Vector projection: scalar projection and vector projection of one vector onto another

What this dot point is asking

NESA wants you to find the shadow of one vector on another: the scalar projection (the signed length of that shadow) and the vector projection (the shadow as a vector), and to interpret them as the part of one vector that lies along the direction of the other. The same idea, run once more, splits any vector into a part parallel to a chosen direction and a part perpendicular to it, which is the move behind resolving forces and a string of geometric proofs.

The answer

Picture both vectors starting at a point $O$. To project $\mathbf{a}$ onto $\mathbf{b}$, shine a light straight down onto the line of $\mathbf{b}$ and read off the shadow of $\mathbf{a}$. Formally, drop a perpendicular from the head of $\mathbf{a}$ to the line through $\mathbf{b}$: the shadow runs from $O$ to the foot of that perpendicular. Building this up one stage at a time is the clearest way to see where every formula comes from.

Stage 1, the two vectors. Draw $\mathbf{a}$ and $\mathbf{b}$ from the same point $O$. We are going to project $\mathbf{a}$ onto the direction of $\mathbf{b}$, so $\mathbf{b}$ sets the direction and $\mathbf{a}$ is the vector being projected.

Stage 2, drop a perpendicular. From the head of $\mathbf{a}$, drop a line straight down onto the line of $\mathbf{b}$. It meets that line at a right angle at the foot $F$. This perpendicular is the key construction; the right-angle marker is what makes the projection "the closest point of $\mathbf{b}$'s line to the head of $\mathbf{a}$".

Stage 3, the vector projection. The arrow from $O$ to $F$, heavy below, is the vector projection of $\mathbf{a}$ onto $\mathbf{b}$. It points along $\mathbf{b}$ (or directly opposite, if the angle is obtuse) and reaches exactly as far as the shadow of $\mathbf{a}$.

Stage 4, the parallel and perpendicular split. The original $\mathbf{a}$, the projection (from $O$ to $F$) and the dropped perpendicular (from $F$ to the head of $\mathbf{a}$) form a right triangle. So $\mathbf{a}$ is the sum of a parallel part (the vector projection, along $\mathbf{b}$) and a perpendicular part (at right angles to $\mathbf{b}$). That decomposition is what "resolve $\mathbf{a}$ into components" means.

Scalar projection

The scalar projection of $\mathbf{a}$ onto $\mathbf{b}$ is the signed length of that shadow (the length $OF$, with a sign):

where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$. The two forms agree because $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$, and dividing by $|\mathbf{b}|$ leaves $|\mathbf{a}|\cos\theta$, which is the adjacent side of the right triangle in Stage 4. The sign is positive when the angle is acute (the shadow points along $\mathbf{b}$) and negative when it is obtuse (the shadow points the opposite way).

Vector projection

The vector projection of $\mathbf{a}$ onto $\mathbf{b}$ is the shadow as an actual vector (the arrow $\overrightarrow{OF}$):

Read it as "scalar projection times the unit vector $\hat{\mathbf{b}}$": take the signed length, then point it along $\mathbf{b}$. Writing $\mathbf{b}/|\mathbf{b}|^2$ rolls the two steps into one, which is why the formula has $|\mathbf{b}|^2$ (one $|\mathbf{b}|$ turns the length into the right size, the other normalises $\mathbf{b}$).

The parallel and perpendicular decomposition

Stage 4 says it in symbols. Any vector $\mathbf{a}$ splits uniquely into a part along $\mathbf{b}$ and a part at right angles to $\mathbf{b}$:

The slick way to get the perpendicular part is to find the vector projection first, then subtract it from $\mathbf{a}$; what is left over must be perpendicular to $\mathbf{b}$. You can always check by dotting the leftover with $\mathbf{b}$ and getting $0$.

Two edge cases worth knowing

• If $\mathbf{a}$ is already parallel to $\mathbf{b}$, the shadow is all of $\mathbf{a}$: $\operatorname{proj}_{\mathbf{b}} \mathbf{a} = \mathbf{a}$ and the perpendicular part is $\mathbf{0}$.
• If $\mathbf{a}$ is perpendicular to $\mathbf{b}$, then $\mathbf{a} \cdot \mathbf{b} = 0$, so both projections are zero: there is no shadow.

A common simplification: if $\mathbf{b}$ is a unit vector ($|\mathbf{b}| = 1$), the formulas collapse to scalar projection $= \mathbf{a} \cdot \mathbf{b}$ and vector projection $= (\mathbf{a} \cdot \mathbf{b})\,\mathbf{b}$.

How exam questions ask about projection

• "Find the scalar projection of $\mathbf{a}$ onto $\mathbf{b}$" or "the component of $\mathbf{a}$ in the direction of $\mathbf{b}$": compute $\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}$ (a number, possibly negative).
• "Find the vector projection of $\mathbf{a}$ onto $\mathbf{b}$" or "the projection vector": compute $\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\mathbf{b}$ (a vector).
• "Resolve $\mathbf{a}$ into components parallel and perpendicular to $\mathbf{b}$": the parallel part is the vector projection; the perpendicular part is $\mathbf{a}$ minus it.
• "Find the component of the force in the direction of motion": a projection of the force onto the direction vector (resolve forces).
• "Find the distance from a point to a line": often the magnitude of the perpendicular part of a displacement onto the line's direction.