When you push a lawnmower, only part of your push moves it forward; the rest presses it into the ground. The dot product splits a vector into the part along a direction and the part across it.
Key idea
The vector projection of $\mathbf{u}$ onto $\mathbf{v}$ is $$\operatorname{proj}_{\mathbf{v}}\mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|^2}\,\mathbf{v},$$ its signed length (scalar component) is $\frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|}$, and $\mathbf{u} - \operatorname{proj}_{\mathbf{v}}\mathbf{u}$ is perpendicular to $\mathbf{v}$. Work done by a constant force $\mathbf{F}$ over a displacement $\mathbf{d}$ is $W = \mathbf{F} \cdot \mathbf{d} = |\mathbf{F}||\mathbf{d}|\cos\theta$.
Why it works
Shine a light straight down onto the line of $\mathbf{v}$: the shadow of $\mathbf{u}$ has length $|\mathbf{u}|\cos\theta$. Multiply that length by the unit vector $\frac{\mathbf{v}}{|\mathbf{v}|}$ to give it direction: $$|\mathbf{u}|\cos\theta \cdot \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{|\mathbf{u}||\mathbf{v}|\cos\theta}{|\mathbf{v}|^2}\mathbf{v} = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|^2}\mathbf{v}.$$ Work counts only the force along the motion, which is that same shadow length times the distance.
Worked examples
Worked example: A projection
Find $\operatorname{proj}_{\mathbf{v}}\mathbf{u}$ for $\mathbf{u} = \langle 4, 2 \rangle$ and $\mathbf{v} = \langle 3, 4 \rangle$.
- $\mathbf{u} \cdot \mathbf{v} = 12 + 8 = 20$ and $|\mathbf{v}|^2 = 25$.
- $\frac{20}{25}\langle 3, 4 \rangle = \frac{4}{5}\langle 3, 4 \rangle$.
Answer: $\left\langle \frac{12}{5}, \frac{16}{5} \right\rangle$
Worked example: Splitting a vector
Write $\mathbf{u} = \langle 4, 2 \rangle$ as a part along $\mathbf{v} = \langle 3, 4 \rangle$ plus a part perpendicular to it.
- Along: $\left\langle \frac{12}{5}, \frac{16}{5} \right\rangle$ from the last example.
- Across: $\langle 4, 2 \rangle - \left\langle \frac{12}{5}, \frac{16}{5} \right\rangle = \left\langle \frac{8}{5}, -\frac{6}{5} \right\rangle$.
- Check: $\frac{8}{5} \cdot 3 - \frac{6}{5} \cdot 4 = 0$, so it is perpendicular to $\mathbf{v}$.
Answer: $\mathbf{u} = \left\langle \frac{12}{5}, \frac{16}{5} \right\rangle + \left\langle \frac{8}{5}, -\frac{6}{5} \right\rangle$
Worked example: Work
You pull a wagon $100$ ft with a force of $50$ lb along a handle at $60^\circ$ to the ground. How much work do you do?
- $$W = |\mathbf{F}||\mathbf{d}|\cos\theta = 50 \cdot 100 \cdot \cos 60^\circ.$$
- $= 5000 \cdot \frac{1}{2}$.
Answer: $2500$ ft-lb
Common mistake
Dividing by $|\mathbf{v}|$ instead of $|\mathbf{v}|^2$ in the vector projection. Check with $\mathbf{u} = \mathbf{v}$: the projection of a vector onto itself must be the vector itself, and $\frac{\mathbf{v} \cdot \mathbf{v}}{|\mathbf{v}|^2} = 1$ gives exactly that.