Vector Projection

Place $\mathbf{u}$ and $\mathbf{v}$ tail-to-tail with angle $\theta$ between them. Drop a perpendicular from the tip of $\mathbf{v}$ to the line of $\mathbf{u}$. The foot of that perpendicular defines a vector along $\mathbf{u}$ — the projection of $\mathbf{v}$ onto $\mathbf{u}$.

Step 1 — scalar projection. Basic trigonometry gives the signed length of the shadow:

Using $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos\theta$ from L03 and dividing by $|\mathbf{u}|$ produces the right-hand form. This is a scalar — a single signed number.

Step 2 — vector projection. Multiply the scalar projection by the unit vector $\hat{\mathbf{u}} = \mathbf{u}/|\mathbf{u}|$ to give it direction:

The $|\mathbf{u}|^2$ in the denominator is doing two jobs at once: one $|\mathbf{u}|$ converts the dot product to the scalar projection, and the second $|\mathbf{u}|$ normalises $\mathbf{u}$ to a unit vector.

Geometric meaning — three boundary cases.

• $\theta = 0°$ ($\mathbf{v}$ parallel to $\mathbf{u}$): $\text{proj}_{\mathbf{u}} \mathbf{v} = \mathbf{v}$ — the shadow is the whole vector.
• $\theta = 90°$ ($\mathbf{v} \perp \mathbf{u}$): $\mathbf{u} \cdot \mathbf{v} = 0$, so $\text{proj}_{\mathbf{u}} \mathbf{v} = \mathbf{0}$ — no shadow.
• $\theta = 180°$ (opposite direction): the scalar projection is $-|\mathbf{v}|$, and the vector projection points opposite to $\mathbf{u}$.

Scalar projection: $\text{comp}_{\mathbf{u}} \mathbf{v} = (\mathbf{u} \cdot \mathbf{v})/|\mathbf{u}| = |\mathbf{v}|\cos\theta$ · Vector projection: $\text{proj}_{\mathbf{u}} \mathbf{v} = ((\mathbf{u} \cdot \mathbf{v})/|\mathbf{u}|^2)\,\mathbf{u}$ · Why $|\mathbf{u}|^2$: one factor for the scalar length, one to normalise $\mathbf{u}$

Pause — copy the scalar and vector projection formulas, the reason for the $|\mathbf{u}|^2$ denominator (one factor for length, one to normalise) into your book.

We just saw the scalar projection $\text{comp}_{\mathbf{u}}\mathbf{v} = (\mathbf{u}\cdot\mathbf{v})/|\mathbf{u}|$ and vector projection $\text{proj}_{\mathbf{u}}\mathbf{v} = ((\mathbf{u}\cdot\mathbf{v})/|\mathbf{u}|^2)\mathbf{u}$. That raises a question: how do we split $\mathbf{v}$ into the part along $\mathbf{u}$ and the part perpendicular to it? This card answers it → $\mathbf{v} = \mathbf{v}_{\parallel} + \mathbf{v}_{\perp}$ with $\mathbf{v}_{\perp} = \mathbf{v} - \text{proj}_{\mathbf{u}}\mathbf{v}$; verify by checking $\mathbf{v}_{\perp}\cdot\mathbf{u} = 0$.

The vector projection lets you split any $\mathbf{v}$ uniquely into a piece along $\mathbf{u}$ and a piece perpendicular to $\mathbf{u}$:

The two pieces are guaranteed to be perpendicular: a direct calculation shows $\mathbf{v}_{\perp} \cdot \mathbf{u} = \mathbf{v} \cdot \mathbf{u} - \text{proj}_{\mathbf{u}} \mathbf{v} \cdot \mathbf{u} = \mathbf{u} \cdot \mathbf{v} - \mathbf{u} \cdot \mathbf{v} = 0$.

Why this matters in HSC questions.

• $|\mathbf{v}_{\perp}|$ is the perpendicular distance from the tip of $\mathbf{v}$ to the line through $\mathbf{u}$.
• In physics, decomposing a force $\mathbf{F}$ along and perpendicular to a surface or displacement is exactly this calculation.
• Any orthogonal basis problem reduces to repeated projection (the Gram–Schmidt idea, beyond HSC scope but powered by today's formula).

Decomposition: $\mathbf{v} = \mathbf{v}_{\parallel} + \mathbf{v}_{\perp}$, with $\mathbf{v}_{\parallel} = \text{proj}_{\mathbf{u}} \mathbf{v}$ · $\mathbf{v}_{\perp} = \mathbf{v} - \text{proj}_{\mathbf{u}} \mathbf{v}$ · Verify: $\mathbf{v}_{\perp} \cdot \mathbf{u} = 0$ — should always hold · $|\mathbf{v}_{\perp}|$ is the perpendicular distance from tip of $\mathbf{v}$ to line of $\mathbf{u}$

Pause — copy the decomposition $\mathbf{v} = \mathbf{v}_{\parallel} + \mathbf{v}_{\perp}$, the formula $\mathbf{v}_{\perp} = \mathbf{v} - \text{proj}_{\mathbf{u}}\mathbf{v}$, the verification $\mathbf{v}_{\perp}\cdot\mathbf{u} = 0$, and the geometric meaning of $|\mathbf{v}_{\perp}|$ into your book.