Dot product — formula, cosine formula, Cauchy-Schwarz inequality proof

4.5.2 · Maths › Linear Algebra (Full)

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1. Algebraic definition

WHAT hai yeh actually? Matching components ko multiply karo, unhe add karo.

WHY yahi aur kuch nahi? Kyunki yeh hume length free mein deta hai. $\mathbf{b}=\mathbf{a}$ rakh do: $\mathbf{a}\cdot \mathbf{a}=a_1^2+a_2^2+\cdots +a_n^2=\Vert \mathbf{a}{\Vert }^2$ Yeh toh bas Pythagorean theorem hai! Toh dot product distance ka notion contain karta hai. Yahi woh design goal hai jiske liye ise banaya gaya tha.

Basic properties (sab sum se prove ho sakte hain)

• Commutative: $\mathbf{a}\cdot \mathbf{b}=\mathbf{b}\cdot \mathbf{a}$ (kyunki $a_i{b}_i=b_i{a}_i$).
• Distributive: $\mathbf{a}\cdot (\mathbf{b}+\mathbf{c})=\mathbf{a}\cdot \mathbf{b}+\mathbf{a}\cdot \mathbf{c}$.
• Scalar pull-out: $(\lambda \mathbf{a})\cdot \mathbf{b}=\lambda (\mathbf{a}\cdot \mathbf{b})$.
• Positive: $\mathbf{a}\cdot \mathbf{a}\ge 0$, aur $=0$ tabhi jab $\mathbf{a}=0$.

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2. Cosine (geometric) formula — derive karo, memorize mat karo

HOW milta hai yeh? $\mathbf{a}$, $\mathbf{b}$ aur side $\mathbf{a}-\mathbf{b}$ se bane triangle par Law of Cosines use karo.

Us triangle par law of cosines: $\Vert \mathbf{a}-\mathbf{b}{\Vert }^2=\Vert \mathbf{a}{\Vert }^2+\Vert \mathbf{b}{\Vert }^2-2\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert \cos \theta$ Yeh step kyun? Vector $\mathbf{a}-\mathbf{b}$ woh side hai jo angle $\theta$ ke opposite hai.

Ab left side ko algebraically dot product use karke expand karo: $\Vert \mathbf{a}-\mathbf{b}{\Vert }^2=(\mathbf{a}-\mathbf{b})\cdot (\mathbf{a}-\mathbf{b})=\mathbf{a}\cdot \mathbf{a}-2\mathbf{a}\cdot \mathbf{b}+\mathbf{b}\cdot \mathbf{b}$ $=\Vert \mathbf{a}{\Vert }^2-2\mathbf{a}\cdot \mathbf{b}+\Vert \mathbf{b}{\Vert }^2$ Yeh step kyun? Hum distributivity aur $\mathbf{a}\cdot \mathbf{a}=\Vert \mathbf{a}{\Vert }^2$ use karte hain.

$\Vert \mathbf{a}-\mathbf{b}{\Vert }^2$ ke dono expressions ko equal set karo. $\Vert \mathbf{a}{\Vert }^2$ aur $\Vert \mathbf{b}{\Vert }^2$ cancel ho jaate hain: $-2\mathbf{a}\cdot \mathbf{b}=-2\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert \cos \theta ⟹\boxed{\mathbf{a}\cdot \mathbf{b}=\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert \cos \theta }$

Angle extract karna: $\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert },\mathbf{a}\perp \mathbf{b}⟺\mathbf{a}\cdot \mathbf{b}=0$

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3. Cauchy–Schwarz inequality

WHY yeh geometrically "obviously" true hai? Kyunki $\mathbf{a}\cdot \mathbf{b}=\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert \cos \theta$ aur $|\cos \theta |\le 1$. Ho gaya!

Lekin woh geometric "proof" secretly assume karta hai ki humara pehle se ek angle $\theta$ hai — jo ${\mathbb{R}}^2,{\mathbb{R}}^3$ mein theek hai lekin abstract spaces mein obvious nahi. Toh hum ise purely algebraically prove karte hain, jo phir angle define karna justify karta hai.

Scratch se derivation (discriminant trick)

Kisi bhi real $t$ ke liye real function consider karo: $f(t)=\Vert \mathbf{a}-t\mathbf{b}{\Vert }^2\ge 0$ Yahaan se kyun shuru karein? Ek squared length kabhi negative nahi hoti — woh ek fact hi poora engine hai.

Ise expand karo: $f(t)=(\mathbf{a}-t\mathbf{b})\cdot (\mathbf{a}-t\mathbf{b})=\Vert \mathbf{a}{\Vert }^2-2t(\mathbf{a}\cdot \mathbf{b})+t^2\Vert \mathbf{b}{\Vert }^2$

Yeh $t$ mein ek quadratic hai $A{t}^2+Bt+C$ form ka jahan $A=\Vert \mathbf{b}{\Vert }^2,B=-2(\mathbf{a}\cdot \mathbf{b}),C=\Vert \mathbf{a}{\Vert }^2.$

Kyunki $f(t)\ge 0$ har $t$ ke liye hai, parabola kabhi axis ke neeche cross nahi karta, toh iska discriminant $B^2-4AC\le 0$ satisfy karna chahiye: $4(\mathbf{a}\cdot \mathbf{b}{)}^2-4\Vert \mathbf{b}{\Vert }^2\Vert \mathbf{a}{\Vert }^2\le 0$ Yeh step kyun? Ek non-negative upward parabola mein at most ek real root hota hai → discriminant $\le 0$.

4 se divide karo aur rearrange karo: $(\mathbf{a}\cdot \mathbf{b}{)}^2\le \Vert \mathbf{a}{\Vert }^2\Vert \mathbf{b}{\Vert }^2$ Square roots lo (dono sides $\ge 0$): $|\mathbf{a}\cdot \mathbf{b}|\le \Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert ■$

Equality case: $B^2-4AC=0$ matlab $f(t)=0$ kisi $t=t_0$ ke liye, yaani $\Vert \mathbf{a}-t_0\mathbf{b}\Vert =0\Rightarrow \mathbf{a}=t_0\mathbf{b}$. Toh equality $⟺$ vectors parallel hain. ✔

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4. Worked examples

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5. Common mistakes (steel-manned)

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6. Flashcards

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Recall Feynman: ek 12-saal ke bachche ko explain karo

Socho do arrows ek hi jagah se shuru ho rahe hain. Dot product ek "teamwork score" hai. Agar dono arrows same direction mein point kar rahe hain, toh woh ek great team hain → bada positive score. Agar ek left point kare aur doosra right, toh woh ladte hain → negative score. Agar woh perfect right angle par hain, toh woh ek doosre ko ignore karte hain → score exactly zero. Score paane ke liye bas matching parts multiply karo aur unhe add karo. Aur ek rule hai (Cauchy–Schwarz) jo kehta hai teamwork score kabhi do arrows ki lengths ke product se zyada nahi ho sakta — jitna arrows mein actually hai, usse zyada nahi nikal sakta.

Concept Map

Deep Dive

• D1 · Foundations — har symbol zero se
• D2 · Visual walkthrough — derivation pictures mein
• D3 · Worked examples — har scenario
• D4 · Exercises — graded, full solutions
• D5 · Question bank — concept traps