Question bank — Lines and planes in 3D — vector equations

Lines $\mathbf{r}=\mathbf{a}+t\mathbf{d}$ aur $\mathbf{r}=\mathbf{a}+t(2\mathbf{d})$ alag lines hain. ::: False. Direction ko double karne se sirf ye badalta hai ki dial $t$ tumhe kitni tezi se chalata hai, kaunse points reach hote hain nahin. Points ka set identical hai, toh ye same line hai. Parallel direction vectors wali do lines necessarily same line hoti hain. ::: False. Parallel directions unhe parallel lines banata hai, lekin wo alag-alag shift ho sakti hain (jaise do rails). Wo tabhi coincide karti hain jab additionally koi ek point bhi share karein. $2x-y+3z=5$ jaisi ek linear equation 3D mein ek line describe karti hai. ::: False. 3D mein ek linear equation sirf ek degree of freedom pin karti hai, ek 2D flat sheet — yaani plane — bacha deti hai. 3D mein ek line ke liye do aisi equations chahiye (ya ek vector equation). Vector $\mathbf{r}=\mathbf{a}+t\mathbf{d}$ kisi $t$ ke liye origin se guzarta hai sirf tab jab origin line par ho. ::: True. Origin reach karne ka matlab hai $\mathbf{a}+t\mathbf{d}=0$ ka koi solution $t$ ho; ye solution tab exist karta hai jab origin line ka ek point ho — ye guaranteed nahin hai. $ax+by+cz=d$ mein puri equation ko $2$ se scale karne par alag plane milti hai. ::: False. $2ax+2by+2cz=2d$ ka same solution set hai; normal sirf length mein double hua, direction same rahi. Same plane. Parametric plane $\mathbf{r}=\mathbf{a}+s\mathbf{u}+t\mathbf{v}$ ke liye $\mathbf{u}$ aur $\mathbf{v}$ perpendicular hona zaroori hai. ::: False. Unhe sirf non-parallel hona chahiye taaki wo do genuinely alag directions span karein. Skewed (non-perpendicular) spanning vectors bhi poora plane sweep kar lete hain. Ek point se plane tak ki distance negative ho sakti hai. ::: False. Distance ek length hai, isliye formula mein absolute value $|\mathbf{p}\cdot \mathbf{n}-d|$ use hoti hai jahaan $d=\mathbf{a}\cdot \mathbf{n}$. Unsigned quantity $\mathbf{p}\cdot \mathbf{n}-d$ negative ho sakti hai — wo sirf ye batati hai ki tum plane ke kis side par ho. Agar line ka direction vector $\mathbf{d}=(2,4,0)$ hai, toh line horizontal hai (constant height $z$). ::: True. $\mathbf{d}$ ka $z$-component $0$ hai, isliye line ke saath chalne par $z$ kabhi nahin badlega. Har point us fixed height par hoga jahan se $\mathbf{a}$ shuru hua tha. Plane mein har point $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$ satisfy karta hai, aur sirf plane ke points karte hain. ::: True. In-plane displacements $\mathbf{r}-\mathbf{a}$, $\mathbf{n}$ ke perpendicular hote hain (dot $=0$); plane se bahar koi bhi point ek aisa displacement deta hai jiska $\mathbf{n}$ ke saath nonzero component hota hai, isliye dot product zero nahin hota.

"Plane $3x+y-2z=7$, toh iska normal $(3,1,-2,7)$ hai." ::: Normal sirf $(3,1,-2)$ hai — $x,y,z$ ke coefficients. $7$ constant $d=\mathbf{a}\cdot \mathbf{n}$ hai; ye plane ko normal ke saath slide karta hai lekin iska koi direction nahin, isliye ye normal vector ka hissa nahin ho sakta. "$(1,0,2)$ aur $(3,4,2)$ se guzarne wali line: symmetric form $\frac{x-1}{2}=\frac{y}{4}=\frac{z-2}{0}$." ::: Kabhi bhi $0$ se divide nahin kar sakte. Kyunki $\mathbf{d}$ mein $c=0$ hai, $z=2$ alag se likhto aur sirf $x$ aur $y$ ke liye symmetric form use karo: $\frac{x-1}{2}=\frac{y}{4},z=2$. "Spanning vectors $\mathbf{u},\mathbf{v}$ se plane ka normal nikalne ke liye $\mathbf{u}\cdot \mathbf{v}$ compute karo." ::: Dot product ek number return karta hai, direction nahin. Normal ko ek aisa vector hona chahiye jo dono $\mathbf{u}$ aur $\mathbf{v}$ ke perpendicular ho, aur yahi exactly $\mathbf{u}\times \mathbf{v}$ (cross product) produce karta hai — Figure s02 mein perpendicular arrow dekho. "Line aur plane parallel hain kyunki line ki direction plane ke normal ke equal hai." ::: Ye ulta hai. Agar direction normal ke equal hai, toh line plane ko right angle par chedhti hai. Line tab parallel hoti hai plane ke, jab uski direction normal ke perpendicular ho (dot product zero). "Point–to–line distance $=\frac{(\mathbf{p}-\mathbf{a})\times \mathbf{d}}{|\mathbf{d}|}$." ::: Numerator ek magnitude hona chahiye: $\frac{|(\mathbf{p}-\mathbf{a})\times \mathbf{d}|}{|\mathbf{d}|}$. Cross product ek vector hai; distance ek single non-negative number hai, isliye tum uski length lete ho. "Point $A=(2,-1,3)$ plane $x+2y+2z=6$ par hai? Check: $2+2(-1)+2(3)=2-2+6=8\ne 6$." ::: Arithmetic slip hai: $2-2+6=6$, $8$ nahin. Ye $d=6$ ke equal hai, isliye $A$ actually plane par hai — jaisa hona chahiye tha, kyunki plane $A$ se hokar banaya gaya tha.

Line equation ek point $\mathbf{a}$ aur direction $\mathbf{d}$ ko alag kyun rakhti hai instead of $y=mx+c$ jaisa slope use karne ke? ::: Slope division karta hai aur vertical lines ke liye break ho jata hai; position ko direction se alag karne par division nahin hoti aur ye kisi bhi dimension mein kaam karta hai, including 3D jahaan "slope" properly defined bhi nahin hai. Normal form $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$ sirf ek equation se poora 2D plane kaise capture kar leti hai? ::: Ek dot-product-equals-zero condition exactly ek degree of freedom remove karti hai ("$\mathbf{n}$ ke along kitna dur" direction), do in-plane freedoms intact reh jaate hain — precisely ek plane. Hum plane ka normal banane ke liye cross product use kyun karte hain, dot product nahin? ::: Hume ek aisa vector chahiye jo do given directions ke perpendicular ho ek saath. Cross product exactly wahi orthogonal vector output karne ke liye defined hai (Figure s02); dot product ek scalar output karta hai aur kisi direction mein point nahin kar sakta. $|(\mathbf{p}-\mathbf{a})\times \mathbf{d}|$ ko $|\mathbf{d}|$ se divide karne par line tak perpendicular distance kyun milti hai? ::: Cross product ki length parallelogram area $|\mathbf{p}-\mathbf{a}||\mathbf{d}|\sin \theta$ hai (Figure s01). $|\mathbf{d}|$ divide karne par $|\mathbf{p}-\mathbf{a}|\sin \theta$ bachta hai, jo exactly $P$ se line tak ki perpendicular height hai. Point–to–plane formula mein $|\mathbf{n}|$ se divide kyun karte hain? ::: Tumhe measure karna hai ki displacement unit normal ke along kitna project karta hai; $\mathbf{n}$ ko uski apni length se divide karke $\hat{\mathbf{n}}$ banaya jata hai, taaki $|(\mathbf{p}-\mathbf{a})\cdot \hat{\mathbf{n}}|$ ek true length ho, $\mathbf{n}$ ki size se scale na ho. Do points se guzarne wali line $(1-t)\mathbf{a}+t\mathbf{b}$ likhna itna convenient kyun hai? ::: Weights $(1-t)$ aur $t$ hamesha $1$ sum karte hain, isliye $t=0$ par $A$ milta hai, $t=1$ par $B$ milta hai, aur $0\le t\le 1$ exactly unke beech ka segment trace karta hai. Do alag $(\mathbf{a},\mathbf{d})$ pairs same line describe kyun kar sakte hain? ::: Ek line points ka set hai, koi formula nahin. Us par ka koi bhi point $\mathbf{a}$ ke roop mein kaam karta hai, aur direction ka koi bhi scalar multiple $\mathbf{d}$ ke roop mein, isliye infinitely many descriptions ek geometric object deti hain.

Agar $\mathbf{d}=0$ ho toh line ka vector equation kya hoga? ::: Wo collapse ho jata hai: har $t$ sirf ek point $\mathbf{a}$ deta hai. Zero direction ka koi "chalne ka raasta" nahin hai, isliye tum ek point pate ho, line nahin — isliye definition mein $\mathbf{d}\ne 0$ zaroori hai. Kya hoga agar tum $\mathbf{r}=\mathbf{a}+s\mathbf{u}+t\mathbf{v}$ plane banane ki koshish karo jab $\mathbf{u}$ parallel ho $\mathbf{v}$ ke? ::: Dono spanning arrows same direction mein point karte hain, isliye combinations sirf ek line sweep karte hain, plane nahin. Unka cross product $\mathbf{u}\times \mathbf{v}=0$ failure signal karta hai — koi normal exist nahin karta. Kya teen points plane determine karne mein fail ho sakte hain? ::: Haan — agar teen points collinear hain (sab ek line par), toh infinitely many planes us line ko contain karti hain. Tab $\mathbf{u}=B-A$ aur $\mathbf{v}=C-A$ parallel hote hain aur $\mathbf{u}\times \mathbf{v}=0$. Point–to–line distance formula kya deta hai jab $P$ line par hi ho? ::: Tab $\mathbf{p}-\mathbf{a}$ $\mathbf{d}$ ke parallel hai, isliye $\sin \theta =0$ aur cross product $0$ hai. Distance $=0$ — bilkul sahi, point line par hai. Plane-distance formula mein $\mathbf{p}\cdot \mathbf{n}-d=0$ (jahaan $d=\mathbf{a}\cdot \mathbf{n}$) ka kya matlab hai? ::: Point $P$ plane equation exactly satisfy karta hai, isliye wo plane par hi hai aur uski distance $0$ hai. Us expression ka sign (jab nonzero ho) batata hai ki tum plane ke kis side par ho. Agar ek plane origin se guzarti hai, toh uska constant $d$ kya hoga? ::: $\mathbf{a}=0$ ke saath $d=\mathbf{a}\cdot \mathbf{n}$ deta hai $d=0$. Toh equation hai $ax+by+cz=0$ — origin se guzarne wale plane mein hamesha zero constant term hoti hai. Ek line poori tarah ek plane ke andar hai — line ki direction $\mathbf{d}$ aur plane ke normal $\mathbf{n}$ ke baare mein kya true hona chahiye? ::: $\mathbf{d}\cdot \mathbf{n}=0$ (direction, normal ke perpendicular hai, taaki line plane se bahar na niklein) aur line ka ek point plane equation satisfy karta ho. Dono conditions zaroori hain; pehli akeli sirf unhe parallel banati hai.