Noesis
1.1.9Measurement, Vectors & Kinematics

Resolution of vectors — into components (any axes)

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🇮🇳 Yeh note Hinglish mein padho →

WHY do we resolve vectors at all?

WHAT is a component? (rectangular case first)

HOW we derive this (from scratch)

Drop a perpendicular from the tip of A⃗\vec A to the x-axis. This makes a right triangle whose hypotenuse is AA.

  • The side adjacent to θ\theta lies along x. By definition of cosine: cos⁡θ=adjacenthyp=AxA\cos\theta = \dfrac{\text{adjacent}}{\text{hyp}} = \dfrac{A_x}{A} → Ax=Acos⁡θA_x = A\cos\theta. (Why? cosine literally measures the shadow of the hypotenuse along the adjacent direction.)
  • The side opposite to θ\theta lies along y: sin⁡θ=AyA\sin\theta = \dfrac{A_y}{A} → Ay=Asin⁡θA_y = A\sin\theta.

Recovering magnitude & angle (inverse problem): Pythagoras on the same triangle gives A=Ax2+Ay2,tan⁡θ=AyAx.A = \sqrt{A_x^2 + A_y^2}, \qquad \tan\theta = \frac{A_y}{A_x}.

Figure — Resolution of vectors — into components (any axes)

Resolving onto ANY axis (not just x–y)

WHY the dot product? Derive it: if n^\hat n makes angle α\alpha with x and A⃗\vec A makes θ\theta, then ϕ=θ−α\phi=\theta-\alpha. Expand: Acos⁡(θ−α)=Acos⁡θcos⁡α+Asin⁡θsin⁡α=Axcos⁡α+Aysin⁡α.A\cos(\theta-\alpha)=A\cos\theta\cos\alpha+A\sin\theta\sin\alpha = A_x\cos\alpha+A_y\sin\alpha. But n^=(cos⁡α,sin⁡α)\hat n=(\cos\alpha,\sin\alpha), so the right side is exactly A⃗⋅n^\vec A\cdot\hat n. The dot product IS the projection — proven, not assumed.

Oblique (non-perpendicular) axes

Worked Examples

Recall Feynman: explain to a 12-year-old

Imagine a kite string pulling at a slant. Part of that pull tugs you sideways and part tugs you up. Resolving the vector just means measuring "how much sideways?" and "how much up?" separately. The slanted pull is the same as those two separate pulls happening together. If you want to know the pull along a different direction — say toward a tree — you tilt your "shadow ruler" toward the tree and read how long the shadow is. That shadow is the component.

Flashcards

Rectangular components of A⃗\vec A at angle θ\theta to x-axis
Ax=Acos⁡θ, Ay=Asin⁡θA_x=A\cos\theta,\ A_y=A\sin\theta
Why does the adjacent side use cosine?
cos⁡θ=adjacent/hyp\cos\theta=\text{adjacent}/\text{hyp}, so adjacent =Acos⁡θ=A\cos\theta
Recover magnitude from components
A=Ax2+Ay2A=\sqrt{A_x^2+A_y^2}
Recover direction from components
θ=tan⁡−1(Ay/Ax)\theta=\tan^{-1}(A_y/A_x)
Component of A⃗\vec A along unit vector n^\hat n
A⃗⋅n^=Acos⁡ϕ\vec A\cdot\hat n = A\cos\phi (the projection)
Why is dot product the projection?
Expanding Acos⁡(θ−α)=Axcos⁡α+Aysin⁡α=A⃗⋅n^A\cos(\theta-\alpha)=A_x\cos\alpha+A_y\sin\alpha=\vec A\cdot\hat n
On an incline of angle θ\theta, weight along slope
Wsin⁡θW\sin\theta
On an incline of angle θ\theta, weight ⟂ to slope
Wcos⁡θW\cos\theta
When does projection equal the parallelogram component?
Only when the axes are mutually perpendicular
Why resolve vectors at all?
Components along a fixed axis add like ordinary numbers, turning geometry into arithmetic

Connections

Concept Map

motivates

swaps for

enables

right triangle

cosine adjacent

sine opposite

Pythagoras

Pythagoras

generalizes to

shadow equals

proven via

if axes not 90 deg

solve

Graphical vector addition clumsy

Resolve into components

Two-number picture

Add pieces as numbers

Rectangular components

Ax = A cos theta

Ay = A sin theta

Magnitude and angle recovered

Projection onto any axis n

Dot product A dot n

Expand cos theta minus alpha

Oblique resolution

Simultaneous equations for a and b

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, ek vector ek tirchi (slanted) arrow hoti hai. "Resolution" ka matlab hai us tirchi arrow ko do seedhe hisson mein todna — kitna part x-direction mein hai aur kitna y-direction mein. Iske liye hum right-angle triangle banate hain: jo side angle ko touch karti hai usme cos⁡\cos lagता hai (Ax=Acos⁡θA_x=A\cos\theta), aur jo side opposite hoti hai usme sin⁡\sin (Ay=Asin⁡θA_y=A\sin\theta). Yaad rakho: "Adjacent–Cos, Opposite–Sin".

Yeh kyun zaroori hai? Kyunki tedhe vectors ko jodna mushkil hai, par ek hi axis ke numbers ko aap normal numbers ki tarah plus-minus kar sakte ho. Isliye har force ko components mein todo, x-wale alag jodo, y-wale alag jodo, phir Pythagoras se wapas magnitude nikaalo: A=Ax2+Ay2A=\sqrt{A_x^2+A_y^2}.

Ab twist: axes hamesha x–y zaroori nahi. Inclined plane (dhalan) pe hum axis slope ke saath tilt karte hain, taaki weight ka ek part slope ke neeche khinche (Wsin⁡θW\sin\theta) aur doosra surface ke andar (Wcos⁡θW\cos\theta). Kisi bhi direction n^\hat n ke along component nikaalna ho to bas dot product lagao: A⃗⋅n^\vec A\cdot\hat n — yeh ussi direction ki "parchhai" (shadow) deta hai.

Ek warning: agar axes 90° pe nahi hain (oblique), to seedha projection mat lo! Tab equations banao A⃗=au^+bv^\vec A=a\hat u+b\hat v aur solve karo, kyunki projection sirf perpendicular axes pe hi component ke barabar hota hai. Yeh chhoti si baat exam mein bade marks bachati hai.

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Connections

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