Projection of vectors: Worked examples — Hinglish deep dive

4.5.4 · D3 · Maths › Linear Algebra (Full) › Projection of vectors

Scenario matrix

Har projection problem exactly inhi cells mein se ek mein aati hai. Last column us example ka naam deta hai jo isse cover karta hai.

#	Case class	Kya special hai	Sign of $a \cdot b$	Covered by
C1	Acute angle ( $0^{\circ} < θ < 9 0^{\circ}$ )	shadow $b$ ke saath point karta hai	positive	Ex 1
C2	Right angle ( $θ = 9 0^{\circ}$ )	shadow zero hai	zero	Ex 2
C3	Obtuse angle ( $9 0^{\circ} < θ < 18 0^{\circ}$ )	shadow backward point karta hai	negative	Ex 3
C4	Non-unit / long $b$	$b \cdot b$ se divide karna zaroori, $	\vec b	$ se nahi
C5	Degenerate: $b = 0$	division by zero — undefined	$0/0$	Ex 5
C5′	Degenerate: $a = 0$	project karne ko kuch nahi — result hai $0$	zero	Ex 5b
C6	Collinear ( $θ = 0^{\circ}$ ya $18 0^{\circ}$ )	shadow = poora vector, $\pm$	$\pm	\vec a
C7	3-D decomposition	parallel + perpendicular mein split karo	any	Ex 7
C8	Real-world word problem (ramp / force)	units attach karo, meaning samjho	positive	Ex 8
C9	Exam twist (projection se unknown solve karo)	ulta kaam karo	given	Ex 9

Ab hum sab walk karenge.

Building blocks jo hum reuse karenge (ek baar define, zero se)

Figure s01 neeche exactly yahi setup draw karta hai: magenta arrow $a$ , violet arrow $b$ , aur orange shadow jo seedha $b$ ki line pe giraaya gaya hai. Dashed navy line woh perpendicular "sun ray" hai jo define karti hai shadow kahan khatam hoti hai — ise dhyaan mein rakhna, har example is ek picture ka ek version hai.

Ex 1 — Acute angle (cell C1)

Worked example Sharp angle: shadow aage point karta hai

$a = (4, 3)$ , $b = (2, 0)$ . $a$ ka $b$ pe scalar aur vector projections nikalo.

Forecast: $b$ $+ x$ ke saath point karta hai, aur $a$ upar-aur-right lean karta hai. Guess: shadow $a$ ka $x$ -part hai, toh scalar $= 4$ aur vector $= (4, 0)$ . Confirm karte hain. (Yeh exactly figure s01 ki geometry hai — orange shadow $b$ ke saath.)

Step 1. $a \cdot b = 4 (2) + 3 (0) = 8$ . Yeh step kyun? Dot product dono formulas ka numerator hai; pehle ise compute karo.

Step 2. $b \cdot b = 2^{2} + 0^{2} = 4$ , toh $∣ b ∣ = 4 = 2$ . Yeh step kyun? Scalar ko $∣ b ∣$ chahiye; vector ko $b \cdot b$ chahiye. Dono abhi le lo.

Step 3. Scalar projection $= \frac{8}{2} = 4$ . Yeh step kyun? $\frac{a \cdot b}{∣ b ∣}$ apply karo. Positive ⇒ shadow $b$ ke saath hai (acute angle, cell C1).

Step 4. Vector projection $= \frac{8}{4} (2, 0) = 2 (2, 0) = (4, 0)$ . Yeh step kyun? $\frac{a \cdot b}{b \cdot b} b$ apply karo. Direction $b$ ki hai, length shadow ki hai.

Verify: $(4, 0)$ $x$ -axis pe lie karta hai (correct direction), aur iska length $4$ scalar projection ke barabar hai. $y$ -part $3$ gaya — exactly shadow ka idea. ✓

Ex 2 — Right angle (cell C2)

Worked example Perpendicular: shadow collapse ho jaata hai

$a = (0, 5)$ , $b = (3, 0)$ . $a$ ko $b$ pe project karo.

Forecast: $a$ seedha upar point karta hai, $b$ seedha right — woh perpendicular hain. Ek vertical stick seedhe-neeche sun ke neeche koi horizontal shadow nahi daalta. Guess: sab zero hai.

Step 1. $a \cdot b = 0 (3) + 5 (0) = 0$ . Yeh step kyun? Zero dot product hi perpendicular ki definition hai ( $cos 9 0^{\circ} = 0$ ).

Step 2. $b \cdot b = 3^{2} + 0^{2} = 9$ , toh $∣ b ∣ = 9 = 3$ . Yeh step kyun? Chahe humein zero answer ki expect ho, denominators explicitly likhna zaroori hai taaki kuch hidden na ho: scalar $∣ b ∣ = 3$ use karega, vector $b \cdot b = 9$ use karega.

Step 3. Scalar projection $= \frac{0}{3} = 0$ . Yeh step kyun? Shadow ki length genuinely zero hai.

Step 4. Vector projection $= \frac{0}{9} (3, 0) = (0, 0)$ . Yeh step kyun? Zero length × koi bhi direction = zero vector.

Verify: Perpendicular part $a_{⊥} = a - 0 = (0, 5) = a$ — yaani saara $a$ $b$ ke perpendicular hai, jo picture se match karta hai. ✓

Ex 3 — Obtuse angle (cell C3)

Worked example Shadow peeche girta hai

$a = (- 3, 1)$ , $b = (4, 0)$ . $a$ ko $b$ pe project karo.

Forecast: $a$ left ki taraf lean karta hai, $b$ right point karta hai — unke beech ka angle obtuse hai. Guess: negative scalar, aur left point karta vector. Figure s02 yahi case dikhata hai — dekho orange shadow origin ke peeche girta hai.

Step 1. $a \cdot b = (- 3) (4) + 1 (0) = - 12$ . Yeh step kyun? Negative dot product matlab $cos θ < 0$ , yaani $θ > 9 0^{\circ}$ (cell C3). Sign pehle se hi story bata raha hai.

Step 2. $b \cdot b = 4^{2} + 0^{2} = 16$ , toh $∣ b ∣ = 16 = 4$ . Yeh step kyun? Scalar $∣ b ∣ = 4$ se divide karta hai; vector $b \cdot b = 16$ se divide karta hai.

Step 3. Scalar projection $= \frac{- 12}{4} = - 3$ . Yeh step kyun? Minus encode karta hai "shadow back side pe hai"; ise drop mat karo.

Step 4. Vector projection $= \frac{- 12}{16} (4, 0) = - \frac{3}{4} (4, 0) = (- 3, 0)$ . Yeh step kyun? Negative scalar × direction $(4, 0)$ ise flip karta hai $- x$ ki taraf point karne ke liye, $b$ se opposite.

Verify: $(- 3, 0)$ $b$ ke against point karta hai aur length $3 = ∣ - 3 ∣$ hai = scalar projection ki magnitude. Sign aur picture agree karte hain. ✓

Ex 4 — Non-unit, long $b$ (cell C4)

Worked example

$b$ ko longer banana answer change nahi karna chahiye $a = (2, 3)$ , $b = (0, 10)$ . $b$ pe project karo, phir $b^{'} = (0, 1)$ (same direction, shorter) ke saath redo karo. Kya vector answers match karte hain?

Forecast: Dono $b$ vectors seedhe upar point karte hain. $a$ ka vertical direction pe shadow iska $y$ -part hona chahiye, $(0, 3)$ , chahe $b$ kitna bhi lamba draw kiya ho. Guess: identical vector projections.

Step 1 ( $b = (0, 10)$ ke saath). $a \cdot b = 2 (0) + 3 (10) = 30$ , aur $b \cdot b = 0 + 100 = 100$ . Yeh step kyun? Dono $b$ ki length ke saath badhte hain; dekho kaise woh cancel karte hain.

Step 2. Vector projection $= \frac{30}{100} (0, 10) = 0.3 (0, 10) = (0, 3)$ . Yeh step kyun? Denominator mein $b \cdot b$ $b$ ki extra length ko khatam kar deta hai. Yahi wajah hai ki vector form ke liye hum $∣ b ∣$ ki jagah $b \cdot b$ se divide karte hain.

Step 3 ( $b^{'} = (0, 1)$ ke saath). $a \cdot b^{'} = 3$ , $b^{'} \cdot b^{'} = 1$ , projection $= \frac{3}{1} (0, 1) = (0, 3)$ . Yeh step kyun? Scale-independence confirm karta hai.

Verify: Dono vector projections $(0, 3)$ hain — bilkul same, toh $b$ ko das guna lamba karne se kuch nahi badla. (Scalar projections bhi dono $3$ nikle hain: $∣ b ∣ = 10$ ke saath $a \cdot b = 30$ deta hai $30/10 = 3$ , aur $3/1 = 3$ ; dono directions same hain, toh unki measured lengths bhi agree karti hain.) Length-invariance confirmed. ✓

Ex 5 — Degenerate: zero vector pe project karna (cell C5)

Worked example Jab formula toot jaata hai

$a = (5, 2)$ , $b = (0, 0)$ . $a$ ko $b$ pe project karo.

Forecast: $b$ ki koi length nahi aur koi direction nahi — koi line nahi jis pe shadow cast kiya jaaye. Guess: undefined.

Step 1. $a \cdot b = 5 (0) + 2 (0) = 0$ aur $b \cdot b = 0$ . Yeh step kyun? Dekho formula aage kya demand karta hai.

Step 2. Scalar $= \frac{0}{∣ b ∣} = \frac{0}{0}$ aur vector $= \frac{0}{0} b$ . Yeh step kyun? Division by zero — expression undefined hai, zero nahi.

Step 3 — sahi statement. Projection exist nahi karta jab $b = 0$ , kyunki " $b$ ki direction" meaningless hai. Hamesha $b \neq = 0$ demand karo. Yeh step kyun? Zero-length arrow ka koi unit vector $\hat{b}$ nahi hota, toh project karne ke liye koi direction nahi.

Verify: $\hat{b} = b /∣ b ∣ = (0, 0) /0$ banane ki koshish karo — impossible. Poora construction $\hat{b}$ ke exist karne pe depend karta hai. ✓ (Undefined confirmed.)

Ex 5b — Degenerate: zero vector ko project karna (cell C5′)

Worked example DOOSRA zero case

$a = (0, 0)$ , $b = (3, 4)$ . $a$ ko $b$ pe project karo.

Forecast: Is baar $b$ ek perfectly good direction hai, lekin $a$ kuch bhi nahi — origin pe ek point. Zero-length stick zero-length shadow daalta hai. Guess: scalar $0$ , vector $0$ (aur is baar yeh sach mein defined hai).

Step 1. $a \cdot b = 0 (3) + 0 (4) = 0$ . Yeh step kyun? Zero vector ke saath koi bhi dot product $0$ hota hai.

Step 2. $b \cdot b = 3^{2} + 4^{2} = 25$ , toh $∣ b ∣ = 25 = 5$ — ek nonzero denominator. Yeh step kyun? Kyunki $b \neq = 0$ , division legal hai (Ex 5 se alag).

Step 3. Scalar $= \frac{0}{5} = 0$ ; vector $= \frac{0}{25} (3, 4) = (0, 0)$ . Yeh step kyun? $0/25$ ek genuine zero hai, $0/0$ nahi. Answer exist karta hai aur zero vector ke barabar hai.

Verify: Ex 5 se compare karo: yahan denominators $5$ aur $25$ nonzero hain, toh hum legally divide karte hain aur well-defined $0$ paate hain. Zero input $a$ → zero output; zero direction $b$ → undefined. Yeh do bahut alag degenerate cases hain. ✓

Ex 6 — Collinear vectors (cell C6)

Worked example Same line, dono directions

(a) $a = (6, 0)$ , $b = (2, 0)$ (same direction, $θ = 0^{\circ}$ ). (b) $a = (6, 0)$ , $b = (- 2, 0)$ (opposite direction, $θ = 18 0^{\circ}$ ).

Forecast: $a$ already $b$ ki line pe lie karta hai. Iska shadow khud hi hona chahiye, $(6, 0)$ , dono cases mein — kyunki vector projection $a$ ki line pe wapas aata hai dono taraf. Scalar (b) mein sign flip karna chahiye.

Step 1 (a). $a \cdot b = 12$ , $b \cdot b = 4$ toh $∣ b ∣ = 2$ . Vector $= \frac{12}{4} (2, 0) = (6, 0)$ . Scalar $= \frac{12}{2} = 6$ . Yeh step kyun? $θ = 0^{\circ}$ , $cos 0^{\circ} = 1$ , toh scalar $= ∣ a ∣ = 6$ : shadow poori length hai.

Step 2 (b). $a \cdot b = - 12$ , $b \cdot b = 4$ toh $∣ b ∣ = 2$ . Vector $= \frac{- 12}{4} (- 2, 0) = (6, 0)$ . Scalar $= \frac{- 12}{2} = - 6$ . Yeh step kyun? $θ = 18 0^{\circ}$ , $cos 18 0^{\circ} = - 1$ , toh scalar $= - ∣ a ∣ = - 6$ ; lekin negative scalar × negative direction $(- 2, 0)$ vector ko wapas $(6, 0)$ pe le jaata hai.

Verify: Dono cases mein vector projection $(6, 0) = a$ hai (poori tarah recover hua, jaise collinear vector ke liye hona chahiye). Sirf scalar direction sign carry karta hai. ✓

Ex 7 — 3-D orthogonal decomposition (cell C7)

Worked example Ek vector ko "along" + "perpendicular" mein split karna

$a = (3, 2, 6)$ , $b = (0, 0, 2)$ . $a = proj_{b} a + a_{⊥}$ mein split karo.

Forecast: $b$ $z$ -axis ke saath point karta hai. Parallel part $a$ ki height $(0, 0, 6)$ honi chahiye; bacha hua ground part $(3, 2, 0)$ .

Step 1. $a \cdot b = 3 (0) + 2 (0) + 6 (2) = 12$ , $b \cdot b = 0 + 0 + 4 = 4$ . Yeh step kyun? Same machinery — dot products ko dimension ki parwah nahi.

Step 2. $proj_{b} a = \frac{12}{4} (0, 0, 2) = 3 (0, 0, 2) = (0, 0, 6)$ . Yeh step kyun? Parallel piece — $a$ ki "height".

Step 3. $a_{⊥} = a - proj_{b} a = (3, 2, 6) - (0, 0, 6) = (3, 2, 0)$ . Yeh step kyun? Jo bhi bacha woh $b$ ke perpendicular hona chahiye (yeh Orthogonal decomposition hai; repeat karo, yeh Gram-Schmidt process ko power deta hai).

Verify: $a_{⊥} \cdot b = 3 (0) + 2 (0) + 0 (2) = 0$ ✓ (perpendicular). Aur $(0, 0, 6) + (3, 2, 0) = (3, 2, 6) = a$ ✓ (pieces reassemble ho jaate hain). ✓

Ex 8 — Word problem: force along a ramp (cell C8)

Worked example Kitna force slope ke saath push karta hai?

Ek box ek ramp pe $F = (0, - 10)$ N force feel karta hai (gravity, seedha neeche point karta). Ramp direction hai $b = (4, 3)$ (rightward aur incline ke upar). Kitna force ramp ke saath act karta hai, aur iske saath ka vector kya hai?

Forecast: Gravity neeche point kati hai, ramp upar-right jaata hai, toh gravity ka component $b$ ke saath up-ramp direction ko oppose karta hai — expect karo negative scalar (force box ko slope ke neeche kheechtaa hai). Figure s03 box, neeche-point karta gravity, aur iska orange shadow ramp ke neeche slide karta draw karta hai. Work done by a force se link, jahan $W = F \cdot d$ exactly yahi projection idea hai.

Step 1. $F \cdot b = 0 (4) + (- 10) (3) = - 30$ . Yeh step kyun? Dot product $b$ ki direction mein force ki signed amount hai.

Step 2. $b \cdot b = 4^{2} + 3^{2} = 25$ , toh $∣ b ∣ = 25 = 5$ . Yeh step kyun? Scalar (newtons mein) ke liye $∣ b ∣$ chahiye aur vector ke liye $b \cdot b$ .

Step 3. Scalar projection $= \frac{- 30}{5} = - 6$ N. Yeh step kyun? $6$ N gravity ramp line ke saath act karta hai; minus matlab woh slope ke neeche point karta hai, box ko neeche kheenchta hai. Units: N (ek force), correct.

Step 4. Vector projection $= \frac{- 30}{25} (4, 3) = - 1.2 (4, 3) = (- 4.8, - 3.6)$ N. Yeh step kyun? Direction ramp ke saath neeche-aur-left point karti hai — box neeche slide karta hai.

Verify: Vector ki magnitude $= (- 4.8)^{2} + (- 3.6)^{2} = 23.04 + 12.96 = 36 = 6$ N, scalar magnitude se match karta hai. Direction $(- 4.8, - 3.6)$ hai $- 1.2 \times (4, 3)$ : exactly up-ramp direction ke opposite. Physically sensible. ✓

Ex 9 — Exam twist: unknown solve karo (cell C9)

Worked example Diye gaye projection se ulta kaam karo

$a = (t, 4)$ unknown $t$ ke saath. Diya gaya hai ki $a$ ka $b = (3, 0)$ pe scalar projection $5$ ke barabar hai. $t$ nikalo. Phir woh value of $t$ nikalo jo projection ko zero banata hai.

Forecast: $x$ -axis pe scalar projection basically $a$ ka $x$ -part hai. Agar yeh $5$ hona chahiye, toh guess karo $t = 5$ ; agar zero, toh guess karo $t = 0$ .

Step 1. $b \cdot b = 3^{2} + 0^{2} = 9$ , toh $∣ b ∣ = 3$ . Scalar projection $= \frac{a \cdot b}{∣ b ∣} = \frac{3 t + 0}{3} = t$ . Yeh step kyun? Solve karne se pehle unknown ke terms mein given quantity express karo.

Step 2 (projection $= 5$ ). $t = 5$ set karo. Yeh step kyun? Equation seedha $t = 5$ mein collapse ho gayi. Clean.

Step 3 (projection $= 0$ ). $t = 0$ set karo. Yeh step kyun? Zero projection matlab $a$ $b$ ke perpendicular hai; $b$ $x$ ke saath hone ki wajah se, woh $a$ ka $x$ -part $= 0$ force karta hai, yaani $t = 0$ , deta hai $a = (0, 4)$ (seedha upar — $x$ -axis ke perpendicular ✓).

Verify: $t = 5$ ke liye: $a \cdot b = 3 (5) = 15$ , scalar $= 15/3 = 5$ ✓. $t = 0$ ke liye: $a \cdot b = 0$ , scalar $= 0$ ✓, aur $(0, 4) ⊥ (3, 0)$ . ✓

Active recall

Recall Kaun sa cell kaun sa hai?

Obtuse angle scalar projection ka kaun sa sign deta hai? ::: Negative — shadow peeche girta hai. $b = 0$ pe project karne se kya milta hai? ::: Undefined — koi direction nahi, $0/0$ . Zero vector $a = 0$ ko nonzero $b$ pe project karne se kya milta hai? ::: Zero vector $(0, 0)$ — well-defined, kyunki denominator $b \cdot b \neq = 0$ . Right-angle projection ki value? ::: Exactly $0$ (ek genuine zero, undefined nahi). $b$ ko das guna longer banana vector projection ko kaise change karta hai? ::: Bilkul nahi — $b \cdot b$ extra length cancel kar deta hai. Collinear opposite vectors: scalar sign? ::: Negative ( $cos 18 0^{\circ} = - 1$ ), lekin vector projection phir bhi $a$ ki line pe aata hai. Force $(0, - 10)$ along ramp $(4, 3)$ : scalar projection? ::: $- 6$ N (slope ke neeche). $comp_{b} a$ ka words mein kya matlab hai? ::: $b$ ke saath $a$ ka component — ek signed number (shadow ki length).

Connections

Projection of vectors — parent: formulas aur shadow intuition.
Dot product — yahan har example ka numerator.
Unit vectors — kyun $b = 0$ (koi $\hat{b}$ nahi) Ex 5 mein sab kuch todta hai.
Orthogonal decomposition — Ex 7 ka split.
Gram-Schmidt process — Ex 7 ka move repeat karta hai orthonormal bases banane ke liye.
Least squares regression — data ka ek subspace pe projection (same C4 scale-invariance).
Work done by a force — Ex 8 ek projection hai physics ke kapdon mein.

Worked examples — Projection of vectors