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

False. Dot product overlap measure karne ke liye bana hai (ek single amount of "same-direction-ness"), isliye iska output ek scalar hai. Cross product woh operation hai jo vector return karta hai.

False. Iska matlab hai woh perpendicular hain. Do nonzero vectors orthogonal ho sakte hain, jaise $(1,0)\cdot (0,1)=0$ — dekho Orthogonality and orthonormal bases.

False. $\mathbf{a}\cdot \mathbf{a}=a_1^2+\cdots +a_n^2$, squares ka sum hai, isliye yeh $\ge 0$ hai aur $0$ sirf zero vector ke liye hota hai. Yahi positivity woh reason hai ki length $\Vert \mathbf{a}\Vert =\sqrt{\mathbf{a}\cdot \mathbf{a}}$ meaningful hai.

True. Har term $a_i{b}_i=b_i{a}_i$ hai kyunki ordinary number multiplication commute karta hai, isliye poora sum unchanged rehta hai.

False. Discriminant proof sirf $\Vert \mathbf{a}-t\mathbf{b}{\Vert }^2\ge 0$ use karta hai, isliye yeh kisi bhi inner product space mein valid hai, chahe wo infinite-dimensional ho jahan angle ki koi picture na ho.

True. Equality se $\cos \theta =+1$ force hota hai, yani $\theta =0^{\circ }$, isliye $\mathbf{b}$ ek positive scalar multiple hai $\mathbf{a}$ ka. (Yaad rakhो $\theta$ upar box mein cosine formula se define hota hai.)

True. Absolute value sign ignore karta hai; opposite directions mein $\cos \theta =-1$ hota hai, isliye $|\cos \theta |=1$ aur equality phir bhi hold karta hai. Same aur opposite dono directions "parallel" equality case hain.

False. Yeh sirf kehta hai $(\mathbf{a}-\mathbf{b})\cdot \mathbf{c}=0$, yani $\mathbf{a}-\mathbf{b}$, $\mathbf{c}$ ke perpendicular hai. Sabhi $\mathbf{c}$ ke liye equality $\mathbf{a}=\mathbf{b}$ force karegi.

True. $(2\mathbf{a})\cdot (2\mathbf{b})=4(\mathbf{a}\cdot \mathbf{b})$ aur dono lengths double ho jaati hain, isliye $\cos \theta =\frac{4(\mathbf{a}\cdot \mathbf{b})}{2\Vert \mathbf{a}\Vert \cdot 2\Vert \mathbf{b}\Vert }$ unchanged rehta hai. Scaling length badalta hai, direction nahin.

Yeh reasoning quietly assume karti hai ki $\mathbf{a}$ aur $\mathbf{b}$ ke beech ek angle $\theta$ already exist karta hai. Abstract spaces mein angle sirf Cauchy–Schwarz se define hota hai, isliye yeh circular hai — pehle discriminant proof chahiye.

Middle term drop ho gaya. Correctly $(\mathbf{a}-\mathbf{b})\cdot (\mathbf{a}-\mathbf{b})=\Vert \mathbf{a}{\Vert }^2-2\mathbf{a}\cdot \mathbf{b}+\Vert \mathbf{b}{\Vert }^2$; yeh $-2\mathbf{a}\cdot \mathbf{b}$ bilkul wahi term hai jo angle information carry karti hai.

Dono expressions defined bhi nahin hain. $\mathbf{a}\cdot \mathbf{b}$ ek scalar hai, aur ek scalar ko vector se dot karna meaningless hai. Dot product do vectors leta hai, isliye usse associativity puch hi nahin sakte.

Galat chain. ${\mathbb{R}}^2$ mein, $(1,0)\perp (0,1)$ aur $(0,1)\perp (1,0)$, phir bhi $(1,0)\cdot (1,0)=1\ne 0$. Perpendicularity transitive nahin hai.

Expand karne par $f(t)=\Vert \mathbf{b}{\Vert }^2{t}^2-2(\mathbf{a}\cdot \mathbf{b})t+\Vert \mathbf{a}{\Vert }^2$ milta hai, jiska leading coefficient $\Vert \mathbf{b}{\Vert }^2$ hai. Agar $\mathbf{b}=0$ ho to yeh leading coefficient $0$ hai, isliye $f$ quadratic nahin hai aur discriminant step invalid hai. Lekin tab C–S ke dono sides $0$ hain, isliye inequality trivially hold karta hai — $\mathbf{b}=0$ case alag handle karna padta hai.

Formula har angle $0^{\circ }\le \theta \le 180^{\circ }$ ke liye hold karta hai. Obtuse angles mein $\cos \theta <0$ hota hai, jo negative dot product se match karta hai, aur $\theta =90^{\circ }$ exactly $0$ deta hai.

Yeh $\sum a_i^2$ deta hai, squared components ka sum, jo exactly $\Vert \mathbf{a}{\Vert }^2$ hai. Isliye dot product ko squared distance contain karne ke liye design kiya gaya tha.

Negative value ka matlab hai $\cos \theta <0$, yani vectors $90^{\circ }$ se zyada apart point karte hain — woh partly oppose karte hain ek doosre ko. Sign encode karta hai "same-ish direction" (+) versus "opposing" (−).

Squared length $t$ mein ek polynomial hai (koi square roots nahin) aur guaranteed $\ge 0$ hai. Ek quadratic ki yahi non-negativity poora engine hai jo discriminant condition force karti hai.

Discriminant $=0$ correspond karta hai $f(t_0)=0$ kisi $t_0$ ke liye, matlab $\mathbf{a}=t_0\mathbf{b}$ (parallel) — exactly equality case. "$\le$" us boundary situation ko theorem ke andar rakhta hai.

Perpendicular ka matlab $\cos \theta =0$ hai, aur kyunki $\mathbf{a}\cdot \mathbf{b}=\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert \cos \theta$, product exactly tab $0$ hota hai jab woh orthogonal hote hain (nonzero vectors ke liye). Ek multiply-and-add answer de deta hai — dekho Orthogonality and orthonormal bases.

Teen vectors $\mathbf{a},\mathbf{b},\mathbf{a}-\mathbf{b}$ ek triangle banate hain jahan $\theta$ pehle do ke beech hai aur $\mathbf{a}-\mathbf{b}$ $\theta$ ke opposite hai. Law of Cosines exactly woh tool hai jo triangle ki teesri side ko do sides aur included angle se relate karta hai.

$\Vert \mathbf{a}+\mathbf{b}{\Vert }^2=\Vert \mathbf{a}{\Vert }^2+2\mathbf{a}\cdot \mathbf{b}+\Vert \mathbf{b}{\Vert }^2$ expand karke aur $\mathbf{a}\cdot \mathbf{b}\le \Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert$ bound karke (jo C–S hai) $\Vert \mathbf{a}+\mathbf{b}\Vert \le \Vert \mathbf{a}\Vert +\Vert \mathbf{b}\Vert$ milta hai.

Yeh $0$ hai, jisse $\Vert 0\Vert =0$ milta hai. Angle undefined hai kyunki $\cos \theta =\frac{0}{0\cdot \Vert \mathbf{b}\Vert }$ zero se divide karta hai — zero vector ki koi direction nahin hoti.

Haan, trivially. $|\mathbf{a}\cdot 0|=0$ aur $\Vert \mathbf{a}\Vert \Vert 0\Vert =0$ dono hain, isliye yeh $0\le 0$ padha jaata hai — ek equality, jo consistent hai $0$ ke kisi bhi vector ka (trivial) scalar multiple hone se.

$\cos \theta =\frac{\Vert \mathbf{a}{\Vert }^2}{\Vert \mathbf{a}{\Vert }^2}=1$, isliye $\theta =0^{\circ }$. Identical vectors same direction mein point karte hain, maximum-overlap case.

$\cos \theta =-1$ isliye $\theta =180^{\circ }$, aur $|\mathbf{a}\cdot \mathbf{b}|=\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert$ — equality case, kyunki $\mathbf{b}$ ek scalar multiple hai $\mathbf{a}$ ka.

$\mathbf{b}$ ke matching slots mein itna negative "pull" hona chahiye ki weighted overlap net negative ho — vectors $90^{\circ }$ se zyada apart hain, isliye $\mathbf{b}$ broadly $\mathbf{a}$ ke against point karta hai.

Yeh ordinary multiplication $a_1{b}_1$ ban jaata hai, aur C–S padhta hai $|a_1{b}_1|\le |a_1||b_1|$, jo sirf $|xy|=|x||y|$ hai — hamesha equality, kyunki koi bhi do 1-D vectors parallel hote hain.