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1.1.8Measurement, Vectors & Kinematics

Vector addition — triangle law, parallelogram law

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🇮🇳 Yeh note Hinglish mein padho →

WHAT are we adding?

Two key facts that follow immediately from the "journey" idea:

  • Addition is commutative: A⃗+B⃗=B⃗+A⃗\vec{A}+\vec{B}=\vec{B}+\vec{A} (the parallelogram makes this obvious).
  • You can only add like quantities (force + force, velocity + velocity) — never a force to a velocity.

HOW: The Triangle Law

WHY head-to-tail? Because the tail of B⃗\vec{B} is where the first journey ended. Placing B⃗\vec{B} there literally means "continue from where you stopped." The closing side is therefore the net trip.

Figure — Vector addition — triangle law, parallelogram law

HOW: The Parallelogram Law

WHY it's the same thing: In the parallelogram, the opposite side equals B⃗\vec{B} (parallelogram → opposite sides equal & parallel). So the diagonal closes a triangle whose two sides are A⃗\vec{A} and B⃗\vec{B} head-to-tail. Parallelogram law = triangle law wearing a different costume.

DERIVING the magnitude & direction (from scratch)

Let the angle between A⃗\vec{A} and B⃗\vec{B} (tail-to-tail) be θ\theta. Place A⃗\vec{A} along the base; build the parallelogram.

Step 1 — Set up coordinates. Put the common tail OO at the origin, A⃗\vec{A} along the x-axis.

  • Tip of A⃗\vec{A}: (A,  0)(A,\;0).
  • B⃗\vec{B} makes angle θ\theta with A⃗\vec{A}, so its components are (Bcos⁡θ,  Bsin⁡θ)(B\cos\theta,\;B\sin\theta).

Why this step? Components let us add x and y independently, turning geometry into arithmetic.

Step 2 — Drop B⃗\vec{B} at the tip of A⃗\vec{A} (head-to-tail). The head of R⃗\vec{R} lands at: Rx=A+Bcos⁡θ,Ry=Bsin⁡θ.R_x = A + B\cos\theta,\qquad R_y = B\sin\theta.

Step 3 — Pythagoras (the head is horizontal RxR_x, vertical RyR_y from OO): R=Rx2+Ry2=(A+Bcos⁡θ)2+(Bsin⁡θ)2.R=\sqrt{R_x^2+R_y^2}=\sqrt{(A+B\cos\theta)^2+(B\sin\theta)^2}.

Step 4 — Expand and simplify. R2=A2+2ABcos⁡θ+B2cos⁡2θ+B2sin⁡2θ.R^2=A^2+2AB\cos\theta+B^2\cos^2\theta+B^2\sin^2\theta. Since cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1:

Step 5 — Direction. The resultant makes angle α\alpha with A⃗\vec{A}: tan⁡α=RyRx=Bsin⁡θA+Bcos⁡θ\boxed{\tan\alpha=\frac{R_y}{R_x}=\frac{B\sin\theta}{A+B\cos\theta}}

Why this step? tan⁡α=oppositeadjacent\tan\alpha = \dfrac{\text{opposite}}{\text{adjacent}} for the right triangle formed by the perpendicular from the tip of B⃗\vec{B}.

Worked Examples

Common Mistakes (Steel-manned)

Flashcards

Triangle law statement
Two vectors as two sides of a triangle in the same order (head-to-tail); resultant is the third side in the opposite order.
Parallelogram law statement
Two vectors as adjacent sides from a common point; resultant is the diagonal through that point.
Magnitude of resultant of A⃗,B⃗\vec A,\vec B at angle θ\theta
R=A2+B2+2ABcos⁡θR=\sqrt{A^2+B^2+2AB\cos\theta}
Direction of resultant (angle with A⃗\vec A)
tan⁡α=Bsin⁡θA+Bcos⁡θ\tan\alpha=\dfrac{B\sin\theta}{A+B\cos\theta}
Maximum possible resultant
A+BA+B, when θ=0∘\theta=0^\circ (parallel).
Minimum possible resultant
∣A−B∣|A-B|, when θ=180∘\theta=180^\circ (anti-parallel).
Resultant when θ=90∘\theta=90^\circ
A2+B2\sqrt{A^2+B^2}.
What is θ\theta in the formula?
The angle between the two vectors measured tail-to-tail.
Why are triangle & parallelogram laws equivalent?
Opposite sides of a parallelogram are equal & parallel, so the diagonal closes a head-to-tail triangle of the same two vectors.
For two equal vectors, where does the resultant point?
Along the angle bisector (by symmetry).
Recall Feynman: explain to a 12-year-old

Imagine you walk 3 steps east, then 4 steps north. You don't end up 7 steps from home — you end up 5 steps away, on a diagonal! Adding "arrows" (vectors) means walking one after another and asking "how far and which way is home from where I started?" The triangle law just draws those two walks as two sides of a triangle, and the way home is the third side. The parallelogram law is the same trick using a slanted box instead.

Connections

  • Vectors — components and unit vectors (the i,j method generalizes Step 1–2 above)
  • Subtraction of vectors and the difference vector (A⃗−B⃗\vec A-\vec B uses θ→180∘−θ\theta\to180^\circ-\theta trick)
  • Resolution of a vector into components
  • Relative velocity (real-world vector addition: boat + river)
  • Law of cosines (the magnitude formula is the cosine rule in disguise)
  • Scalar (dot) product (gives cos⁡θ\cos\theta that lives inside the cross term)

Concept Map

leads to

is

requires

method 1

method 2

equivalent to

set coordinates

Rx = A + B cos theta

Ry = B sin theta

Pythagoras

Pythagoras

simplify with identity

Vector as a journey

Vector addition = resultant

Commutative A+B=B+A

Only like quantities

Triangle law head-to-tail

Parallelogram law tail-to-tail

Components Rx and Ry

Horizontal part

Vertical part

Magnitude R

R = sqrt of A2 + B2 + 2AB cos theta

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, vector add karna matlab "do journeys ko jodna". Maan lo tum pehle 3 step east chale, phir 4 step north — tum 7 step door nahi pahunchte, balki diagonal me sirf 5 step door! Yahi vector addition ka asli funda hai. Triangle law bolta hai: pehle vector ke head par doosre vector ka tail rakho (head-to-tail), aur start se end tak jo arrow banta hai wahi resultant hai. Parallelogram law bhi yahi cheez hai, bas dono vectors ko ek hi point se (tail-to-tail) shuru karke ek tilted box banate ho, aur uska diagonal resultant hota hai.

Magnitude nikalne ka formula R=A2+B2+2ABcos⁡θR=\sqrt{A^2+B^2+2AB\cos\theta} aasmaan se nahi aaya — humne bas components liye (AA ko x-axis pe, BB ko Bcos⁡θ,Bsin⁡θB\cos\theta, B\sin\theta), phir Pythagoras lagaya, aur cos⁡2+sin⁡2=1\cos^2+\sin^2=1 use karke simplify kiya. Bas! Yaad rakho yahan θ\theta hamesha dono vectors ke beech ka angle hai, tail-to-tail wala — naa ki horizontal se.

Direction ke liye tan⁡α=Bsin⁡θA+Bcos⁡θ\tan\alpha=\dfrac{B\sin\theta}{A+B\cos\theta}. Quick checks: same direction (θ=0\theta=0) pe R=A+BR=A+B (maximum), opposite (θ=180\theta=180) pe R=∣A−B∣R=|A-B| (minimum), aur 90∘90^\circ pe simple A2+B2\sqrt{A^2+B^2}. Isliye resultant hamesha ∣A−B∣|A-B| se A+BA+B ke beech rehta hai.

Sabse common galti: 33 aur 44 ko seedha jod ke 77 likh dena. Yeh tabhi sahi hai jab vectors parallel ho. Aur ek aur trap — triangle law me galti se tip-to-tip jod dena (woh subtraction de deta hai). Toh mantra yaad rakho: "Head-to-Tail to find the trail; Tail-to-Tail, the Diagonal prevails." Exam me yeh formula aur ye checks bahut kaam aayenge!

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Connections

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