Projections
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 2.3: “The Dot Product” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/2-3-the-dot-product |
| Textbook used in class | Stewart, Calculus, Section 12.3: “The Dot Product” (Examples 6, 7) |
Opening Scenario
A sled is pushed along a flat road with a force at an angle. The only part of the force that moves the sled forward is the component along the road; the component perpendicular to the road just presses down. This “shadow” of one vector onto another is the projection.
Quick Reference
Scalar projection of $\mathbf{a}$ onto $\mathbf{b}$ (the signed length of the shadow): \[ \text{comp}_{\mathbf{b}}\,\mathbf{a} = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}. \]
Vector projection of $\mathbf{a}$ onto $\mathbf{b}$ (the shadow as a vector in the direction of $\mathbf{b}$): \[ \text{proj}_{\mathbf{b}}\,\mathbf{a} = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\,\mathbf{b}. \]
Orthogonal decomposition: $\mathbf{a} = \text{proj}_{\mathbf{b}}\,\mathbf{a} + (\mathbf{a} - \text{proj}_{\mathbf{b}}\,\mathbf{a})$, where the second component is perpendicular to $\mathbf{b}$.
Work done by a constant force $\mathbf{F}$ along displacement $\mathbf{d}$: \[ W = \mathbf{F}\cdot\mathbf{d} = |\mathbf{F}||\mathbf{d}|\cos\theta = |\mathbf{d}|\,\text{comp}_{\mathbf{d}}\,\mathbf{F}. \]
Key Concepts
1. Scalar and Vector Projection
Example 1. Find the scalar and vector projections of $\mathbf{b} = \langle 1, 1, 2\rangle$ onto $\mathbf{a} = \langle -2, 3, 1\rangle$. (Stewart 12.3, Example 6.)
\[ \mathbf{a}\cdot\mathbf{b} = -2 + 3 + 2 = 3, \quad |\mathbf{a}|^2 = 4 + 9 + 1 = 14. \]
Scalar projection: $\text{comp}_{\mathbf{a}}\,\mathbf{b} = 3/\sqrt{14}$.
Vector projection: $\text{proj}_{\mathbf{a}}\,\mathbf{b} = \dfrac{3}{14}\langle -2, 3, 1\rangle = \left\langle -\dfrac{3}{7}, \dfrac{9}{14}, \dfrac{3}{14}\right\rangle$.
2. Orthogonal Decomposition
Example 2. Decompose $\mathbf{b} = \langle 1, 1, 2\rangle$ into components parallel and perpendicular to $\mathbf{a} = \langle -2, 3, 1\rangle$.
Parallel component: $\mathbf{b}_\parallel = \text{proj}_{\mathbf{a}}\,\mathbf{b} = \langle -3/7, 9/14, 3/14\rangle$ (from Example 1).
Perpendicular component: $\mathbf{b}_\perp = \mathbf{b} - \mathbf{b}_\parallel = \langle 1 + 3/7, 1 - 9/14, 2 - 3/14\rangle = \langle 10/7, 5/14, 25/14\rangle$.
Verify: $\mathbf{b}_\perp \cdot \mathbf{a} = (10/7)(-2) + (5/14)(3) + (25/14)(1) = -20/7 + 15/14 + 25/14 = -20/7 + 40/14 = -20/7 + 20/7 = 0$. Perpendicular, confirmed.
3. Work as a Dot Product
Example 3. A force $\mathbf{F} = \langle 3, 4, 5\rangle$ N acts on a particle that moves from $(0,0,0)$ to $(2, 1, 3)$ m. Find the work done. (Stewart 12.3, Example 7.)
Displacement $\mathbf{d} = \langle 2, 1, 3\rangle$. Work $= \mathbf{F}\cdot\mathbf{d} = 6 + 4 + 15 = 25$ J.
confusing the scalar projection with the vector projection. The scalar projection $\text{comp}_{\mathbf{b}}\,\mathbf{a}$ is a number (possibly negative). The vector projection $\text{proj}_{\mathbf{b}}\,\mathbf{a}$ is a vector in the direction of $\mathbf{b}$. To go from scalar to vector, multiply by the unit vector $\hat{\mathbf{b}} = \mathbf{b}/|\mathbf{b}|$. A common error is to report a number when a vector is required, or vice versa.
Common Errors Summary
| Error | Correction |
|---|---|
| Using $|\mathbf{b}|$ instead of $|\mathbf{b}|^2$ in the vector projection | $\text{proj}_{\mathbf{b}}\,\mathbf{a} = (\mathbf{a}\cdot\mathbf{b}/|\mathbf{b}|^2)\,\mathbf{b}$; dividing by $|\mathbf{b}|$ gives the scalar projection |
| Taking the projection onto $\mathbf{a}$ when the question asks for onto $\mathbf{b}$ | The denominator uses the vector you are projecting onto; the dot product uses both |
Common Misconceptions
the scalar projection and the vector projection are the same object.
This is the input-output-confusion error. The scalar projection $\text{comp}_{\mathbf{b}}\,\mathbf{a} = \mathbf{a}\cdot\mathbf{b}/|\mathbf{b}|$ is a signed number giving the length of the shadow. The vector projection $\text{proj}_{\mathbf{b}}\,\mathbf{a} = (\mathbf{a}\cdot\mathbf{b}/|\mathbf{b}|^2)\,\mathbf{b}$ is a vector in the direction of $\mathbf{b}$. Reporting a number when a vector is required, or vice versa, is a unit error. The denominators also differ: $|\mathbf{b}|$ for the scalar, $|\mathbf{b}|^2$ for the vector.
Leveled Practice
Problem 1. Find $\text{proj}_{\mathbf{b}}\,\mathbf{a}$ for $\mathbf{a} = \langle 2, -1, 3\rangle$ and $\mathbf{b} = \langle 1, 2, 1\rangle$.
Show answer
$\mathbf{a}\cdot\mathbf{b} = 2 - 2 + 3 = 3$, $|\mathbf{b}|^2 = 1 + 4 + 1 = 6$. $\text{proj}_{\mathbf{b}}\,\mathbf{a} = \frac{3}{6}\langle 1,2,1\rangle = \langle 1/2, 1, 1/2\rangle$.