Compute the scalar and vector projection of one vector onto another and interpret it geometrically

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.