Exercises — Lines and planes in 3D — vector equations

(a) $t=0$ set karo: tum start point $\mathbf{a}=(2,-1,4)$ pe baithe ho. Kyun: $t\mathbf{d}$, $t=0$ pe vanish ho jaata hai, sirf position part bachta hai.

(b) Direction woh vector hai jo $t$ se multiply hota hai: $\mathbf{d}=(3,0,-2)$.

(c) Hum poochte hain: kya $t$ ki ek value hai jo $Q$ ke har coordinate ko hit karti hai? $5=2+3t\Rightarrow t=1,-1=-1+0\cdot t✓\text{ (any }t),2=4-2t\Rightarrow t=1.$ Saare coordinates $t=1$ pe agree karte hain, isliye haan, $Q$ line pe hai ($t=1$ pe).

(a) $x,y,z$ ke coefficients hi normal hain (parent, "seedha padh lo"): $\mathbf{n}=(2,-1,2)$. Constant $7$ normal ka part nahi hai — yeh sirf plane ko $\mathbf{n}$ ke along slide karta hai.

(b) Easy coordinates pick karo aur last wala solve karo. $x=0,y=0$ lete hain: toh $2z=7\Rightarrow z=3.5$. Isliye $(0,0,3.5)$ kaam karta hai. Check: $2(0)-0+2(3.5)=7$ ✓.

Direction = points ke beech displacement (parent: "do points se line"): $\mathbf{d}=B-A=(4-1,2-2,5-(-1))=(3,0,6).$ Vector form: $\mathbf{r}=(1,2,-1)+t(3,0,6)$. Parametric: $x=1+3t,y=2,z=-1+6t$. Symmetric: har ek ko $t$ ke liye solve karo. Lekin $y$ ka direction-component $0$ hai — hum $\frac{y-2}{0}$ nahi likh sakte. Isliye hum $y=2$ alag state karte hain: $\frac{x-1}{3}=\frac{z+1}{6},y=2.$ Split kyun: zero component se divide karna illegal hai; woh coordinate simply frozen hai.

(a) $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$ use karo. $\mathbf{r}=(x,y,z)$ likho aur dot product ko term by term expand karo: $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=2(x-1)-2(y-1)+1(z-1)=0.$ Yeh standard form mein kyun split hota hai: dot product linear hai — usey distribute karo, aur yeh "moving" part $\mathbf{r}\cdot \mathbf{n}=2x-2y+z$ aur "fixed" part $\mathbf{a}\cdot \mathbf{n}$ mein separate ho jaata hai. $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=\mathbf{r}\cdot \mathbf{n}-\mathbf{a}\cdot \mathbf{n}=0$ rearrange karne se $\mathbf{r}\cdot \mathbf{n}=\mathbf{a}\cdot \mathbf{n}$ milta hai. Woh right-hand constant isliye $d=\mathbf{a}\cdot \mathbf{n}$ hone pe majboor hai — yeh sirf woh fixed number hai jise moving part equal karna hai: $2x-2y+z=d,d=\mathbf{a}\cdot \mathbf{n}=2-2+1=1.$ Toh plane hai $2x-2y+z=1$.

(b) Point-to-plane distance $=\frac{|\mathbf{p}\cdot \mathbf{n}-d|}{|\mathbf{n}|}$. Yahan $\mathbf{p}=O$, isliye $\mathbf{p}\cdot \mathbf{n}=0$: $\text{dist}=\frac{|0-1|}{\sqrt{2^2+(-2{)}^2+1^2}}=\frac{1}{\sqrt{9}}=\frac{1}{3}.$

Step 1 — parallel? Directions hain ${\mathbf{d}}_1=(1,1,0)$, ${\mathbf{d}}_2=(0,0,1)$. Ek doosre ka scalar multiple nahi hai (koi $k$ nahi hai jo $(1,1,0)=k(0,0,1)$ de), isliye parallel nahi.

Step 2 — kya woh milte hain? Do general points ko equal set karo aur dekho ki kya koi $(t,s)$ saare teen coordinates solve karta hai (yeh ek chhota system hai): $1+t=0,t=1,0=s.$ Pehle equation se $t=-1$; doosre se $t=1$. Contradiction — koi ek $t$ kaam nahi karta. Isliye equations ka koi solution nahi hai.

Conclusion: parallel nahi aur kabhi nahi milti $\Rightarrow$ lines skew hain (woh different heights pe pass karti hain, jaise ek overpass road ke upar — figure dekho).

Idea: line ke point ke coordinates hain $(1+t,1+2t,1+3t)$. Yeh plane pe lie karta hai exactly jab plane ke equation ko satisfy kare — isliye substitute karo (dekho Intersection of lines and planes): $(1+t)+(1+2t)+(1+3t)=12\Rightarrow 3+6t=12\Rightarrow t=\frac{3}{2}.$ $t=\frac{3}{2}$ wapas line mein daalo: $\left(1+\frac{3}{2},1+3,1+\frac{9}{2}\right)=\left(\frac{5}{2},4,\frac{11}{2}\right).$ Check: $\frac{5}{2}+4+\frac{11}{2}=\frac{5+8+11}{2}=\frac{24}{2}=12$ ✓.

(a) Plane ke andar do spanning directions banao, phir normal ke liye unka cross product lo (parent: $\mathbf{n}=\mathbf{u}\times \mathbf{v}$): $\mathbf{u}=B-A=(1,1,0),\mathbf{v}=C-A=(0,1,1).$ $\mathbf{n}=\mathbf{u}\times \mathbf{v}=(1\cdot 1-0\cdot 1,0\cdot 0-1\cdot 1,1\cdot 1-1\cdot 0)=(1,-1,1).$ Plane: $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$ jahan $\mathbf{a}=A=(1,0,0)$: $1(x-1)-1(y)+1(z)=0\Rightarrow x-y+z=1.$ Har point check karo: $A:1-0+0=1$ ✓, $B:2-1+0=1$ ✓, $C:1-1+1=1$ ✓.

(b) $P=(3,3,3)$ se distance $\frac{|\mathbf{p}\cdot \mathbf{n}-d|}{|\mathbf{n}|}$ use karke jahan $d=1$: $\mathbf{p}\cdot \mathbf{n}=3-3+3=3,|\mathbf{n}|=\sqrt{1+1+1}=\sqrt{3}.$ $\text{dist}=\frac{|3-1|}{\sqrt{3}}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\approx \mathrm{1.155.}$

$\text{dist}=\frac{|(\mathbf{p}-\mathbf{a})\times \mathbf{d}|}{|\mathbf{d}|}$ use karo. Cross product perpendicular distance kyun deta hai: us parallelogram ka picture banao jiske sides $\mathbf{p}-\mathbf{a}$ (line ke point $\mathbf{a}$ se target $P$ tak) aur $\mathbf{d}$ (line ki direction) hain. Cross product ka magnitude us parallelogram ka area hai: $|(\mathbf{p}-\mathbf{a})\times \mathbf{d}|=|\mathbf{p}-\mathbf{a}||\mathbf{d}|\sin \theta$, jahan $\theta$ do sides ke beech ka angle hai. Lekin area yeh bhi equals karta hai $\text{base}\times \text{height}$, jahan base $|\mathbf{d}|$ hai aur height = $P$ se line tak ka perpendicular gap. Toh area ko base $|\mathbf{d}|$ se divide karne se exactly woh perpendicular height milti hai — precisely woh distance jo hum chahte hain. $\mathbf{p}-\mathbf{a}=(1,1,1),\mathbf{d}=(1,0,0).$ $(\mathbf{p}-\mathbf{a})\times \mathbf{d}=(1\cdot 0-1\cdot 0,1\cdot 1-1\cdot 0,1\cdot 0-1\cdot 1)=(0,1,-1).$ $|(0,1,-1)|=\sqrt{0+1+1}=\sqrt{2},|\mathbf{d}|=1.$ $\text{dist}=\frac{\sqrt{2}}{1}=\sqrt{2}\approx \mathrm{1.414.}$ Sanity: $x$-axis pe $(1,1,1)$ se closest point $(1,0,0)$ hai; gap hai $(0,1,1)$ jiska length $\sqrt{2}$ hai ✓.

(a) Intersection. Ek line point hai $(1+2t,2-t,3+2t)$. $\Pi$ mein substitute karo: $2(1+2t)+(2-t)-2(3+2t)=3.$ $2+4t+2-t-6-4t=3\Rightarrow (-2)+(-t)=3\Rightarrow -t=5\Rightarrow t=-5.$ Point: $(1+2(-5),2-(-5),3+2(-5))=(-9,7,-7)$. Check: $2(-9)+7-2(-7)=-18+7+14=3$ ✓.

(b) Line aur plane ke beech angle. Pehle, is angle ka matlab kya hai? $\varphi$ ko line aur plane ke beech ka acute angle maano — yaani line ko surface ki taraf neeche tilt karo aur line aur plane pe uske shadow ke beech ka chhota wedge measure karo. Jab $\varphi =0^{\circ }$ hota hai line plane mein flat lie karti hai; jab $\varphi =90^{\circ }$ hota hai line seedha andar ghus jaati hai, normal ke parallel. Yahi woh quantity hai jo hum compute kar rahe hain.

Line ka arrow along point karta hai; plane ka normal across point karta hai (parent mistake (d) warn karta hai inhe mix na karo). Angle $\varphi$ satisfy karta hai $\sin \varphi =\frac{|\mathbf{d}\cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}.$ $\sin$ kyun, $\cos$ nahi? Dot product $\mathbf{d}$ aur normal $\mathbf{n}$ ke beech angle deta hai. Lekin plane apne normal se $90^{\circ }$ door hoti hai, isliye line-to-plane angle $\varphi$ line-to-normal angle ka complement hai — aur ek angle ka $\cos$ uske complement ka $\sin$ equal karta hai. $\mathbf{d}=(2,-1,2),\mathbf{n}=(2,1,-2):\mathbf{d}\cdot \mathbf{n}=4-1-4=-1.$ $|\mathbf{d}|=\sqrt{4+1+4}=3,|\mathbf{n}|=\sqrt{4+1+4}=3.$ $\sin \varphi =\frac{|-1|}{3\cdot 3}=\frac{1}{9}\Rightarrow \varphi =\arcsin \frac{1}{9}\approx 6.38^{\circ }.$

(c) $O$ se perpendicular ka foot aur distance. $O$ se normal $\mathbf{n}$ ke along ek line drop karo: $\mathbf{r}=(0,0,0)+\lambda (2,1,-2)$. Yeh $\Pi$ se milti hai jab $2(2\lambda )+(1\lambda )-2(-2\lambda )=3\Rightarrow 4\lambda +\lambda +4\lambda =3\Rightarrow 9\lambda =3\Rightarrow \lambda =\frac{1}{3}.$ Foot: $\left(\frac{2}{3},\frac{1}{3},-\frac{2}{3}\right)$. Distance = $O\to$foot ki length $=|\lambda ||\mathbf{n}|=\frac{1}{3}\cdot 3=1$. Direct formula se cross-check: $\frac{|\mathbf{p}\cdot \mathbf{n}-d|}{|\mathbf{n}|}=\frac{|0-3|}{3}=1$ ✓.

Line point $(1+t,t,1-t)$ ko plane mein substitute karo: $(1+t)-(t)+(1-t)=2\Rightarrow 2-t=2\Rightarrow t=0.$ Ruko — yeh ek single $t$ deta hai. Re-examine karo: $1+t-t+1-t=2-t$. $2-t=2$ set karne se $t=0$ milta hai, ek point $(1,0,1)$.

Lekin interesting degenerate case tab hota hai jab line ki direction normal ke perpendicular ho (toh line plane ke parallel chalti hai). Check karo: $\mathbf{d}\cdot \mathbf{n}=(1,1,-1)\cdot (1,-1,1)=1-1-1=-1\ne 0$, isliye line parallel nahi hai — yeh plane ko ek baar pierce karti hai, $(1,0,1)$ pe.

General rule (saare cases):

• $\mathbf{d}\cdot \mathbf{n}\ne 0$ → line plane ko exactly ek point mein pierce karti hai.
• $\mathbf{d}\cdot \mathbf{n}=0$ aur start point plane ke bahar hai → line parallel hai, kabhi touch nahi karti → koi solution nahi.
• $\mathbf{d}\cdot \mathbf{n}=0$ aur start point plane pe hai → line plane ke andar lie karti hai → infinitely many points.

Yahan pehla case apply hota hai: ek point, $(1,0,1)$.