Worked examples — Lines and planes in 3D — vector equations

4.5.5 · D3 · Maths › Linear Algebra (Full) › Lines and planes in 3D — vector equations

Kuch bhi karne se pehle, teen symbols jinhe tumhare paas hona chahiye, kyunki har example inhi par tikta hai:

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Scenario matrix

Is topic ke har problem ka koi na koi ek cell hota hai. Har cell ko cover karne wala example aakhri column mein naam se diya gaya hai.

#	Case class	Trap / twist jo test hoti hai	Covered by
C1	2 points se line, saare directions nonzero	plain symmetric form	Ex 1
C2	Line mein $\mathbf{d}$ ka zero component	tum 0 se divide NAHI kar sakte	Ex 2
C3	Point + normal se plane, negative constant	$d$ ka sign, distance ka sign	Ex 3
C4	3 points se plane	cross product se normal banao	Ex 4
C5	Degenerate 3 points (collinear)	cross product $=0$ ⇒ koi plane nahi	Ex 5
C6	Point-to-plane distance, point "doosri taraf"	absolute value dono sides handle karta hai	Ex 6
C7	Full 3D mein Point-to-line distance	cross product kyun, sirf projection kyun nahi	Ex 7
C8	Line plane ke parallel vs use kaatna	$\mathbf{d}$ aur $\mathbf{n}$ ka dot: zero hai ya nahi	Ex 8
C9	Word problem (drone flight path)	English → vectors mein translate karo	Ex 9
C10	Exam twist: do lines — parallel, intersecting, ya skew?	limiting/degenerate classification	Ex 10

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Example 1 — C1: Do points se line, clean case

Step 1 — direction. $\mathbf{d}=B-A=(5-2,7-1,10-4)=(3,6,6)$. Yeh step kyun? Ek line ki direction iske ek point se doosre tak ka displacement hai — literally $A$ se $B$ tak ka arrow.

Step 2 — vector form. $A$ par anchor karo, $\mathbf{r}$ line par ek generic point ka position vector hai: $\mathbf{r}=(2,1,4)+t(3,6,6)$. Yeh step kyun? Har point hai "$A$ par shuru karo, $\mathbf{d}$ ke $t$ copies chalo" — parameter $t$ poori line sweep karta hai, aur $\mathbf{r}$ wahan hai jahan tum land karte ho.

Step 3 — parametric form ($\mathbf{r}=(x,y,z)$ ka har coordinate read karo): $x=2+3t,y=1+6t,z=4+6t.$ Yeh step kyun? Vector equation actually teen equations ka stack hai — har coordinate ke liye ek. Inhe alag karne se hum ek single value of $t$ daalke ek concrete point $(x,y,z)$ read kar sakte hain.

Step 4 — symmetric form. Har ek ko $t$ ke liye solve karo aur barabar rakh do (allowed kyunki saare $3,6,6\ne 0$): $\frac{x-2}{3}=\frac{y-1}{6}=\frac{z-4}{6}.$ Yeh step kyun? Har point ek $t$ share karta hai; teen "$t$ expressions" ko equal rakhne se $t$ hat jaata hai.

Verify: Parametric mein $t=1$ daalo: $(2+3,1+6,4+6)=(5,7,10)=B$. ✓ Aur $t=0$ se $A$ milta hai. ✓

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Example 2 — C2: Zero component, divide-by-zero trap

Step 1 — zero dhundo. $\mathbf{d}=(2,4,0)$: teesra slot $0$ hai. Yeh step kyun? Symmetric form har component se divide karta hai. $0$ se division undefined hai — formula yahaan silently toot jaata hai.

Step 2 — zero direction ka MATLAB kya hai. Agar $z=2+0\cdot t=2$ sabhi $t$ ke liye, toh line kabhi height nahi badlati. Yeh poori tarah flat sheet $z=2$ par rehti hai. Yeh step kyun? Zero ko geometrically interpret karne se pata chalta hai ki likhna kya hai.

Step 3 — correct symmetric form. Symmetric form sirf un coordinates par use karo jinki direction nonzero hai, aur frozen wale ko alag batao: $\frac{x-1}{2}=\frac{y-0}{4},z=2.$ Yeh step kyun? Jo do coordinates ki direction nonzero hai woh ek hi $t$ share karte hain, isliye unke ratios equal hain; frozen coordinate mein bilkul bhi $t$ nahi hai, isliye ise ek fraction mein thonsne ki jagah standalone constraint ke roop mein likhna chahiye.

Verify: $t=1$ lo: $x=3,y=4,z=2$. Check karo $\frac{3-1}{2}=1$ aur $\frac{4-0}{4}=1$ — equal ✓ aur $z=2$ ✓.

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Example 3 — C3: Point + normal se plane, negative constant

Step 1 — normal form. $A$ ko uske position vector $\mathbf{a}=(2,-1,3)$ ke roop mein likho. Phir $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$ se $1(x-2)-2(y+1)+2(z-3)=0$ milta hai. Yeh step kyun? Koi bhi in-plane arrow $\mathbf{r}-\mathbf{a}$ perpendicular hota hai $\mathbf{n}$ se, isliye uska $\mathbf{n}$ ke saath dot product zero hai.

Step 2 — simplify karo. $x-2-2y-2+2z-6=0\Rightarrow x-2y+2z=10$. Yeh step kyun? Constant hai $d=\mathbf{a}\cdot \mathbf{n}=2+2+6=10$. Yahaan yeh positive aaya bawajood $\mathbf{n}$ mein negative ke.

Step 3 — $P=(0,0,0)$ se distance, jiska position vector $\mathbf{p}=(0,0,0)$ hai: $\text{dist}=\frac{|\mathbf{p}\cdot \mathbf{n}-d|}{|\mathbf{n}|}=\frac{|0-10|}{\sqrt{1+4+4}}=\frac{10}{3}.$ Absolute value kyun? Distance kabhi negative nahi hoti; $|\cdot |$ handle karta hai ki $P$ plane ke kisi bhi side ho.

Verify: $\mathbf{n}=(1,-2,2)$, $|\mathbf{n}|=\sqrt{9}=3$; $10/3\approx 3.33$. Aur $A$ plane ko satisfy karta hai: $2-2(-1)+2(3)=2+2+6=10$ ✓.

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Example 4 — C4: Teen points se plane

Step 1 — do spanning directions. $\mathbf{u}=B-A=(1,-1,0)$, $\mathbf{v}=C-A=(0,1,2)$. Yeh step kyun? Ek plane hai "ek point plus do independent chalne ki directions." Yeh woh do directions hain.

Step 2 — cross product se normal. Cross product dono ke perpendicular arrow deta hai: $\mathbf{n}=\mathbf{u}\times \mathbf{v}=((-1)(2)-(0)(1),(0)(0)-(1)(2),(1)(1)-(-1)(0))=(-2,-2,1).$ Yeh step kyun? Definition se $\mathbf{u}\times \mathbf{v}\perp \mathbf{u}$ aur $\perp \mathbf{v}$ — exactly woh normal jo chahiye.

Step 3 — Cartesian equation. Point $A$ use karke: $-2(x-1)-2(y-1)+1(z-1)=0\Rightarrow -2x-2y+z=-3$, yaani $2x+2y-z=3.$ Yeh step kyun? Ek baar normal milne ke baad, plane sirf "woh saare points jinका $A$ se arrow $\mathbf{n}$ ke perpendicular ho" hai — yaani $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$. Us dot product ko multiply karke terms collect karne se geometry ek single tidy Cartesian equation mein aa jaati hai jiske coefficients NORMAL hain.

Verify: Har point ise satisfy kare. $A$: $2+2-1=3$ ✓. $B$: $4+0-1=3$ ✓. $C$: $2+4-3=3$ ✓.

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Example 5 — C5: Degenerate — teen points jo plane nahi banate

Step 1 — spanning directions. $\mathbf{u}=B-A=(1,1,1)$, $\mathbf{v}=C-A=(2,2,2)$. Yeh step kyun? Ex 4 wali hi recipe — lekin do directions dekho.

Step 2 — cross product. $\mathbf{n}=\mathbf{u}\times \mathbf{v}=(1\cdot 2-1\cdot 2,1\cdot 2-1\cdot 2,1\cdot 2-1\cdot 2)=(0,0,0).$ Yeh step kyun? Cross product tab $0$ hota hai jab do directions parallel hon. Yahaan $\mathbf{v}=2\mathbf{u}$.

Step 3 — interpret karo. Zero normal ka matlab hai "koi unique perpendicular exist nahi karta." Teen points collinear hain — woh saare single line $\mathbf{r}=t(1,1,1)$ par hain. Teen collinear points infinitely many planes define karte hain (unka ek poora hinge), isliye koi ek plane nahi hai. Yeh step kyun? Algebra ($\mathbf{n}=0$) ek degenerate geometric setup ke liye warning light hai.

Verify: Kya $C$ line through $A,B$ par hai? $C=(2,2,2)=2\cdot (1,1,1)$ — haan, $t=2$. Collinear confirm ✓.

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Example 6 — C6: Point-to-plane distance, dono sides

Step 1 — set up karo. $\mathbf{n}=(2,2,-1)$, $d=3$, $|\mathbf{n}|=\sqrt{4+4+1}=3$. Yeh step kyun? Distance formula ko teen cheezein chahiye: normal $\mathbf{n}$ (coefficients se seedha padho), constant $d$ (right-hand side), aur normal ki length $|\mathbf{n}|$ (ek raw dot-product difference ko actual distance mein convert karne ke liye). Inhe pehle ikattha karne se dono point-computations clean rehti hain.

Step 2 — $P$ ke liye signed value. $\mathbf{p}=(3,3,3)$ ke saath: $\mathbf{p}\cdot \mathbf{n}-d=(6+6-3)-3=6$. Toh ${\text{dist}}_P=\frac{|6|}{3}=2$. Yeh step kyun? Numerator $\mathbf{p}\cdot \mathbf{n}-d$ measure karta hai ki $P$ ki "$\mathbf{n}$ ke along height" plane ki height se kitni zyada hai. Uska sign batata hai ki kaun si side hai.

Step 3 — $Q$ ke liye signed value. $\mathbf{q}=(1,1,1)$ ke saath: $\mathbf{q}\cdot \mathbf{n}-d=(2+2-1)-3=-3$. Toh ${\text{dist}}_Q=\frac{|-3|}{3}=1$. Yeh step kyun? $Q$ ka signed value negative hai jabki $P$ ka positive tha ⇒ woh plane ke opposite sides par hain.

Verify: Signs $+6$ vs $-3$ differ karte hain ⇒ opposite sides ✓. Distances $2$ aur $1$ dono positive hain jaise distances honi chahiye ✓.

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Example 7 — C7: 3D mein Point-to-line distance

Step 1 — line ke ek point se $P$ tak ka vector. $A=(0,0,0)$ lo (position vector $\mathbf{a}$) aur $\mathbf{p}=(1,2,3)$: $\mathbf{p}-\mathbf{a}=(1,2,3)$. Yeh step kyun? Ek line se distance perpendicular drop hai $P$ se; hum kisi bhi jaane-maane line point se shuru karte hain.

Step 2 — cross product kyun. Hum chahte hain $|\mathbf{p}-\mathbf{a}|\sin \theta$ (perpendicular height), jahan $\theta$ line ke saath angle hai. Cross product magnitude exactly $|\mathbf{p}-\mathbf{a}||\mathbf{d}|\sin \theta$ deta hai, isliye $|\mathbf{d}|$ se divide karne par height isolate ho jaati hai: $\text{dist}=\frac{|(\mathbf{p}-\mathbf{a})\times \mathbf{d}|}{|\mathbf{d}|}.$ Sirf project kyun nahi? Projection along-line component deta hai; hume woh chahiye jo perpendicular mein bacha — cross product directly woh capture karta hai.

Step 3 — compute karo. $(1,2,3)\times (1,0,0)=(2\cdot 0-3\cdot 0,3\cdot 1-1\cdot 0,1\cdot 0-2\cdot 1)=(0,3,-2)$. Magnitude $\sqrt{0+9+4}=\sqrt{13}$. Aur $|\mathbf{d}|=1$. Toh $\text{dist}=\sqrt{13}$. Yeh step kyun? Yeh Step 2 ka formula hamari numbers ke saath execute karta hai: cross product banao, uski length lo, aur $|\mathbf{d}|$ se divide karo. Kyunki yahaan $|\mathbf{d}|=1$ hai, cross-product length HI distance hai — koi rescaling nahi chahiye.

Verify: $(1,2,3)$ se $x$-axis tak distance $\sqrt{y^2+z^2}=\sqrt{4+9}=\sqrt{13}$ ✓ honi chahiye — $x$-coordinate irrelevant hai, jaise expected.

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Example 8 — C8: Line plane ke parallel vs use kaatna

Step 1 — deciding test. Ek line plane ke parallel hai jab uski direction normal ke perpendicular ho, yaani $\mathbf{d}\cdot \mathbf{n}=0$ (direction mein $\mathbf{n}$ ke along koi component nahi, isliye kabhi "height" gain nahi karta). Yeh step kyun? $\mathbf{n}$ ke along chalana hi plane ko leave/approach karne ka ek tarika hai; $\mathbf{d}\cdot \mathbf{n}$ us motion ko measure karta hai.

Step 2 — $L_1$ test karo. ${\mathbf{d}}_1\cdot \mathbf{n}=(2)(1)+(1)(2)+(-2)(2)=2+2-4=0$. Toh $L_1$ plane ke parallel hai. Yeh step kyun? Hum Step-1 test $L_1$ ki specific direction $(2,1,-2)$ par apply karte hain. Exactly $0$ milna confirm karta hai ki direction mein along-normal koi component nahi, isliye line $\mathbf{n}$ ke relative constant height par rehti hai — yeh kabhi plane approach ya leave nahi kar sakti. Kyunki $L_1$ origin se guzarti hai aur $0+0+0=0\ne 6$, woh origin plane se off hai, isliye $L_1$ kabhi use touch nahi karti.

Step 3 — $L_2$ test karo. ${\mathbf{d}}_2\cdot \mathbf{n}=(1)(1)+0+0=1\ne 0$. Nonzero ⇒ $L_2$ plane cross karti hai. Kahan: $x+0+0=6\Rightarrow t=6$, intersection $(6,0,0)$ par. Yeh step kyun? Nonzero dot ka matlab hai line steadily $\mathbf{n}$ ke along height change karti hai, isliye use kisi bhi given height par exactly ek baar hit karna hi hoga — dekho Intersection of lines and planes.

Verify: $L_1$: ${\mathbf{d}}_1\cdot \mathbf{n}=0$ ✓ (parallel). $L_2$ intersection $(6,0,0)$: $6+0+0=6$ ✓.

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Example 9 — C9: Word problem (drone flight)

Step 1 — path model karo. Direction $\mathbf{d}=B-A=(30,40,60)$. Path: $\mathbf{r}=(0,0,0)+t(30,40,60)$, toh $x=30t,y=40t,z=60t$, jahan $\mathbf{r}$ drone ki position hai "clock reading" $t$ par. Yeh step kyun? "Do jaane-maane points se seedha flight" exactly ek line hai — point + direction.

Step 2 — fence model karo aur $t$ nikalo. Fence hai $z=50$. Set karo $60t=50\Rightarrow t=\frac{5}{6}$. Yeh step kyun? Cross karne ka matlab hai $z$-coordinate fence height ke barabar ho; single unknown $t$ ke liye solve karo.

Step 3 — crossing point. $x=30\cdot \frac{5}{6}=25$, $y=40\cdot \frac{5}{6}=\frac{100}{3}\approx 33.33$, $z=50$. Point $=(25,33.33,50)$ m. Yeh step kyun? Value $t=\frac{5}{6}$ woh "clock reading" hai jis par drone fence tak pahunchta hai; ise path ke $x$ aur $y$ equations mein wapas daalne se woh actual location space mein milti hai jahan crossing hoti hai.

Step 4 — fence tak travel ki distance. $|\mathbf{d}|=\sqrt{30^2+40^2+60^2}=\sqrt{900+1600+3600}=\sqrt{6100}=10\sqrt{61}$. Distance $=t|\mathbf{d}|=\frac{5}{6}\cdot 10\sqrt{61}=\frac{25\sqrt{61}}{3}\approx 65.1$ m. Yeh step kyun? Distance flown hai (raaste ka fraction $=t$) $\times$ (ek direction step ki length).

Verify: $t=\frac{5}{6}$ par, $z=60\cdot \frac{5}{6}=50$ ✓ (fence par). $\frac{25\sqrt{61}}{3}\approx 65.1$ m, aur $\sqrt{25^2+33.33^2+50^2}\approx \sqrt{625+1111+2500}=\sqrt{4236}\approx 65.1$ ✓.

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Example 10 — C10: Exam twist — parallel, intersecting, ya skew?

Step 1 — parallel check (dono pairs). Directions parallel ⟺ ek doosre ka scalar multiple hai. Pair A: $(2,2,0)=2(1,1,0)$ — parallel directions. Pair B: $(1,0,0)$ vs $(0,1,0)$ — nahi parallel. Yeh step kyun? Parallel directions teen possibilities ko do mein collapse karta hai (same line, ya parallel-and-distinct).

Step 2 — Pair A resolve karo. Directions parallel hain: kya woh same line hain? Check karo ki $L_2$ ka point $(0,5,0)$ $L_1$ par hai ya nahi: chahiye $(0,5,0)=(1+t,t,0)$. Row 1 se, $t=-1$; phir row 2 deta hai $t=5$. Contradiction ⇒ point $L_1$ par nahi. Toh parallel aur distinct (kabhi nahi milenge). Yeh step kyun? Same direction + shared point ⇒ identical line; same direction + koi shared point nahi ⇒ parallel distinct.

Step 3 — Pair B resolve karo (parallel nahi). Meeting point ke liye solve karo: $L_3$ deta hai $(t,0,0)$; $L_4$ deta hai $(0,s,1)$. Match karo: $t=0$, $0=s$, aur $z$: $0=1$. Aakhri impossible hai ⇒ koi intersection nahi. Not parallel + no intersection = skew. Yeh step kyun? Jab directions parallel nahi hoti, ya toh equations ka solution hota hai (milte hain) ya woh inconsistent hote hain (skew).

Verify: Pair A directions: $2\cdot (1,1,0)=(2,2,0)$ ✓ parallel; point test $t=-1$ vs $t=5$ inconsistent ✓ distinct. Pair B: $z$-equation $0=1$ false hai ✓ ⇒ skew.

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Recall Active recall — answers cover karo

• Direction component $0$ hai: symmetric form kaise likhoge? ::: us coordinate ko $z=\text{const}$ ki tarah alag karo; ratios sirf nonzero components ke liye use karo.
• Do spanning directions ka cross product $0$ hai — iska kya matlab? ::: points collinear hain; koi unique plane exist nahi karta.
• Ek dot product line-vs-plane parallelism decide karta hai — kaun sa? ::: $\mathbf{d}\cdot \mathbf{n}=0$ matlab parallel; nonzero matlab cross karta hai.
• Do 3D lines, parallel nahi, kabhi nahi miltin — unhe kya kehte hain? ::: skew.
• Ek point plane ke kis side par hai? ::: $\mathbf{p}\cdot \mathbf{n}-d$ ka sign: positive vs negative = opposite sides.

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Connections

• Lines and planes in 3D — vector equations — parent toolkit jise yeh examples exercise karte hain.
• Cross product — normals banata hai (Ex 4, 5) aur point-to-line distance (Ex 7).
• Vectors and scalar (dot) product — parallel/side tests (Ex 3, 6, 8).
• Vector projection — distance formulas ke peeche geometry.
• Intersection of lines and planes — Ex 8, 9, 10.
• Systems of linear equations — meeting parameters ke liye solve karna (Ex 8–10).
• Parametrisation and parameters — $t,s$ ke roles.