow{AB} \cdot \overrightarrow{AC} = 0 + 24 + 0 = 24$ and $|\overrightarrow{AC}| = 10$.

(a) Scalar projection: $\dfrac{24}{10} = 2.4$.

(b) The vector projection is $\dfrac{24}{100}[0, 6, 8] = [0, 1.44, 1.92]$. Starting at $A$:

$$D = (1 + 0,\ 0 + 1.44,\ 2 + 1.92) = (1, 1.44, 3.92)$$

(c) The distance is the length of the perpendicular part, $\overrightarrow{DB} = \overrightarrow{AB} - [0, 1.44, 1.92] = [3, 2.56, -1.92]$:

$$|\overrightarrow{DB}| = \sqrt{9 + 6.5536 + 3.6864} = \sqrt{19.24} \approx 4.39$$

Check: $[3, 2.56, -1.92] \cdot [0, 6, 8] = 0 + 15.36 - 15.36 = 0$. ✓

9. (Challenge) Show that for any non-zero scalar $k$, the vector projection of $\vec{a}$ onto $k\vec{b}$ is the same as the vector projection of $\vec{a}$ onto $\vec{b}$. What happens to the scalar projection when $k$ is negative? Explain why this makes sense.

Solution

$$\operatorname{proj}_{k\vec{b}}\vec{a} = \frac{\vec{a} \cdot (k\vec{b})}{(k\vec{b}) \cdot (k\vec{b})}\,(k\vec{b}) = \frac{k(\vec{a} \cdot \vec{b})}{k^2(\vec{b} \cdot \vec{b})}\,k\vec{b} = \frac{\vec{a} \cdot \vec{b}}{\vec{b} \cdot \vec{b}}\,\vec{b} = \operatorname{proj}_{\vec{b}}\vec{a}$$

The scalar projection onto $k\vec{b}$ is

$$\frac{\vec{a} \cdot (k\vec{b})}{|k\vec{b}|} = \frac{k}{|k|} \cdot \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$

For $k \lt 0$, $\dfrac{k}{|k|} = -1$, so the scalar projection changes sign. That makes sense: $k\vec{b}$ lies on the same line as $\vec{b}$, so the shadow of $\vec{a}$ on that line is the same and the vector projection doesn’t change. But the scalar projection measures the shadow as positive or negative relative to the direction of the vector you project onto, and a negative $k$ flips that direction.