Visual walkthrough — Scalars vs vectors — definition, examples

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Step 0 — The one piece of trigonometry we will need: cosine

WHY we introduce it now. Our derivation compares a slanted arrow to a straight baseline. The single number that captures "how much this angle leans along that baseline" is the cosine. We meet it once here so it is never a surprise later.

PICTURE. A right triangle with the angle $\alpha$, its adjacent side and hypotenuse marked, and the three landmark values below.

Recall Why cosine can be negative

Past $90^{\circ }$ the direction starts leaning backward along the baseline. "Forward" was positive, so "backward" is negative. That sign is the whole engine of the minus-to-plus flip you will see in Step 5.

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Step 1 — What is a vector, drawn as an arrow?

WHAT. We take our two quantities — call them $\vec{A}$ and $\vec{B}$ — and draw each as an arrow.

WHY. Because the whole difficulty (from the parent's walking problem) is that direction changes the answer. An arrow is the one picture that stores both size and direction at once, so it is the honest way to represent the quantity.

PICTURE. Two arrows starting from the same dot (their tails touch). The angle between them is $\theta$ (the Greek letter "theta", just a name for that opening angle).

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Step 2 — What does "adding" two vectors even mean?

WHAT. To add, we slide $\vec{B}$ without turning it, until its tail sits on the head of $\vec{A}$. Then we draw one new arrow $\vec{R}$ from the tail of $\vec{A}$ to the head of the moved $\vec{B}$. Its length is $R=|\vec{R}|$.

WHY. "Do $\vec{A}$, then continue with $\vec{B}$" is head-to-tail placement — you finish the first walk and start the second from there. The straight arrow home is the net effect, the resultant $\vec{R}$.

PICTURE. The head-to-tail chain forming a triangle; $\vec{R}$ closes it.

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Step 3 — The angle inside the triangle is NOT $\theta$

WHAT. In the head-to-tail triangle, the angle at the join (where $\vec{A}$'s head meets $\vec{B}$'s tail) is $180^{\circ }-\theta$, not $\theta$.

WHY. When we slid $\vec{B}$ to the head of $\vec{A}$, we kept it pointing the same way. But now $\vec{A}$ arrives into that corner and $\vec{B}$ leaves out of it. The two arrows lie along a straight-through path, so the interior corner is the supplement of the original opening angle: it fills up the rest of the straight $180^{\circ }$.

PICTURE. A zoom on the corner: the original $\theta$ and its supplement $180^{\circ }-\theta$ shown together adding to a straight line.

This single fact — supplement, not $\theta$ — is the reason a minus appears mid-derivation and then flips back to a plus. Watch for it.

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Step 4 — Why we reach for the Law of Cosines (and not Pythagoras)

WHAT. We name the pieces of our triangle to match the law:

• side $a=A$, side $b=B$,
• side $c=R$ (the unknown we want),
• angle $C=180^{\circ }-\theta$ (the interior corner from Step 3, which sits opposite $R$).

WHY. The formula demands the angle opposite the side you're solving for. $\vec{R}$ closes the triangle, so the corner facing it — the $180^{\circ }-\theta$ from Step 3 — is exactly the angle to plug in.

PICTURE. The triangle relabelled with $a,b,c,C$ mapped onto $A,B,R,(180^{\circ }-\theta )$.

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Step 5 — Substitute, and turn the minus into a plus

WHAT. Put our labels into the Law of Cosines. Here $R$ still means the length $|\vec{R}|$: $R^2=A^2+B^2-2AB\cos (180^{\circ }-\theta ).$

Term by term, right where each lives:

• $R^2$ ::: square of the resultant length $|\vec{R}|$ (what we're solving for)
• $A^2$, $B^2$ ::: the "Pythagoras part" — what you'd get if the triangle had a right angle
• $-2AB\cos (180^{\circ }-\theta )$ ::: the correction, still holding the supplement angle from Step 3

WHY the next move. That $\cos (180^{\circ }-\theta )$ is awkward — it uses the interior angle, but we want the answer in terms of the original angle $\theta$ between the vectors. We use the identity flagged in Step 0 to translate: $\cos (180^{\circ }-\theta )=-\cos \theta .$

WHAT (finish). Substitute the identity: $R^2=A^2+B^2-2AB(-\cos \theta )=A^2+B^2+2AB\cos \theta .$

The minus times the minus becomes a plus. Take the square root (lengths are never negative, so we keep the $+$ root), and recall $R\equiv |\vec{R}|$: $\boxed{|\vec{R}|=\sqrt{A^2+B^2+2AB\cos \theta }}$

PICTURE. The sign-flip shown as a reflection of the correction term, "$-$ meets $-$ becomes $+$".

That is the parent's formula — no longer stated, now earned.

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Step 6 — Check every case (the formula must never surprise you)

A good formula covers all inputs. Let us feed it every angle and watch it behave. Throughout, $R=|\vec{R}|$.

Case $\theta =0^{\circ }$ (same direction). $\cos 0^{\circ }=1$: $R=\sqrt{A^2+B^2+2AB}=\sqrt{(A+B{)}^2}=A+B.$ The arrows stack in a line — maximum length, plain arithmetic. This is exactly why aligned vectors "act like scalars".

Case $\theta =90^{\circ }$ (perpendicular). $\cos 90^{\circ }=0$: $R=\sqrt{A^2+B^2}.$ The correction dies; we fall back to pure Pythagoras. (This is Example 1's $3,4,5$ triangle in the parent.)

Case $\theta =180^{\circ }$ (opposite directions). $\cos 180^{\circ }=-1$: $R=\sqrt{A^2+B^2-2AB}=\sqrt{(A-B{)}^2}=|A-B|.$ The arrows fight; the result is their difference — minimum length. If $A=B$ they cancel to $0$.

Case $\theta$ between (e.g. $120^{\circ }$). $\cos 120^{\circ }=-\frac{1}{2}$, giving $R=\sqrt{A^2+B^2-AB}$ — a value smoothly between the extremes.

PICTURE. One figure showing $R$ shrinking from $A+B$ down to $|A-B|$ as $\theta$ opens from $0^{\circ }$ to $180^{\circ }$.

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The one-picture summary

Every idea on this page, stacked in one frame: two arrows tail-to-tail ($\theta$), slid head-to-tail (corner $180^{\circ }-\theta$), closed by $\vec{R}$, with the law of cosines and the sign-flip turning it into $\sqrt{A^2+B^2+2AB\cos \theta }$, plus the min/max dial along the bottom.

Recall Feynman retelling — the whole walk in plain words

First, one small tool: on a right triangle, cosine of an angle is just "how much the slanted side leans forward" — a number between $-1$ and $1$, negative when it leans backward. Keep that in your pocket. Now picture two arrows growing out of the same spot. To add them, I pick up the second arrow, keep it pointing the same way, and stick its tail onto the tip of the first — like taking one walk, then a second walk from where I stopped. The straight line from my start to my finish is the answer arrow, and its length is what I call $R$ (same thing as $|\vec{R}|$, just fewer bars). That answer is the third side of a triangle. There's a rule for finding a triangle's third side when you know the two other sides and the corner between them — the law of cosines. But there's a twist: the corner inside my triangle isn't the angle I started with; it's what's left over from a straight line, $180^{\circ }$ minus my angle. When I feed that in, the cosine flips sign (forward becomes backward), a minus meets a minus, and out pops a plus. That's the famous $+2AB\cos \theta$. Then I just try every angle. Point them the same way and they stack to their full sum. Point them at right angles and it's plain Pythagoras. Point them opposite and they cancel down to their difference. The answer arrow can never be longer than the sum or shorter than the difference — it just dials smoothly between those two as I open the angle. That dial is the whole reason vectors aren't scalars.

Recall Quick self-test

Why does the correction term switch from $-$ (law of cosines) to $+$ (final formula)? ::: Because the triangle's interior angle is $180^{\circ }-\theta$, and $\cos (180^{\circ }-\theta )=-\cos \theta$ flips the sign. What angle gives the longest resultant? ::: $\theta =0^{\circ }$, giving $R=A+B$. What angle recovers plain Pythagoras? ::: $\theta =90^{\circ }$, since $\cos 90^{\circ }=0$. Two vectors of equal length pointing opposite — resultant? ::: Zero (the $|A-B|$ case with $A=B$). What is the relation between $R$ and $|\vec{R}|$? ::: They are the same number — the length of $\vec{R}$; the bars are optional once it's clear we mean length.

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Connections

• Vector Addition — Triangle & Parallelogram Law — the home of this result; the parallelogram view is the same triangle mirrored.
• Components of a Vector — the alternative route to the same $R$, splitting each arrow into perpendicular scalar parts.
• Distance vs Displacement — the $\theta =90^{\circ }$ special case in action.
• Speed vs Velocity — where the $\theta =180^{\circ }$ cancellation shows up on a round trip.
• Dot and Cross Products — the $\cos \theta$ here is exactly the dot product in disguise.
• Units and Dimensions — $A$, $B$, $R$ all carry the same unit through every step.