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

KYA karte hain: matching slots multiply karo, add karo. $\mathbf{a}\cdot \mathbf{b}=(3)(4)+(-1)(2)+(2)(-1)=12-2-2=\boxed{8}.$ KYUN: component form ka matlab hi hai "vectors ko line up karo aur products ka sum karo." Abhi kuch geometric nahi chahiye.

KYUN yeh tool: length aati hai vector ko khud se dot karne se — yahi dot product ke andar chupi hui Pythagorean theorem hai. $\Vert \mathbf{v}\Vert =\sqrt{\mathbf{v}\cdot \mathbf{v}}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=\boxed{10}.$

KYUN: do vectors perpendicular hote hain bilkul tabhi jab unka dot product $0$ ho (kyunki $\cos 90^{\circ }=0$). $\mathbf{a}\cdot \mathbf{b}=(2)(3)+(-3)(2)=6-6=0.$ Dot product zero hai → $\boxed{\text{yes, perpendicular}}$. General idea ke liye Orthogonality and orthonormal bases dekho.

KYUN cosine formula: hume ek angle chahiye, aur components se angle tak ka ek hi bridge hai $\mathbf{a}\cdot \mathbf{b}=\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert \cos \theta$. Isse $\cos \theta$ ke liye solve karo.

• $\mathbf{a}\cdot \mathbf{b}=(1)(1)+(\sqrt{3})(0)=1.$
• $\Vert \mathbf{a}\Vert =\sqrt{1+3}=2,\Vert \mathbf{b}\Vert =1.$ $\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert }=\frac{1}{2\cdot 1}=\frac{1}{2}\Rightarrow \theta =\boxed{60^{\circ }}.$ KAISA dikhta hai: figure mein, yellow arrow $\mathbf{b}$ $x$-axis pe flat leta hai aur blue arrow $\mathbf{a}$ steeply upar uthta hai; unke beech ka red wedge wahi $60^{\circ }$ hai jo humne solve kiya. Dhyan do ki arrow lengths ($\Vert \mathbf{b}\Vert =1$, $\Vert \mathbf{a}\Vert =2$) angle ko nahi badaltein — sirf unki directions badaltein hain, isliye length factors $\cos \theta$ se cancel ho gaye.

KYUN phir se cosine formula: humse $\cos \theta$ maanga ja raha hai, toh hum wahi bridge $\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert }$ chalate hain — ab 3D mein, jahan hum angle easily draw nahi kar sakte, toh algebra hi iska ek maatra handle hai.

• Numerator (overlap): $\mathbf{a}\cdot \mathbf{b}=(2)(0)+(1)(3)+(-2)(4)=0+3-8=-5.$ Kyun: matching slots multiply karo aur add karo — wahi machine jaise L1.1 mein.
• Denominator (lengths): $\Vert \mathbf{a}\Vert =\sqrt{4+1+4}=3,\Vert \mathbf{b}\Vert =\sqrt{0+9+16}=5.$ Kyun: har length $\sqrt{\text{vector}\cdot \text{itself}}$ hai, 3D mein Pythagorean idea. $\cos \theta =\frac{-5}{3\cdot 5}=\frac{-5}{15}=-\frac{1}{3}.$ KYUN sign matter karta hai: numerator sign carry karta hai, aur negative cosine ka matlab hai $\theta >90^{\circ }$ — vectors ek doosre se door jhuk rahe hain (obtuse angle). $\boxed{\cos \theta =-\frac{1}{3}}$ (toh $\theta \approx 109.47^{\circ }$). Sanity check: $\left|-\frac{1}{3}\right|\le 1$, jaise koi bhi genuine cosine hona chahiye — yeh reassurance hai ki humne lengths se theek se divide kiya.

PROJECTION kya hota hai: wo shadow jo $\mathbf{a}$ $\mathbf{b}$ ki direction mein daalta hai. Uski signed length $\frac{\mathbf{a}\cdot \mathbf{b}}{\Vert \mathbf{b}\Vert }$ hai — Vector projection dekho. KYUN woh formula: $\mathbf{a}\cdot \mathbf{b}=\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert \cos \theta$, aur $\Vert \mathbf{a}\Vert \cos \theta$ bilkul adjacent side (shadow) hai. $\Vert \mathbf{b}\Vert$ se divide karne se $\mathbf{b}$ ki length strip ho jaati hai toh sirf shadow bachta hai. Pehle edge case — KYUN $\Vert \mathbf{b}\Vert \ne 0$ zaroori hai: formula $\Vert \mathbf{b}\Vert$ se divide karta hai, toh yeh sirf tab defined hai jab $\mathbf{b}\ne 0$. Zero vector pe project karna meaningless hai — zero vector ki koi direction nahi jis par shadow daala jaaye, aur formula $0$ se division maangega. Project karne se pehle hamesha confirm karo $\mathbf{b}\ne 0$. Yahan $\Vert \mathbf{b}\Vert =1\ne 0$ hai, toh hum safe hain. $\frac{\mathbf{a}\cdot \mathbf{b}}{\Vert \mathbf{b}\Vert }=\frac{(3)(1)+(4)(0)}{1}=\boxed{3}.$ FIGURE mein kya dekhna hai: blue arrow $\mathbf{a}=(3,4)$ hai; $\mathbf{b}$ ki direction ($x$-axis) pe seedha ek white dashed vertical line neeche girао. Jahan woh land kare (red dot at $x=3$) woh length $3$ ke green shadow ki tip mark karta hai. Shadow $\mathbf{a}$ se chhota hai ($\Vert \mathbf{a}\Vert =5$) kyunki $\mathbf{a}$ ka kuch hissa upar point karta hai, $\mathbf{b}$ se door — projection sirf woh part rakhta hai jo $\mathbf{b}$ ke saath hai.

KYUN subtract karein: equation kehti hai $\mathbf{a}$ aur $\mathbf{b}$ ka har cheez ke saath identical overlap hai. Distributivity se rewrite karo: $\mathbf{a}\cdot \mathbf{c}-\mathbf{b}\cdot \mathbf{c}=0\Rightarrow (\mathbf{a}-\mathbf{b})\cdot \mathbf{c}=0\text{for all }\mathbf{c}.$ Key move: "for all $\mathbf{c}$" mein $\mathbf{c}=\mathbf{a}-\mathbf{b}$ ka choice bhi shamil hai. Plug karo: $(\mathbf{a}-\mathbf{b})\cdot (\mathbf{a}-\mathbf{b})=\Vert \mathbf{a}-\mathbf{b}{\Vert }^2=0.$ Ek squared length sirf zero vector ke liye zero hoti hai, toh $\mathbf{a}-\mathbf{b}=0$, yaani $\boxed{\mathbf{a}=\mathbf{b}}$. $■$

KYUN dot product se expand karein: har length-squared ek dot product hai, aur dot products distribute karte hain, toh hum bas multiply out kar sakte hain. $\Vert \mathbf{a}+\mathbf{b}{\Vert }^2=(\mathbf{a}+\mathbf{b})\cdot (\mathbf{a}+\mathbf{b})=\Vert \mathbf{a}{\Vert }^2+2\mathbf{a}\cdot \mathbf{b}+\Vert \mathbf{b}{\Vert }^2.$ $\Vert \mathbf{a}-\mathbf{b}{\Vert }^2=(\mathbf{a}-\mathbf{b})\cdot (\mathbf{a}-\mathbf{b})=\Vert \mathbf{a}{\Vert }^2-2\mathbf{a}\cdot \mathbf{b}+\Vert \mathbf{b}{\Vert }^2.$ Add karne par kya hota hai: $\pm 2\mathbf{a}\cdot \mathbf{b}$ terms cancel ho jaate hain: $\Vert \mathbf{a}+\mathbf{b}{\Vert }^2+\Vert \mathbf{a}-\mathbf{b}{\Vert }^2=2\Vert \mathbf{a}{\Vert }^2+2\Vert \mathbf{b}{\Vert }^2.■$ Picture: ek parallelogram ke do diagonals uski chaar sides se relate karte hain — yeh usi geometry ka algebraic version hai.

KYUN yeh identity: $\Vert \mathbf{a}-\mathbf{b}{\Vert }^2$ un cheezein mein expand hoti hai jo hum pehle se jaante hain. $\Vert \mathbf{a}-\mathbf{b}{\Vert }^2=\Vert \mathbf{a}{\Vert }^2-2\mathbf{a}\cdot \mathbf{b}+\Vert \mathbf{b}{\Vert }^2=25-12+9=22.$ $\Vert \mathbf{a}-\mathbf{b}\Vert =\sqrt{22}\approx \boxed{4.69}.$ (Yeh bilkul Law of cosines disguise mein hai — angle ke opposite ki side.)

KYUN carefully expand karein: scalars $2$ aur $-3$ dot product se bahar aa jaate hain, toh unhe track karo. $\Vert 2\mathbf{a}-3\mathbf{b}{\Vert }^2=(2\mathbf{a}-3\mathbf{b})\cdot (2\mathbf{a}-3\mathbf{b})=4\Vert \mathbf{a}{\Vert }^2-12\mathbf{a}\cdot \mathbf{b}+9\Vert \mathbf{b}{\Vert }^2.$ $\Vert \mathbf{a}{\Vert }^2=1,\Vert \mathbf{b}{\Vert }^2=1,\mathbf{a}\cdot \mathbf{b}=\frac{1}{2}$ substitute karo: $=4(1)-12\left(\frac{1}{2}\right)+9(1)=4-6+9=7.$ $\Vert 2\mathbf{a}-3\mathbf{b}\Vert =\sqrt{7}\approx \boxed{2.65}.$

KYUN dot product zero set karein: perpendicular $⟺$ $\mathbf{a}\cdot \mathbf{b}=0$. $\mathbf{a}\cdot \mathbf{b}=3k+2k+(-4)=5k-4=0\Rightarrow k=\frac{4}{5}=\boxed{0.8}.$ Ek linear equation → ek value of $k$. (Agar components ne ek quadratic produce ki hoti, toh hum har root rakhte.)

Square se shuru karo (lengths non-negative hain, toh squaring reversible hai): $\Vert \mathbf{a}+\mathbf{b}{\Vert }^2=\Vert \mathbf{a}{\Vert }^2+2\mathbf{a}\cdot \mathbf{b}+\Vert \mathbf{b}{\Vert }^2.$ KYUN Cauchy–Schwarz aata hai: middle term par ek upper bound chahiye. Cauchy–Schwarz deta hai $\mathbf{a}\cdot \mathbf{b}\le |\mathbf{a}\cdot \mathbf{b}|\le \Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert$. Toh: $\Vert \mathbf{a}+\mathbf{b}{\Vert }^2\le \Vert \mathbf{a}{\Vert }^2+2\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert +\Vert \mathbf{b}{\Vert }^2=(\Vert \mathbf{a}\Vert +\Vert \mathbf{b}\Vert {)}^2.$ Dono sides non-negative hain → square roots lo: $\Vert \mathbf{a}+\mathbf{b}\Vert \le \Vert \mathbf{a}\Vert +\Vert \mathbf{b}\Vert .■$ Yeh "sabse chhota raasta seedha line hota hai" ka vector version hai — Triangle inequality dekho.

KYUN koi angle nahi: ek abstract inner product space mein abhi angle ki koi picture nahi hoti — toh hume sirf algebra se argue karna hoga. Ek hi engine hai: ek squared length kabhi negative nahi hoti.

Step 1 — ek non-negative machine banao. Kisi bhi real number $t$ ke liye, $f(t)=\Vert \mathbf{a}-t\mathbf{b}{\Vert }^2\ge 0.$ Step 2 — $t$ mein ek quadratic mein expand karo: $f(t)=\Vert \mathbf{a}{\Vert }^2-2t(\mathbf{a}\cdot \mathbf{b})+t^2\Vert \mathbf{b}{\Vert }^2=A{t}^2+Bt+C,$ $A=\Vert \mathbf{b}{\Vert }^2,B=-2(\mathbf{a}\cdot \mathbf{b}),C=\Vert \mathbf{a}{\Vert }^2.$ Step 3 — discriminant $\le 0$ force karo. Ek upward parabola ($A\ge 0$) jo kabhi axis ke neeche nahi jaati, usme zyaada se zyaada ek real root ho sakta hai: $B^2-4AC\le 0\Rightarrow 4(\mathbf{a}\cdot \mathbf{b}{)}^2-4\Vert \mathbf{b}{\Vert }^2\Vert \mathbf{a}{\Vert }^2\le 0.$ Step 4 — tidy up karo. $4$ se divide karo, doosra term move karo, square roots lo (dono sides $\ge 0$ hain): $(\mathbf{a}\cdot \mathbf{b}{)}^2\le \Vert \mathbf{a}{\Vert }^2\Vert \mathbf{b}{\Vert }^2\Rightarrow |\mathbf{a}\cdot \mathbf{b}|\le \Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert .■$ Equality case: discriminant $=0$ ka matlab hai $f(t_0)=0$ kisi $t_0$ ke liye, yaani $\Vert \mathbf{a}-t_0\mathbf{b}\Vert =0$, toh $\mathbf{a}=t_0\mathbf{b}$ — vectors parallel hain. (Degenerate check: agar $\mathbf{b}=0$ toh $A=0$, $f$ linear/constant $\ge 0$ hai, aur dono sides $0$ hain — inequality $0\le 0$ ke roop mein hold karti hai.) FIGURE mein kya dekhna hai: blue curve non-parallel pair ke liye $f(t)$ hai — yeh poori tarah $t$-axis ke upar float karti hai (uska minimum positive hai), jo exactly "discriminant $<0$" hai aur strict inequality deta hai. Yellow curve parallel case hai: yeh sirf ek point par axis ko chhooti hai (red dot, $t=t_0$), toh discriminant $0$ hai aur Cauchy–Schwarz equality ban jaati hai. Picture hi proof hai: "axis ke neeche kabhi nahi" discriminant condition force karta hai.

KYUN numerically verify karein: ek general theorem concrete tests se survive karke trust kamaata hai. Hum har piece recompute karte hain aur Cauchy–Schwarz ko room ke saath hold hote dekhte hain.

• Overlap: $\mathbf{a}\cdot \mathbf{b}=(1)(-2)+(2)(1)+(3)(1)=-2+2+3=3.$ Kyun: component machine, hamesha ki tarah.
• Lengths: $\Vert \mathbf{a}\Vert =\sqrt{1+4+9}=\sqrt{14},\Vert \mathbf{b}\Vert =\sqrt{4+1+1}=\sqrt{6}.$ Kyun: $\Vert \mathbf{v}\Vert =\sqrt{\mathbf{v}\cdot \mathbf{v}}$.
• Cauchy–Schwarz ke dono sides: $|\mathbf{a}\cdot \mathbf{b}|=3,\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert =\sqrt{14}\sqrt{6}=\sqrt{84}\approx \mathrm{9.17.}$ Kyun dono check karein? inequality exactly inhi do numbers ko compare karti hai; hum left wale ko right ke neeche dekhna chahte hain.
• Strict ya equal? $3\le 9.17$ hold karta hai, aur yeh strict hai kyunki $\mathbf{a}$ $\mathbf{b}$ ka scalar multiple nahi hai (koi single $t$ nahi hai jo $\mathbf{a}=t\mathbf{b}$ banaye) — yaad karo equality ke liye parallel vectors chahiye, jo L5.1 ke equality case se match karta hai. $\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert }=\frac{3}{\sqrt{84}}\approx 0.327\Rightarrow \theta \approx \boxed{70.9^{\circ }}.$ Sanity: $0.327\le 1$, ek valid cosine — wahi $|\cos \theta |\le 1$ fact jo Cauchy–Schwarz generally guarantee karta hai.

KYUN sum ko square karein: condition mein vectors ka sum hai; us sum ko khud se dot karne par exactly woh pairwise products milte hain jo hume chahiye. $0=\Vert \mathbf{u}+\mathbf{v}+\mathbf{w}{\Vert }^2=(\mathbf{u}+\mathbf{v}+\mathbf{w})\cdot (\mathbf{u}+\mathbf{v}+\mathbf{w}).$ Expand karo — teen self-terms lengths hain ($=1$ each kyunki unit vectors hain) aur har cross pair do baar aata hai: $0=\underset{1+1+1=3}{\underbrace{\Vert \mathbf{u}{\Vert }^2+\Vert \mathbf{v}{\Vert }^2+\Vert \mathbf{w}{\Vert }^2}}+2(\mathbf{u}\cdot \mathbf{v}+\mathbf{v}\cdot \mathbf{w}+\mathbf{w}\cdot \mathbf{u}).$ Solve karo: $\mathbf{u}\cdot \mathbf{v}+\mathbf{v}\cdot \mathbf{w}+\mathbf{w}\cdot \mathbf{u}=-\frac{3}{2}=\boxed{-1.5}.$ Sanity: vectors $120^{\circ }$ apart point karte hain (Mercedes logo ki tarah), aur $3\cos 120^{\circ }=3(-\frac{1}{2})=-\frac{3}{2}$. ✔

Component dot ::: matching slots multiply karo, phir add karo → ek scalar Length ::: $\Vert \mathbf{a}\Vert =\sqrt{\mathbf{a}\cdot \mathbf{a}}$ Angle ::: $\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert }$ Perpendicular test ::: $\mathbf{a}\cdot \mathbf{b}=0$ Projection length onto $\mathbf{b}$ ::: $\frac{\mathbf{a}\cdot \mathbf{b}}{\Vert \mathbf{b}\Vert }$, valid sirf tab jab $\mathbf{b}\ne 0$ Cauchy–Schwarz ::: $|\mathbf{a}\cdot \mathbf{b}|\le \Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert$, equality iff parallel C–S proof engine ::: $f(t)=\Vert \mathbf{a}-t\mathbf{b}{\Vert }^2\ge 0$ → discriminant $\le 0$ Squaring a vector sum ::: har cross pair do baar count hota hai