Vector addition — triangle law, parallelogram law: Foundations

This page assumes nothing. The parent note ends with a scary-looking formula and a direction rule, full of letters and squiggles. Our whole job here is to earn every one of those symbols — the arrow-hat, the lengths, the plus sign, the angle, the sine/cosine/tangent, the components, the square root — one at a time, in an order where each new symbol only leans on ones already built. By the last section you'll be able to read the parent's formula out loud in plain words.

1. The arrow-hat: $A$

Figure 1.

Figure — Vector addition — triangle law, parallelogram law

Look at the blue arrow in Figure 1. Its tail is where it starts (the fat dot), its head is the pointy end. To fully describe it you must say both "how long" and "which way it points". A plain number like $7$ can't do that — $7$ what? $7$ steps where?

2. The plus sign for arrows: $A + B$

So the symbol " $+$ " changes meaning depending on what sits on either side of it:

between two plain numbers ( $3 + 4$ ) it means ordinary counting, giving $7$ ;
between two arrows ( $A + B$ ) it means "do one journey then the other," which gives the diagonal $5$ -arrow, not $7$ .

3. The playing field: the $x$ and $y$ axes

Figure 2.

Think of it as graph paper stretching in four directions (Figure 2). Every point in the plane now has an address $(x, y)$ : " $x$ steps across, $y$ steps up." Crucially, the two axes can be walked either way — so a component can be a negative number (pointing left or down). We will need that fact soon.

4. Magnitude: $A$ (same letter, no hat)

Picture: if $A$ is the blue arrow, then $A$ is the number you'd read off a ruler laid along it. Direction thrown away, length kept.

5. Tail-to-tail vs head-to-tail — the two ways to place two arrows

Figure 3.

Left picture of Figure 3 (head-to-tail): the orange arrow $B$ continues from where the blue arrow ended. That is literally "do journey $A$ , then journey $B$ ." The green closing arrow is the resultant.

Right picture of Figure 3 (tail-to-tail): both arrows start from the same dot, and we complete the full parallelogram — copy $B$ across to the head of $A$ , copy $A$ across to the head of $B$ (dashed sides). The green diagonal from the shared corner is the resultant. You can see the shaded wedge between the two originals; that wedge is the angle we name next.

6. The angle between: $θ$

Look again at the right picture of Figure 3: the shaded wedge at the common tail is $θ$ . We measure it in degrees (a full turn is $36 0^{\circ}$ , a square corner is $9 0^{\circ}$ ). Because we aligned the $x$ -axis along $A$ (Section 3), $θ$ is also the angle $B$ makes with the $x$ -axis.

7. Right triangle, opposite & adjacent — the shape hiding inside

Figure 4.

Why does this matter for adding arrows? Because when we drop a straight-down line from the head of a tilted arrow, we manufacture a right triangle (Figure 4). The two legs of that triangle are exactly the horizontal and vertical parts of the arrow — the components we'll meet next.

8. Sine, cosine, tangent — turning an angle into ratios

Picture it on Figure 4: for a hypotenuse of length $1$ , the shadow it casts on the floor is $cos θ$ , and its height is $sin θ$ . Tilt the arrow flat ( $θ = 0$ ): shadow is full length ( $cos 0 = 1$ ), height is zero ( $sin 0 = 0$ ). Stand it upright ( $θ = 9 0^{\circ}$ ): shadow vanishes ( $cos 9 0^{\circ} = 0$ ), height is full ( $sin 9 0^{\circ} = 1$ ). This is why the parent's sanity checks work.

9. Components of each arrow — and building $R_{x}, R_{y}$

Now we cash in Sections 3, 7 and 8. With the $x$ -axis laid along $A$ :

Figure 5 shows $B$ alone, tilted at $θ$ , with its two shadows drawn.

Figure 5.

Read the components straight off the definitions in Section 8 (hypotenuse $= B$ ):

$A$ lies along the $x$ -axis, so its shadows are $(A, 0)$ — all across, none up.
$B$ is tilted by $θ$ , so its horizontal shadow is $B cos θ$ (adjacent) and its vertical shadow is $B sin θ$ (opposite): $(B cos θ, B sin θ)$ .

Now add the arrows using the operator from Section 2. Because horizontal journeys never disturb vertical ones, we add the across-parts together and the up-parts together, separately:

R_y=\underbrace{0}_{\vec A\text{ up}}+\underbrace{B\sin\theta}_{\vec B\text{ up}}=B\sin\theta.$$ So the resultant's components are $$\boxed{R_x=A+B\cos\theta,\qquad R_y=B\sin\theta.}$$ This is the whole engine of the parent's Steps 1–2 — now *derived*, not just quoted. The little subscripts mean "the across part" ($x$) and "the up part" ($y$). Because of the axes (Section 3), each component **carries a sign**: - $R_x>0$ points right, $R_x<0$ points left, - $R_y>0$ points up, $R_y<0$ points down. The pair of signs $(R_x,R_y)$ tells you which of the four **quadrants** the resultant lives in: | $R_x$ | $R_y$ | quadrant | arrow points... | |:---:|:---:|:---:|:---| | $+$ | $+$ | I | up-right | | $-$ | $+$ | II | up-left | | $-$ | $-$ | III | down-left | | $+$ | $-$ | IV | down-right | **Figure 6.** ![[deepdives/dd-physics-1.1.08-d1-s06.png]] --- ## 10. Pythagoras: gluing components back into a length > [!formula] Pythagoras' theorem > In a right triangle, $\;\text{hypotenuse}^2 = \text{adjacent}^2 + \text{opposite}^2$. For components: > $$R=\sqrt{R_x^2+R_y^2}.$$ > [!intuition] Why we need it > After splitting the resultant into a sideways part $R_x$ and an up part $R_y$, those two parts form the two legs of a right triangle — and the resultant itself is the hypotenuse. Pythagoras is the tool that converts "$R_x$ across, $R_y$ up" back into a single length $R$. The little "$\sqrt{\phantom{x}}$" is the **square root** — it undoes squaring. We square each leg, add, then take the square root to get the length back out. Notice that **squaring erases signs** ($(-4)^2=16$), so $R$ comes out positive no matter which quadrant the arrow is in — exactly as a length should. --- ## 11. Putting the length formula together (the algebra shown in full) Substitute the boxed components from Section 9 into Pythagoras: $$R=\sqrt{(A+B\cos\theta)^2+(B\sin\theta)^2}.$$ **Expand the first bracket** — using $(p+q)^2=p^2+2pq+q^2$ with $p=A,\ q=B\cos\theta$: $$(A+B\cos\theta)^2 = A^2+2AB\cos\theta+B^2\cos^2\theta.$$ **Expand the second bracket:** $$(B\sin\theta)^2 = B^2\sin^2\theta.$$ **Add them:** $$R^2 = A^2+2AB\cos\theta+B^2\cos^2\theta+B^2\sin^2\theta.$$ **Group the two $B^2$ pieces** and factor out $B^2$: $$R^2 = A^2+2AB\cos\theta+B^2\big(\cos^2\theta+\sin^2\theta\big).$$ Now use the one identity every right triangle obeys, $\cos^2\theta+\sin^2\theta=1$ (it is just Pythagoras on a hypotenuse of length $1$): $$R^2 = A^2+B^2+2AB\cos\theta.$$ > [!formula] Magnitude of the resultant > $$\boxed{R=\sqrt{A^2+B^2+2AB\cos\theta}}$$ You can now *see* where the famous $2AB\cos\theta$ term is born: it is the cross-term $2\cdot A\cdot B\cos\theta$ from squaring the bracket $(A+B\cos\theta)$. It measures how much $\vec{B}$ points *along* $\vec{A}$ — big and positive when they align, negative when they oppose. --- ## 12. The direction angle $\alpha$ — and choosing the right quadrant > [!definition] Alpha > $\alpha$ (the Greek letter "alpha") is the angle the **resultant** $\vec{R}$ makes with the **$x$-axis**. Because we deliberately laid the $x$-axis along $\vec{A}$ (Section 3), this is *the very same* as "the angle $\vec{R}$ makes with $\vec{A}$" — which is how the parent page phrases it. The alignment is what makes the two descriptions identical. We find it with the steepness tool from Section 8: $$\tan\alpha=\frac{R_y}{R_x}\quad(\text{vertical part over horizontal part}).$$ > [!mistake] "$\tan\alpha=R_y/R_x$ alone gives the direction." > **Why it feels right:** it's the slope, and slope means direction. **The flaw:** $\tan$ gives the *same* number for an arrow and its exact opposite (e.g. up-right vs down-left) — so the bare ratio can't tell those two apart. **Fix:** read the **signs** of $R_x,R_y$ first to fix the quadrant (Section 9 table), *then* trust the tilt. If $R_x<0$, add $180^\circ$ to what the calculator hands back. > [!mistake] "The formula still works when $R_x=0$." > **Why it feels right:** we just plug numbers in. **The flaw:** if $R_x=0$ the resultant is **straight up or straight down**, and $R_y/R_x$ divides by zero — $\tan\alpha$ is *undefined*, no angle comes out. **Fix:** handle this vertical case by hand: if $R_x=0$ and $R_y>0$ then $\alpha=+90^\circ$ (straight up); if $R_x=0$ and $R_y<0$ then $\alpha=-90^\circ$ (straight down); if both are $0$ the resultant is the zero arrow and has no direction at all. **Figure 7.** ![[deepdives/dd-physics-1.1.08-d1-s07.png]] > [!intuition] Why this matters here > On the parent page $\vec{B}$ sits at a friendly angle so $R_x=A+B\cos\theta$ is positive and everything lands in quadrant I — no correction needed. But the moment you add arrows that point left or down (any real physics problem), you *must* fold in the sign rule and the $R_x=0$ case, or your beautiful $R$ will point the wrong way. --- ## 13. Reading the boxed formula — every symbol earned Now the parent's headline result reads in plain words: $$R=\sqrt{\underbrace{A^2}_{(\text{length of }\vec A)^2}+\underbrace{B^2}_{(\text{length of }\vec B)^2}+\underbrace{2AB\cos\theta}_{\text{overlap term}}}$$ - $R$ — the **length** of the resultant arrow (Section 10), - $A,B$ — lengths of the two arrows (Section 4), - $\cos\theta$ — how much $\vec B$ points *along* $\vec A$ (Section 8), with $\theta$ the tail-to-tail angle (Section 6), - the $2AB\cos\theta$ cross-term — born from expanding $(A+B\cos\theta)^2$ (Section 11), - the square and square-root — Pythagoras (Section 10). And its partner, the direction: $$\tan\alpha=\frac{R_y}{R_x},\qquad \alpha=\text{the tilt of }\vec R\text{ measured from }\vec A\ (\text{Section 12, mind the signs and the }R_x=0\text{ case}).$$ The $2AB\cos\theta$ term is the "do the arrows help or fight each other?" term: when they align ($\cos\theta=1$) it adds the most; when they oppose ($\cos\theta=-1$) it subtracts. You now own every piece. --- ## Prerequisite map ```mermaid graph TD HAT["Arrow-hat vector A"] --> PLUS["Plus operator for arrows"] HAT --> MAG["Magnitude A length only"] AXES["x and y axes with signs"] --> ALIGN["Align x-axis along A"] ALIGN --> COMP["Components Rx and Ry"] HAT --> PLACE["Head-to-tail and tail-to-tail"] PLACE --> THETA["Angle between theta"] THETA --> TRIG["Sine cosine tangent"] MAG --> COMP TRIG --> COMP PLUS --> COMP COMP --> PYTH["Pythagoras gives length"] COMP --> SIGN["Quadrant sign rule"] PYTH --> RES["Resultant magnitude R"] TRIG --> DIR["Direction angle alpha"] SIGN --> DIR RES --> TOPIC["Triangle and parallelogram laws"] DIR --> TOPIC ``` See how components sit at the crossroads: they need the magnitude idea, the trig ratios, the "$+$" operator *and* the signed, $\vec{A}$-aligned axes, and they feed *both* the length and the direction of the answer. This is why [[Resolution of a vector into components]] and [[Vectors — components and unit vectors]] are the natural next stops. --- ## Equipment checklist What does the hat in $\vec{A}$ tell you? ::: It is a vector — an arrow with both a length and a direction. What does the plain letter $A$ (no hat) mean? ::: Just the length (magnitude) of $\vec{A}$ — a non-negative number. What does "$+$" mean between two arrows $\vec{A}+\vec{B}$? ::: A new operation: lay $\vec{B}$'s tail on $\vec{A}$'s head and draw the arrow from start to finish — it keeps direction, so it is not "add the lengths." Why do we lay the $x$-axis along $\vec{A}$? ::: It's a free choice that makes $\vec{A}$ horizontal with components $(A,0)$, so "angle with $\vec{A}$" equals "angle with the $x$-axis." What do the $x$ and $y$ axes give you? ::: A shared agreement on "across" vs "up" and which side is positive — so components get signed numbers. How do you place two arrows for the triangle law? ::: Head-to-tail: tail of the second on the head of the first. How do you place them for the parallelogram law? ::: Tail-to-tail from a common point; complete the parallelogram; the resultant is the diagonal. What exactly is $\theta$ in every formula? ::: The angle between the two vectors, measured tail-to-tail. $\cos\theta$ equals which ratio in a right triangle? ::: adjacent ÷ hypotenuse. $\sin\theta$ equals which ratio? ::: opposite ÷ hypotenuse. $\tan\theta$ equals which ratio, and what does it measure? ::: opposite ÷ adjacent — the steepness (slope), used for direction. What are the components of $\vec{A}$ and $\vec{B}$ once the $x$-axis is along $\vec{A}$? ::: $\vec{A}=(A,0)$ and $\vec{B}=(B\cos\theta,\,B\sin\theta)$. How do $R_x$ and $R_y$ come from those components? ::: Add across-parts and up-parts separately: $R_x=A+B\cos\theta$, $R_y=B\sin\theta$. Where does the $2AB\cos\theta$ term come from? ::: The cross-term when you expand $(A+B\cos\theta)^2$ inside Pythagoras. Which tool re-assembles $R_x$ and $R_y$ into one length? ::: Pythagoras: $R=\sqrt{R_x^2+R_y^2}$. What is $\alpha$, and why isn't $\tan\alpha=R_y/R_x$ enough on its own? ::: The direction of the resultant measured from $\vec{A}$ (the $x$-axis); tan can't tell an arrow from its opposite, so fix the quadrant from the signs (add $180^\circ$ if $R_x<0$). What do you do when $R_x=0$? ::: $R_y/R_x$ is undefined; set $\alpha=+90^\circ$ if $R_y>0$, $\alpha=-90^\circ$ if $R_y<0$, and no direction if both are $0$. In $R=\sqrt{A^2+B^2+2AB\cos\theta}$, what is the $2AB\cos\theta$ term doing? ::: Measuring how much the two arrows help (align) or fight (oppose) each other. ## Connections - [[Vector addition — triangle law, parallelogram law]] (the parent — this page is its ground floor) - [[Resolution of a vector into components]] (Sections 7–9 made into a full method) - [[Vectors — components and unit vectors]] (the $\hat i,\hat j$ language for components) - [[Law of cosines]] (the magnitude formula in triangle-geometry clothing) - [[Scalar (dot) product]] (a compact machine for the $\cos\theta$ overlap term)