Vector projection in HSC Mathematics Extension 2: the scalar projection, the vector projection, resolving a vector into parallel and perpendicular components, and the distance from a point to a line in 3D

What this dot point is asking

NESA wants you to project one vector onto the direction of another. You must compute the scalar projection (a signed number measuring how much of $\mathbf{a}$ points along $\mathbf{b}$), the vector projection (the actual vector component of $\mathbf{a}$ along $\mathbf{b}$), and use these to split $\mathbf{a}$ into a part parallel to $\mathbf{b}$ and a part perpendicular to $\mathbf{b}$. In Extension 2 this runs in three dimensions and is the engine behind one headline application: the shortest distance from a point to a line.

The scalar projection

Given vectors $\mathbf{a}$ and $\mathbf{b}$ with $\mathbf{b} \neq \mathbf{0}$, the scalar projection of $\mathbf{a}$ onto $\mathbf{b}$ measures the signed length of the shadow $\mathbf{a}$ casts on the line through $\mathbf{b}$. Since $\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$, the scalar projection is

It is positive when the angle $\theta$ between the vectors is acute, zero when they are perpendicular, and negative when $\theta$ is obtuse. The sign carries genuine information about direction, so it must not be silently dropped.

The vector projection

The vector projection is the scalar projection multiplied by the unit vector in the direction of $\mathbf{b}$, namely $\hat{\mathbf{b}} = \dfrac{\mathbf{b}}{|\mathbf{b}|}$. Thus

This is a vector, parallel to $\mathbf{b}$, representing the component of $\mathbf{a}$ lying along $\mathbf{b}$. Its magnitude equals the absolute value of the scalar projection, and its direction is along $\mathbf{b}$ (or against it, when the scalar projection is negative). Geometrically, to project $\mathbf{a}$ onto $\mathbf{b}$ you drop a perpendicular from the head of $\mathbf{a}$ to the line carrying $\mathbf{b}$ and read off the arrow from $O$ to the foot. In three dimensions the picture is the same, with the perpendicular now dropping through space.

Resolving a vector into components

Every vector $\mathbf{a}$ can be split into a part parallel to $\mathbf{b}$ and a part perpendicular to $\mathbf{b}$:

The perpendicular component is genuinely perpendicular to $\mathbf{b}$. Checking the scalar product confirms it,

since $\mathbf{b}\cdot\mathbf{b} = |\mathbf{b}|^2$. The slick way to get the perpendicular part is therefore to find the vector projection first, then subtract it from $\mathbf{a}$; what is left over must be perpendicular. This decomposition is the foundation for resolving forces in mechanics and for the distance from a point to a line.

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 $= (\mathbf{a}\cdot\mathbf{b})\,\mathbf{b}$. Because both projections depend only on the direction of $\mathbf{b}$, scaling $\mathbf{b}$ leaves the projection of $\mathbf{a}$ onto it unchanged.

Distance from a point to a line

The perpendicular component is what makes the distance from a point $P$ to a line accessible. The shortest distance is the length of the perpendicular dropped from $P$ to the line, because any slanted segment to the line is the hypotenuse of a right triangle with the perpendicular as one leg, hence longer. Build the construction up one stage at a time.

Stage 1, the line and the point. Take a line through a known point $A$ with direction $\mathbf{d}$, and an external point $P$. We want the shortest distance from $P$ to the line.

Stage 2, drop the perpendicular. From $P$, the perpendicular to the line meets it at the foot $F$. There are two equivalent ways to find $F$. Either project $\overrightarrow{AP}$ onto $\mathbf{d}$ to get $\overrightarrow{AF} = \operatorname{proj}_{\mathbf{d}}\overrightarrow{AP}$, so $F = A + \overrightarrow{AF}$; or write $F = A + \lambda\mathbf{d}$ and impose $\overrightarrow{FP}\cdot\mathbf{d} = 0$, which gives one equation for $\lambda$. Both encode the same condition: the connector to $P$ is perpendicular to the direction.

Stage 3, read off the distance. The shortest distance is the length $|\overrightarrow{FP}|$ of the perpendicular. Equivalently, it is the magnitude of the perpendicular component of $\overrightarrow{AP}$ relative to $\mathbf{d}$, that is $\bigl|\overrightarrow{AP} - \operatorname{proj}_{\mathbf{d}}\overrightarrow{AP}\bigr|$.

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).
• "Show that $\mathbf{a} - \frac{\mathbf{a}\cdot\mathbf{b}}{\mathbf{b}\cdot\mathbf{b}}\mathbf{b}$ is perpendicular to $\mathbf{b}$": this is the standard HSC phrasing of the perpendicular component; dot it with $\mathbf{b}$ and get $0$.
• "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 shortest (perpendicular) distance from the point $P$ to the line": find the foot $F$ via the projection or the $\overrightarrow{FP}\cdot\mathbf{d} = 0$ condition, then report $|\overrightarrow{FP}|$.
• "Find the component of the force in the direction of motion": a projection of the force onto the direction vector.