Recipe

Find the orthogonal projection of one vector onto another

Use when: you have a nonzero vector $u$ and any vector $v$ , and need the piece of $v$ that lies exactly along $u$ 's own direction

Goal: the projection vector $\operatorname{proj}_u v$ , and if asked, confirmation that the leftover residual is perpendicular to $u$

Taught in Orthogonal projection, revisited

Quick recipe

Compute the two dot products, $v\cdot u$ and $u\cdot u$
Divide to get the scalar "how far along $u$ " factor
Multiply that scalar by $u$ to get the projection vector
Confirm the residual is perpendicular to $u$ , not to $v$

1
Multiply matching components of $v$ and $u$ and add them for $v\cdot u$ ; do the same with $u$ against itself for $u\cdot u$ .

Example

For $u=(2,-1)$ and $v=(3,4)$ : $v\cdot u = (3)(2)+(4)(-1) = 2$ , and $u\cdot u = 2^2+(-1)^2=5$ .
- $u$ is the nonzero vector being projected onto — continue normally; $u\cdot u$ is guaranteed positive.
- $u$ is the zero vector — the projection is undefined, since there's no direction to measure "along."
Why: the projection formula divides by $u\cdot u$ , so it only makes sense once $u$ genuinely points somewhere.
2
Divide $v\cdot u$ by $u\cdot u$ . That single number says how many multiples of $u$ the shadow reaches.

Example

$\dfrac{v\cdot u}{u\cdot u} = \dfrac{2}{5}$ .

Why: this ratio is exactly the coefficient the Orthogonal Decomposition Theorem needs to build the closest point on $u$ 's line.
3
Scale $u$ by the fraction from Step 2, one component at a time.

Example

$\operatorname{proj}_u v = \frac25(2,-1) = \left(\frac45,-\frac25\right)$ .

⚡ Shortcut

If $v$ is already a scalar multiple of $u$ , skip the arithmetic: $\operatorname{proj}_u v = v$ itself, since the whole vector already lies on $u$ 's line. If $v$ is already perpendicular to $u$ (so $v\cdot u=0$ ), skip straight to $\operatorname{proj}_u v = \mathbf 0$ , since there's no shadow to cast at all.

Why: the projection is required to lie on $u$ 's line, meaning it must be some scalar multiple of $u$ ; Step 2 found exactly which multiple.
4
Subtract the projection from $v$ to get the residual $w=v-\operatorname{proj}_uv$ , then dot $w$ specifically with $u$ and confirm it comes out to $0$ .

Example

$w = \left(3-\frac45,\ 4+\frac25\right) = \left(\frac{11}{5},\frac{22}{5}\right)$ , and $w\cdot u = \left(\frac{11}{5}\right)(2)+\left(\frac{22}{5}\right)(-1) = \frac{22}{5}-\frac{22}{5}=0$ .
- $w\cdot u = 0$ — the projection is verified correct.
- $w\cdot u \ne 0$ — an arithmetic error happened upstream; redo Steps 1 through 3.
Why: perpendicularity to $u$ specifically, not to $v$ , is the defining property that makes this the closest point on $u$ 's line to $v$ .

Common mistakes

Dotting the residual against $v$ instead of against $u$ when checking perpendicularity.
Writing $\lVert u\rVert$ where the formula actually needs $u\cdot u$ , which is $\lVert u\rVert^2$ .
Assuming $v$ itself must be perpendicular to $u$ whenever a projection is asked for.
Trying to project onto the zero vector without checking $u$ is nonzero first.