Visual walkthrough — Projection of vectors

Step 0 — Two arrows and a floor

Before any formula, we need to agree on what we are even looking at.

An arrow (a vector) is just a length pointing in a direction. We draw two of them starting from the same point $O$:

• $\vec{a}$ — the arrow we want to take the shadow of.
• $\vec{b}$ — the arrow we shine the shadow onto. Think of $\vec{b}$ as marking out a floor-direction, a line to drop shadows onto.

The little symbol $\stackrel{⃗}{}$ over a letter just means "this is an arrow, not a plain number".

Step 1 — Drop the shadow: a right triangle appears

WHAT. Shine light straight down onto the line through $\vec{b}$. The tip of $\vec{a}$ lands somewhere on that line. The segment from $O$ to that landing point is the shadow. Call its signed length $s$.

WHY. "Straight down" means the light rays hit the floor at $90^{\circ }$. That perpendicular drop is the only rule that makes a clean triangle — and a right triangle is a shape whose sides we know how to relate.

PICTURE. Look at the red dashed line dropping from $\vec{a}$'s tip: it meets $\vec{b}$'s line at a perfect right angle (the small square). That gives us a right triangle:

• the slanted side (hypotenuse) is $\vec{a}$ itself, with length $|\vec{a}|$;
• the base lying along $\vec{b}$ is the shadow, length $s$;
• the angle at $O$ between $\vec{a}$ and $\vec{b}$ we call $\theta$ (the Greek letter "theta", just a name for that opening angle).

The vertical bars $|\vec{a}|$ mean length of $\vec{a}$ — a plain positive number, like measuring the arrow with a ruler.

Step 2 — Why $\cos \theta$, and nothing else, gives the shadow

WHAT. In our right triangle, the shadow $s$ is the side next to the angle $\theta$ (the adjacent side), and $|\vec{a}|$ is the hypotenuse. We claim $s=|\vec{a}|\cos \theta .$

WHY this tool? We need a machine that eats the angle $\theta$ and the hypotenuse $|\vec{a}|$ and spits out the adjacent side. That machine is cosine. By its very definition on a right triangle: $\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{s}{|\vec{a}|}.$ We did not pick sine (that gives the opposite side, the height of the shadow above the floor — the part we are throwing away) and not tangent (a ratio of the two sides that ignores the hypotenuse's actual length). Cosine is the unique fit: it measures how much of the hypotenuse survives the lean onto the floor.

PICTURE. As $\theta$ grows from $0$, the arrow leans further from the floor and the green shadow shrinks — exactly what $\cos \theta$ does (it falls from $1$ toward $0$). At $\theta =90^{\circ }$ the arrow points straight up, the shadow vanishes, and $\cos 90^{\circ }=0$. Perfect match.

Multiplying: $s=|\vec{a}|\cos \theta$. Here $|\vec{a}|$ is the arrow's full length and $\cos \theta$ is the fraction of that length that lies along $\vec{b}$.

Step 3 — The sign of the shadow: cover the back-side case

WHAT. What if $\theta$ is bigger than $90^{\circ }$ — $\vec{a}$ leans away from $\vec{b}$? Then the landing point is on the opposite side of $O$, and we say the shadow length $s$ is negative.

WHY. $\cos \theta$ is positive for $\theta <90^{\circ }$, exactly zero at $90^{\circ }$, and negative for $\theta >90^{\circ }$. So $s=|\vec{a}|\cos \theta$ automatically turns negative. This is a feature, not a bug: the minus sign records which side of $O$ the shadow fell on.

PICTURE. Three cases side by side:

• $\theta <90^{\circ }$ (blue): shadow points forward along $\vec{b}$, $s>0$.
• $\theta =90^{\circ }$: $\vec{a}$ is perpendicular, shadow has zero length, $s=0$.
• $\theta >90^{\circ }$ (red): shadow points backward, $s<0$.

Step 4 — Escape the protractor with the dot product

WHAT. The formula $|\vec{a}|\cos \theta$ is honest but useless with only coordinates: nobody hands you $\theta$. We swap it out using the dot product.

WHY this tool? The Dot product is defined by two facts that are secretly the same: $\vec{a}\cdot \vec{b}=a_x{b}_x+a_y{b}_y(\text{multiply matching parts, add})=|\vec{a}||\vec{b}|\cos \theta .$ The right-hand side already contains the $\cos \theta$ we want, glued to the two lengths. So the dot product is the one tool that hides an angle inside pure arithmetic — it lets us compute $\cos \theta$'s effect without ever measuring the angle.

PICTURE. The dot product bar: coordinates go in the left slot, the geometric meaning comes out the right slot — the same number by two routes.

Now isolate the shadow. We want $|\vec{a}|\cos \theta$ alone, so divide both sides of $\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos \theta$ by $|\vec{b}|$: $\underset{\text{pure arithmetic}}{\underbrace{\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}}}=\underset{\text{the shadow }s}{\underbrace{|\vec{a}|\cos \theta }}.$ Every symbol: $\vec{a}\cdot \vec{b}$ is the multiply-and-add number, $|\vec{b}|$ is the length of $\vec{b}$ we divide out because the dot product carried an extra factor of $|\vec{b}|$ we never asked for.

Step 5 — From a length to an arrow

WHAT. So far $s=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}$ is just a number. The vector projection is an actual arrow: it has that length and points along $\vec{b}$.

WHY. Any vector is (how long) × (which way). We have the length $s$. The "which way" is the unit vector $\hat{b}$ — the direction of $\vec{b}$ stripped of its length. Attaching them: ${\text{proj}}_{\vec{b}}\vec{a}=s\hat{b}=\left(\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}\right)\frac{\vec{b}}{|\vec{b}|}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}{|}^2}\vec{b}=\frac{\vec{a}\cdot \vec{b}}{\vec{b}\cdot \vec{b}}\vec{b}.$

Watch the denominator: one $|\vec{b}|$ came from getting the length in Step 4, the second $|\vec{b}|$ came from turning $\vec{b}$ into a unit direction. Together $|\vec{b}|\cdot |\vec{b}|=|\vec{b}{|}^2=\vec{b}\cdot \vec{b}$ (since $\vec{b}\cdot \vec{b}=|\vec{b}{|}^2\cos 0=|\vec{b}{|}^2$).

PICTURE. The green arrow is the finished vector projection, sitting on $\vec{b}$'s line with length $s$, and $\vec{b}$ itself scaled to unit length shown underneath.

Step 6 — The leftover piece must be perpendicular

WHAT. Subtract the shadow-arrow from $\vec{a}$. What remains is the part of $\vec{a}$ that had no direction along $\vec{b}$: ${\vec{a}}_{\perp }=\vec{a}-{\text{proj}}_{\vec{b}}\vec{a}.$

WHY. By construction the projection soaked up all the along-$\vec{b}$ content of $\vec{a}$. Whatever is left can only point at right angles to $\vec{b}$. Let us prove it — take the dot product of the leftover with $\vec{b}$: ${\vec{a}}_{\perp }\cdot \vec{b}=\vec{a}\cdot \vec{b}-\frac{\vec{a}\cdot \vec{b}}{\vec{b}\cdot \vec{b}}(\vec{b}\cdot \vec{b})=\vec{a}\cdot \vec{b}-\vec{a}\cdot \vec{b}=0.$ A dot product of zero means the two arrows meet at $90^{\circ }$. So ${\vec{a}}_{\perp }\perp \vec{b}$. ✓

PICTURE. $\vec{a}$ (orange) is now the exact sum of two perpendicular arrows: the green shadow along $\vec{b}$ plus the red perpendicular remainder. They form a right triangle whose hypotenuse is $\vec{a}$ — the same triangle from Step 1, now labelled with vectors.

This split $\vec{a}={\text{proj}}_{\vec{b}}\vec{a}+{\vec{a}}_{\perp }$ is Orthogonal decomposition, the engine behind Gram-Schmidt process and Least squares regression.

Step 7 — The degenerate case: what if $\vec{b}=\vec{0}$?

WHAT. If $\vec{b}$ is the zero vector (a point, no direction), then $\vec{b}\cdot \vec{b}=0$ and the formula asks us to divide by zero.

WHY it must break. A shadow needs a direction to fall along. The zero vector points nowhere, so "the shadow of $\vec{a}$ onto nothing" has no meaning — the maths correctly refuses (division by zero) rather than inventing a fake answer.

PICTURE. $\vec{b}$ collapsed to the origin: there is no line to drop a shadow onto, and the projection is undefined.

The one-picture summary

Everything on one canvas: the right triangle (Step 1–2), the dot product replacing the angle (Step 4), the green shadow-arrow (Step 5), and the red perpendicular leftover (Step 6).

Active recall

Connections

• Projection of vectors — the parent topic this walkthrough derives.
• Dot product — the multiply-and-add engine that eliminates $\theta$.
• Unit vectors — $\hat{b}$ supplies the direction in Step 5.
• Orthogonal decomposition — the parallel + perpendicular split of Step 6.
• Gram-Schmidt process — repeated projection to orthonormalise.
• Least squares regression — projecting data onto a subspace.
• Work done by a force — $W=\vec{F}\cdot \vec{d}$ uses the projection of force along displacement.