Recipe

Find the angle between two vectors, a projection, or work done by a force

Use when: you have two vectors, or a force and a displacement, and need the angle between them, a scalar or vector projection of one onto the other, or the work a constant force does

Goal: the angle, projection, or work value the problem specifically asks for, all built from the same dot product and magnitudes

Taught in Projections and work

Quick recipe

Compute the dot product
Decide which of the three questions is being asked
Solve for the angle
Compute the scalar projection, then the vector projection

1
Multiply matching components together and add the three results.

Example

For $\mathbf u=\langle3,4,0\rangle$ and $\mathbf v=\langle1,2,2\rangle$ : $\mathbf u\cdot\mathbf v = (3)(1)+(4)(2)+(0)(2) = 3+8+0=11$ .

Why: every quantity below, the angle, the projection, or the work, starts from this one number.
2
Read the problem to see whether it wants an angle, a projection, or work done by a force, since the next step branches on that answer.
- asked for the angle between the vectors — continue to Step 3.
- asked for a projection of one vector onto the other — skip to Step 4.
- asked for the work a constant force does along a displacement — stop here. Relabeling $\mathbf u$ and $\mathbf v$ as $\mathbf F$ and $\mathbf d$ , Step 1's dot product $\mathbf F\cdot\mathbf d$ already is the work.
Why: the three questions share Step 1 completely and then need different follow-up arithmetic, or none at all for work.
3
Find both magnitudes, divide the dot product by their product, then take the inverse cosine.

Example

$|\mathbf u|=\sqrt{9+16+0}=5$ and $|\mathbf v|=\sqrt{1+4+4}=3$ , so $\cos\theta = \dfrac{11}{(5)(3)}=\dfrac{11}{15}$ , giving $\theta=\cos^{-1}\!\left(\dfrac{11}{15}\right)\approx42.8°$ .

Why: the geometric form of the dot product, $|\mathbf u||\mathbf v|\cos\theta$ , is just Step 1's value restated, so solving it for $\theta$ needs nothing beyond both magnitudes.
4
Divide the dot product by $|\mathbf u|$ for the scalar projection. Multiply that result by $\mathbf u/|\mathbf u|$ for the vector projection.

Example

$\text{comp}_{\mathbf u}\mathbf v = \dfrac{11}{5}=2.2$ , and $\text{proj}_{\mathbf u}\mathbf v = \dfrac{11}{25}\langle3,4,0\rangle = \langle1.32,\,1.76,\,0\rangle$ .

Why: dividing by $|\mathbf u|$ once gives the shadow's signed length, and dividing by $|\mathbf u|$ a second time, then multiplying by $\mathbf u$ itself, turns that length into a full vector pointing along $\mathbf u$ .

Common mistakes

Dividing by the magnitude of the wrong vector, the one being projected instead of the one being projected onto.
Reporting only the scalar projection when the question asked for the vector projection.
Treating a negative scalar projection as a mistake instead of recognizing that the shadow falls behind the vector it's cast onto.
Finding the angle first, then multiplying $|\mathbf F||\mathbf d|\cos\theta$ back out for work, when Step 1's dot product already was the answer.