Three Dimensional Geometry: Class 12 Maths Chapter 11 Hinglish mein

Space mein lines: woh kis taraf point karti hain, unhe kaise likhte hain, aur woh ek doosre se kitni door se guzarti hain

Mathematics Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

75 min
29 practice questions
18 MCQs

Is chapter mein aap kya seekhenge

Kisi line ke direction cosines uske direction angles se, kisi bhi direction ratios ke set se, ya us par ke do points se nikaalo
Proportional direction ratios se tay karo ki teen points collinear hain ya nahi, aur do lines parallel hain ya nahi
Known point se guzarne wali aur known direction wali line ko r = a + λb ki tarah likho, aur uske vector, parametric aur Cartesian forms ke beech aana-jaana karo
Do lines ke beech ka acute angle unke direction ratios ya vector equations se nikaalo, aur check karo ki woh perpendicular hain ya parallel
Intersecting, parallel aur skew lines mein farak pehchaano, aur samjhao ki skew lines ke beech ka sabse chhota gap dono ko right angle par kyun milta hai
Do skew lines ke beech, aur do parallel lines ke beech ka shortest distance dot aur cross products se calculate karo

NCERT Mathematics Class 12, Chapter 11, pp. 377–393 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

11.1 Introduction

Kisi metro viaduct ke neeche khade hokar upar dekho. Track tumhare upar ek taraf jaata hai aur road tumhare pairon ke neeche doosri taraf. Yeh kabhi milte nahi, phir bhi parallel bhi nahi hain. Jahan yeh sabse paas aate hain, wahan inke beech kitni doori hai? Iska jawab dene ke liye space mein lines ki ek precise language chahiye.

Pehle hum batayenge ki line kis taraf point karti hai, uske direction cosines aur direction ratios se. Phir aayenge line ki equation vector aur Cartesian form mein, do lines ke beech ka angle, aur na milne wali do lines ke beech ka shortest distance.

Tools tumhare paas pehle se hain: Class XI se space mein points ke coordinates, aur Chapter 10 se vectors aur unke dot aur cross products. Vectors algebra ko chhota rakhte hain; coordinates dikhate hain ki har formula ka matlab kya hai.

Quick check

3 mein se 1

Class XI se warm-up: point

(3, 1, - 3)

point

(1, - 2, 3)

se kitni door hai?

units

Answer dekho

Answer 7 units

Differences 2, 3 aur

- 6

hain, isliye distance

\sqrt{4 + 9 + 36} = \sqrt{49} = 7

hai.

11.2 Line ke Direction Cosines aur Direction Ratios

Origin $O$ se guzarne wali ek line $L$ lo aur uski do directions mein se ek chunkar us par arrow lagao; ab yeh ek directed line hai. Iski direction teen angles se tay hoti hai: positive $x$ -axis ke saath $α$ , positive $y$ -axis ke saath $β$ aur positive $z$ -axis ke saath $γ$ , har ek $0^{\circ}$ aur $180^{\circ}$ ke beech. Yeh iske direction angles hain.

Definition

Direction cosines

Direction angles $α$ , $β$ , $γ$ wali directed line ke liye numbers $l = \cos α$ , $m = \cos β$ aur $n = \cos γ$ uske direction cosines hain.

ExamplePositive $z$ -axis ke direction angles $90^{\circ}, 90^{\circ}, 0^{\circ}$ hain aur direction cosines $0, 0, 1$ .

Angles ki jagah cosines kyun? $L$ par woh point $P$ mark karo jo $O$ se ek unit door hai, us taraf jidhar arrow point karta hai. $P$ ka $x$ -coordinate $O P$ ka $x$ -axis par projection hai, jo $1 \times \cos α$ hai. $y$ aur $z$ ke liye bhi yahi hota hai (figure $y$ ke liye dikhata hai), isliye $P$ bilkul point $(l, m, n)$ hai aur $L$ ke along unit vector $l \hat{i} + m \hat{j} + n \hat{k}$ hai (Fig. 11.1).

3/4Har axis ke liye yahi chalta hai: $P = (l, m, n)$ ek box ka door wala corner hai, aur $O P = 1$ se $l^{2} + m^{2} + n^{2} = 1$ milta hai.

Fig. 11.1Direction cosines directed line par unit point ke coordinates hain.NCERT Fig. 11.1 · p. 378

Teen axes O par milte hain: x viewer ki taraf aata hai, y right ki taraf aur z upar. Ek directed line L, O se guzarti hai, aur P uske along ek unit door wala point hai. Doosre step mein O, P aur P se y-axis par gire perpendicular ke foot se bana right triangle shade kiya gaya hai, O par angle beta mark hai, aur y-axis ke along wale edge par m likha hai, jo cos beta ke barabar hai. Teesre step mein teeno axes ke along l, m aur n edges wala ek dashed box banaya gaya hai, jiska diagonal OP hai, aur labels par P = (l, m, n) aur l square plus m square plus n square barabar 1 likha hai. Aakhri step mein arrow ulta ho jaata hai: O ke doosri taraf (−l, −m, −n) par point P prime dikhta hai, har angle jaise alpha 180 degrees minus alpha ban jaata hai, aur har cosine ka sign badal jaata hai.

$O P$ ek aise box ka diagonal hai jiske edges $l$ , $m$ aur $n$ hain, aur $O P = 1$ . Isliye distance formula woh ek rule deta hai jo direction cosines ka har set follow karta hai.

l^{2} + m^{2} + n^{2} = 1

(11.1)

jahan

$l, m, n$: kisi bhi directed line ke direction cosines

Bina arrow wali line ko do mein se koi bhi arrow diya ja sakta hai, aur dono choices alag direction cosines deti hain. Arrow ghumane par har direction angle $θ$ badalkar $180^{\circ} - θ$ ho jaata hai, aur $\cos (180^{\circ} - θ) = - \cos θ$ , isliye doosra set seedha $(- l, - m, - n)$ hai. Isiliye line ke wahi ek set of direction cosines ki baat karne se pehle hum ek arrow fix karte hain.

Yahan line ka $O$ se guzarna zaroori nahi hai. Direction angles sirf is par depend karte hain ki line kis taraf point karti hai, isliye space mein kahin bhi padi line ke direction cosines wahi hote hain jo origin se guzarne wali uski parallel copy ke, same arrow ke saath.

Direction ratios

Direction cosines (11.1) se aapas mein bandhe hote hain, isliye unhe seedha problem se padhna mushkil hota hai. Aksar direction sirf scale tak hi pata hoti hai. "2 east, 1 north aur 2 upar chalo" usi taraf point karta hai jis taraf "4 east, 2 north aur 4 upar chalo". $l, m, n$ ke proportional aise numbers line ke direction ratios hain; kuch books inhe direction numbers bhi kehti hain.

Definition

Direction ratios

Teen numbers $a, b, c$ , jo sab zero nahi hain, direction cosines $l, m, n$ wali line ke direction ratios hain jab kisi real $k \neq 0$ ke liye $a = k l$ , $b = k m$ aur $c = k n$ ho.

Example $2, - 1, 2$ aur $- 4, 2, - 4$ ek hi line ke direction ratios hain.

DerivationDirection ratios se direction cosines tak

Maan lo humein kisi line ke teen direction ratios $a, b, c$ pata hain. Inse vector $\vec{v} = a \hat{i} + b \hat{j} + c \hat{k}$ banao; yeh line ke along point karta hai, ya to arrow ke saath ya uske ulta.

Kisi vector ko uski length se divide karne par uski direction mein unit vector milta hai, aur line ke along unit vector ke components hi uske direction cosines hain. Isliye direction cosines $\pm \vec{v} / ∣ \vec{v} ∣$ ke components hain.

\begin{aligned} (l, m, n) \\ = \pm \frac{(a, b, c)}{r} \end{aligned}

(11.2)

jahan

$r$: components $a, b, c$ wale vector ki length $\sqrt{a^{2} + b^{2} + c^{2}}$
$\pm$: teeno ke liye ek hi sign: plus agar line $\vec{v}$ ke along directed hai, minus agar uske ulta

To line ke direction cosines ke sirf do sets hote hain, lekin direction ratios ke infinitely many sets: kisi set ko kisi bhi non-zero number se scale karne par doosra set milta hai, aur koi bhi do sets ek doosre ke scalings hain.

InteractiveDirection lab

Axes ki ek three-dimensional picture jismein origin se point (a, b, c) tak ek arrow hai, woh dashed box jiska door wala corner yeh point hai, aur arrow ke teeno axes ke saath bane angles ke arcs. Teen sliders direction ratios a, b aur c ko minus 6 se 6 tak set karte hain, aur ek Reverse button arrow ko ulta karta hai. Ek panel vector ki length, direction cosines fractions aur decimals mein, teeno direction angles degrees mein, aur cosines ke squares ka sum dikhata hai. Do aur sliders, ya picture par drag karna, view ko ghumata aur jhukata hai. Jab tak learner kuch nahi karta, kuch nahi hilta.

Fig. 11.2Teen direction ratios set karo aur direction cosines aur angles padho. Picture ko ghumane ke liye drag karo (ya sliders use karo).

direction lab mein $a, b, c$ ko $2, - 1, 2$ par aur phir $4, - 2, 4$ par set karo: arrow bada hota hai lekin cosines wahi rehte hain. Reverse dabao aur dekho ki har cosine ka sign badalta hai jabki har angle apna supplement ban jaata hai.

Quick check

3 mein se 1

Kya koi line teeno axes ke saath

45^{\circ}

ka angle bana sakti hai?

Haan;

0^{\circ}

180^{\circ}

tak ke koi bhi teen angles possible hainHaan, lekin sirf tab jab line origin se guzreNahi; cosines ke squares ka sum

\frac{3}{2}

aayega, 1 nahiNahi; line axes ke saath zyada se zyada do barabar angles bana sakti hai

Answer dekho

Answer (C) Nahi; cosines ke squares ka sum $\frac{3}{2}$ aayega, 1 nahi

Har cosine

\frac{1}{\sqrt{2}}

hoga, aur teen halves ka sum

\frac{3}{2} \neq 1

hai. Position se farak nahi padta (option B), aur line teen barabar angles bana sakti hai: har ek lagbhag

{54.7}^{\circ}

, jaise tum Practice 11.1 mein dekhoge.

11.2.1 Do points se guzarne wali line ke direction cosines

Do points ek line fix karte hain, aur jab hum tay kar lete hain ki pehle se doosre ki taraf chalenge, to ek direction bhi. Maan lo points $P$ aur $Q$ ke coordinates $(x_{1}, y_{1}, z_{1})$ aur $(x_{2}, y_{2}, z_{2})$ hain.

Vector $\vec{P Q}$ ke components $x_{2} - x_{1}$ , $y_{2} - y_{1}$ aur $z_{2} - z_{1}$ hain. Ise iski length $P Q$ se divide karne par line ke along unit vector milta hai, aur unit vector ke components bilkul direction cosines hi hain.

3/3Right triangle $P M Q$ mein $\cos γ = \frac{z_{2} - z_{1}}{P Q}$ .

Fig. 11.3

P Q

ki direction us box se padhi jaati hai jiske opposite corners par

P

aur

Q

hain.NCERT Fig. 11.2 · p. 379

Neeche left mein origin O par chhote axes hain. Points P aur Q ek segment se jude hain. Doosre step mein axes ke parallel edges wala ek dashed box banaya gaya hai jiske opposite corners P aur Q hain; P se Q tak ka raasta banane wale teen edges highlight kiye gaye hain aur un par x2 minus x1, y2 minus y1 aur z2 minus z1 likha hai. Teesre step mein P ke theek upar wala corner M mark hai, right triangle P, M, Q shade kiya gaya hai jiska right angle M par hai, P par PM aur PQ ke beech angle gamma mark hai, aur label par cos gamma barabar z2 minus z1 upon PQ likha hai.

Fig. 11.3 yahi baat picture mein dikhata hai. Teen components us box ke edges hain jiske opposite corners par $P$ aur $Q$ hain, aur $P Q$ uska lamba diagonal hai. Shaded right triangle mein vertical edge $z_{2} - z_{1}$ angle $γ$ ke saath laga hai aur $P Q$ hypotenuse hai, isliye $\cos γ = \frac{z_{2} - z_{1}}{P Q}$ . Signs apne aap sahi aa jaate hain: agar $Q$ , $P$ se neeche hai, to $n < 0$ aur $γ$ obtuse hai.

\begin{aligned} l & = \frac{x_{2} - x_{1}}{P Q} \\ m & = \frac{y_{2} - y_{1}}{P Q} \\ n & = \frac{z_{2} - z_{1}}{P Q} \end{aligned}

(11.3)

jahan

$P Q$: points ke beech ki distance: teeno differences ke squares ke sum ka square root

Khaas taur par $x_{2} - x_{1}$ , $y_{2} - y_{1}$ , $z_{2} - z_{1}$ line $P Q$ ke direction ratios hain. $x_{1} - x_{2}$ , $y_{1} - y_{2}$ , $z_{1} - z_{2}$ bhi hain, jo $Q$ se $P$ ki taraf directed usi line ko describe karte hain.

Example 1NCERT Example 1 · p. 380

Angles se cosines tak

Ek line ke direction angles $120^{\circ}$ , $45^{\circ}$ aur $60^{\circ}$ hain. Uske direction cosines kya hain? Unhe (11.1) se check karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 2NCERT Example 2 · p. 380

Direction ratios se cosines tak

Direction ratios $4, - 4, 7$ ko direction cosines mein badlo, aur woh set chuno jiske liye line positive $z$ -axis ke saath acute angle banati hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 3NCERT Example 3 · p. 380

Ek zip line

Ek adventure camp positions ko metres mein naapta hai. Uski seedhi zip line ek tower ke top $A (10, 20, 90)$ se pahaadi ke neeche ki taraf ek landing platform $B (70, 50, 70)$ tak jaati hai. $A$ se $B$ ki taraf directed line ke direction cosines aur cable ki length nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4NCERT Example 4 · p. 381

Khud axes

Teeno coordinate axes (positive taraf directed) ke direction cosines nikaalo, aur $x y$ -plane mein $O$ se guzarne wali us line ke bhi jo positive $x$ - aur $y$ -axes ke beech ke angle ko bisect karti hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 5NCERT Example 5 · p. 381

Ek line par teen points

Kya points $A (- 1, 4, 2)$ , $B (1, 1, 3)$ aur $C (5, - 5, 5)$ ek seedhi line par hain, yaani kya woh collinear hain? Direction ratios se tay karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

(2, - 1, 4)

(5, 3, 4)

tak ki line

y

-axis ke saath angle

β

banati hai.

\cos β

nikaalo.

Answer dekho

Answer 0.8

Differences

3, 4, 0

hain aur distance 5 hai, isliye

\cos β = \frac{4}{5} = 0.8

. Line height 4 par hi rehti hai, isliye

z

-axis ke saath uska angle

90^{\circ}

hai.

Practice · 6 questions

Practice 11.1: answers ke saath

Short answerEasyNCERT Exercise 11.1, Q1
Ek directed line positive $x$ -, $y$ - aur $z$ -axes ke saath $135^{\circ}$ , $120^{\circ}$ aur $60^{\circ}$ ke angles banati hai. Uske direction cosines likho aur (11.1) verify karo.
Solution dikhao
Answer $- \frac{1}{\sqrt{2}}$ , $- \frac{1}{2}$ , $\frac{1}{2}$
$\cos 135^{\circ} = - \cos 45^{\circ} = - \frac{1}{\sqrt{2}}$ , $\cos 120^{\circ} = - \frac{1}{2}$ aur $\cos 60^{\circ} = \frac{1}{2}$ .
Check: $\frac{1}{2} + \frac{1}{4} + \frac{1}{4} = 1$ .
Short answerMediumNCERT Exercise 11.1, Q2
Ek cube ka ek corner origin par hai aur teen edges positive axes ke along hain. $O$ se opposite corner tak cube ke diagonal ke direction cosines nikaalo, aur $O$ par har edge ke saath uska angle bhi.
Solution dikhao
Answer $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ ; har edge ke saath lagbhag ${54.7}^{\circ}$
Diagonal $(0, 0, 0)$ se $(s, s, s)$ tak jaata hai, isliye $s, s, s$ (ya seedha $1, 1, 1$ ) direction ratios hain.
Inki length $\sqrt{3}$ hai, isliye $O$ se door ki taraf directed line ke direction cosines har ek $\frac{1}{\sqrt{3}}$ hain.
Barabar cosines matlab barabar angles, isliye diagonal har edge ke saath same angle banata hai: $\cos^{- 1} \frac{1}{\sqrt{3}} \approx {54.7}^{\circ}$ .
Doosri taraf directed hone par saare cosines $- \frac{1}{\sqrt{3}}$ hain.
Short answerEasyNCERT Exercise 11.1, Q3
Us line ke direction cosines nikaalo jiske direction ratios $12, - 15, - 16$ hain.
Solution dikhao
Answer $\pm (\frac{12}{25}, - \frac{3}{5}, - \frac{16}{25})$
$144 + 225 + 256 = 625$ , jiska square root 25 hai.
(11.2) se direction cosines $\pm (\frac{12}{25}, - \frac{15}{25}, - \frac{16}{25})$ hain, yaani $\pm (\frac{12}{25}, - \frac{3}{5}, - \frac{16}{25})$ .
ProofMediumNCERT Exercise 11.1, Q4
Prove karo ki points $(3, - 1, 0)$ , $(5, 2, - 4)$ aur $(- 1, - 7, 8)$ ek line par hain.
Solution dikhao
Points ko $A$ , $B$ , $C$ kaho. $A B$ ke direction ratios $2, 3, - 4$ hain.
$A C$ ke direction ratios $- 4, - 6, 8$ hain, jo pehle set ka $- 2$ guna hai.
To $A B ∥ A C$ , aur dono segments $A$ share karte hain. Isliye $A$ , $B$ , $C$ collinear hain, aur $A$ , $B$ aur $C$ ke beech mein hai.
Short answerMediumNCERT Exercise 11.1, Q5
Ek triangle ke vertices $A (4, 2, 3)$ , $B (- 2, - 1, 1)$ aur $C (0, 0, - 1)$ hain. Uski sides ke direction cosines nikaalo, $A \to B$ , $B \to C$ aur $C \to A$ directed.
Solution dikhao
Answer $A B$ : $- \frac{6}{7}, - \frac{3}{7}, - \frac{2}{7}$ ; $B C$ : $\frac{2}{3}, \frac{1}{3}, - \frac{2}{3}$ ; $C A$ : $\frac{2}{3}, \frac{1}{3}, \frac{2}{3}$
$\vec{A B}$ ke components $- 6, - 3, - 2$ hain aur length $\sqrt{36 + 9 + 4} = 7$ .
$\vec{B C}$ ke components $2, 1, - 2$ hain aur length 3.
$\vec{C A}$ ke components $4, 2, 4$ hain aur length 6.
Har ek ko uski length se divide karne par answers milte hain. Check: teeno vectors ka sum zero vector hai, jaisa closed triangle ke liye hona chahiye.
Quick answerEasy
Inmein se kaunse kisi line ke direction cosines ho sakte hain?
$\frac{1}{2}, \frac{1}{2}, \frac{1}{2}$ $\frac{1}{3}, \frac{2}{3}, \frac{2}{3}$ $1, 1, 0$ $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{3}$
Answer dekho
Answer (B) $\frac{1}{3}, \frac{2}{3}, \frac{2}{3}$
Sirf $\frac{1}{9} + \frac{4}{9} + \frac{4}{9}$ hi 1 ke barabar hai. Baaki $\frac{3}{4}$ , 2 aur $\frac{7}{9}$ dete hain; woh direction ratios ho sakte hain, lekin direction cosines nahi.

11.3 Space mein Line ki Equation

Plane mein ek point aur ek slope line fix kar dete hain. Space mein ek slope kaafi nahi hai, lekin do doosri jaankariyan kaafi hain: line par ek point aur uski direction, ya us par do points. Dono se ek hi tarah ki equation milti hai.

11.3.1 Diye gaye point se guzarne wali aur diye gaye vector ke parallel line ki equation

Position vector $\vec{a}$ wala point $A$ aur ek direction $\vec{b} \neq \vec{0}$ fix karo. Position vector $\vec{r}$ wala point $P$ kab $A$ se $\vec{b}$ ke along jaane wali line par hota hai? Theek tab jab $A$ se $P$ tak ka arrow $\vec{b}$ ka multiple ho. Woh arrow $\vec{r} - \vec{a}$ hai, isliye test hai kisi real number $λ$ ke liye $\vec{r} - \vec{a} = λ \vec{b}$ .

Line ki vector equation

\vec{r} = \vec{a} + λ \vec{b}

(11.4)

jahan

$\vec{a}$: line par ek known point $A$ ka position vector
$\vec{b}$: line ke parallel ek vector
$λ$: ek real number, parameter

Ise ek recipe ki tarah padho: $A$ se shuru karo, phir $\vec{b}$ ke $λ$ steps lo. Har real $λ$ line ka ek point deta hai, aur line ka har point theek ek $λ$ se aata hai, isiliye $λ$ ko parameter kehte hain. $λ = 0$ khud $A$ deta hai, $λ = 1$ ek poora step, aur negative $λ$ peeche ki taraf chalta hai (Fig. 11.4).

3/4 $λ = 2.5$ par $\vec{A P} = 2.5 \vec{b}$ .

Fig. 11.4Line ka har point

λ

ki kisi ek value ke liye

\vec{a} + λ \vec{b}

hai.NCERT Fig. 11.3 · p. 382

Axes origin O par milte hain. a likha position-vector arrow O se point A tak jaata hai, aur b likha ek chhota mota arrow A se line ke along shuru hota hai. Ek purple point P, A se guzarti line par chalta hai. Pehle step mein lambda 0 hai aur P, A par hai. Doosre step mein lambda 1 hai, P, b ki tip par hai aur r likha ek arrow O se P tak jaata hai. Teesre step mein lambda 2.5 hai aur P aur aage hai, A se P tak ka hissa highlight kiya gaya hai. Aakhri step mein lambda minus 1.2 hai aur P, A ke peeche hai, b ki doosri taraf. Label r barabar a plus lambda b poori picture mein bana rehta hai.

$\vec{b}$ ke components line ke direction ratios hain, aur direction ratios ka koi bhi set $\vec{b}$ ke components ki tarah use ho sakta hai. Letters ke baare mein ek warning: $a, b, c$ mein se number $b$ , vector $\vec{b}$ aur uski length $∣ \vec{b} ∣$ teen alag cheezein hain.

DerivationVector form se Cartesian form tak

(11.4) ko ek-ek coordinate karke padho. Agar $A = (x_{1}, y_{1}, z_{1})$ hai aur $\vec{b}$ ke components $a, b, c$ hain, to $\vec{a} + λ \vec{b}$ ka $x$ -coordinate $x_{1} + λ a$ hai, aur $y$ aur $z$ ke liye bhi yahi pattern chalta hai.

To point $(x, y, z)$ line par tab hi hota hai jab ek hi $λ$ ke liye neeche ki teeno parametric equations sahi hon.

\begin{aligned} x & = x_{1} + λ a \\ y & = y_{1} + λ b \\ z & = z_{1} + λ c \end{aligned}

(11.5)

Agar $a, b, c$ mein se koi bhi zero nahi hai, to har equation ko $λ$ ke liye solve kiya ja sakta hai. Ek hi $λ$ teeno ke liye kaam karta hai, isliye teeno expressions barabar hone chahiye, aur isse parameter poori tarah hat jaata hai. Is result ko hum line ki equation ka Cartesian form kehte hain.

\begin{aligned} \frac{x - x_{1}}{a} & = \frac{y - y_{1}}{b} \\ = \frac{z - z_{1}}{c} \end{aligned}

(11.6)

jahan

$(x_{1}, y_{1}, z_{1})$: line par ek point
$a, b, c$: line ke direction ratios (direction cosines $l, m, n$ bhi chalte hain)

Do baatein (11.6) ko safely use karne mein madad karti hain. Pehli, numerators "variable minus constant" hone chahiye, variable ka coefficient 1 ke saath; $\frac{3 - x}{2}$ jaise form ko kuch bhi padhne se pehle $\frac{x - 3}{- 2}$ likho. Doosri, agar koi direction ratio 0 hai, maan lo $c = 0$ , to line height $z = z_{1}$ par hi rehti hai. Hum phir bhi shorthand ke liye $\frac{z - z_{1}}{0}$ likhte hain, jiska matlab sirf itna hai ki $z - z_{1} = 0$ .

Line fix karne ka doosra tareeka, do points se, pehle tareeke mein hi badal jaata hai. Agar $A$ aur $B$ ke position vectors $\vec{a}$ aur $\vec{b}$ hain, to line $A B$ , $A$ se guzarti hai aur $\vec{b} - \vec{a}$ ke parallel hai.

Do points se guzarti line

\vec{r} = \vec{a} + λ (\vec{b} - \vec{a})

(11.7)

Example 6NCERT Example 6 · p. 382

Drone ka raasta

Ek drone point $(- 1, 3, 2)$ se nikalta hai aur $4 \hat{i} - \hat{j} + 5 \hat{k}$ ki direction mein seedhi line mein udta hai. Uske raaste ki vector aur Cartesian equations likho, aur $λ = 2$ wala point nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Bhes badli hui line

Line $\frac{3 - x}{2} = \frac{2 y + 1}{4} = \frac{z}{- 3}$ par ek point, uske direction ratios, aur uski vector equation nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Kaunsa point

\vec{r} = (\hat{i} - 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + \hat{j} - \hat{k})

par hai?

(2, 1, - 1)

(3, 1, 2)

(1, - 2, 4)

(5, 0, 1)

Answer dekho

Answer (D) $(5, 0, 1)$

λ = 2

(1 + 4, - 2 + 2, 3 - 2) = (5, 0, 1)

milta hai.

(2, 1, - 1)

direction vector hai, line ka point nahi;

(3, 1, 2)

ke liye

x

λ = 1

chahiye, lekin tab

y

- 1

hoga.

11.4 Do Lines ke beech Angle

Space mein do lines ka milna zaroori nahi, phir bhi unke beech ek angle hota hai: unki directions ke beech ka angle. Har line ki ek copy ko bina ghumaye tab tak khiskao jab tak dono copies ek point $C$ se na guzrein, aur wahan angle naapo (Fig. 11.5). Copies do angles banati hain, $θ$ aur $180^{\circ} - θ$ ; agreement ke hisaab se hum woh batate hain jo obtuse nahi hai.

3/3Copies $θ$ aur $180^{\circ} - θ$ ke angles banati hain; hum acute wala lete hain.

Fig. 11.5Do lines ke beech ka angle unki directions ke beech ka angle hai, acute liya hua.NCERT Fig. 11.4 · p. 383

Axes origin par milte hain. Line L1, solid blue, aur line L2, dashed orange, ek doosre ke paas se bina mile guzarti hain; b1 aur b2 likhe arrows unki directions dikhate hain. Doosre step mein dono lines ki dotted copies ek hi point C se banayi gayi hain, directions wahi rakhkar. Teesre step mein original lines halki ho jaati hain, copies ke beech angle theta ek solid arc se mark hai, aur doosra angle, 180 degrees minus theta, opposite taraf ek dashed arc se mark hai.

Pehli line ke direction ratios $a_{1}, b_{1}, c_{1}$ aur doosri ke $a_{2}, b_{2}, c_{2}$ lo, taaki ${\vec{b}}_{1} = a_{1} \hat{i} + b_{1} \hat{j} + c_{1} \hat{k}$ aur ${\vec{b}}_{2} = a_{2} \hat{i} + b_{2} \hat{j} + c_{2} \hat{k}$ unke along chalein. Dot product in do vectors ke beech ka angle deta hai. Absolute value lene se obtuse answer apne acute supplement mein badal jaata hai, jo lines ke beech ka angle hai.

Do lines ke beech ka angle

\cos θ = \frac{∣ {\vec{b}}_{1} \cdot {\vec{b}}_{2} ∣}{∣ {\vec{b}}_{1} ∣ ∣ {\vec{b}}_{2} ∣}

(11.8)

jahan

${\vec{b}}_{1} \cdot {\vec{b}}_{2}$: $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}$
$∣ {\vec{b}}_{1} ∣$: $\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}}$ , aur isi tarah $∣ {\vec{b}}_{2} ∣$ ke liye
$θ$: lines ke beech ka acute (ya right) angle

Formula ko sirf directions chahiye. Lines $\vec{r} = {\vec{a}}_{1} + λ {\vec{b}}_{1}$ aur $\vec{r} = {\vec{a}}_{2} + μ {\vec{b}}_{2}$ ke liye ${\vec{b}}_{1}$ aur ${\vec{b}}_{2}$ jaise hain waise hi use karo; Cartesian form wali lines ke liye direction ratios denominators se padho (zaroorat ho to pehle standard form mein likhkar). Jab direction cosines pata hon, to (11.8) mein lengths 1 hoti hain.

\cos θ = ∣ s ∣

(11.9)

jahan

$s$: $l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}$ , dono unit vectors ka dot product

Sine isi tarah cross product se aata hai: $\sin θ = \frac{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}{∣ {\vec{b}}_{1} ∣ ∣ {\vec{b}}_{2} ∣}$ . ${\vec{b}}_{1} \times {\vec{b}}_{2}$ ke components $b_{1} c_{2} - b_{2} c_{1}$ , $c_{1} a_{2} - c_{2} a_{1}$ aur $a_{1} b_{2} - a_{2} b_{1}$ hain. Direction cosines ke saath lengths phir se 1 hoti hain.

\sin^{2} θ = u^{2} + v^{2} + w^{2}

(11.10)

jahan

$u$: $l_{1} m_{2} - l_{2} m_{1}$
$v$: $m_{1} n_{2} - m_{2} n_{1}$
$w$: $n_{1} l_{2} - n_{2} l_{1}$ , dono unit vectors ke cross product ke components

Example 7NCERT Example 7 · p. 384

Vector equations se angle

Lines $\vec{r} = (\hat{i} + \hat{j}) + λ (2 \hat{i} - 2 \hat{j} + \hat{k})$ aur $\vec{r} = (3 \hat{i} - \hat{k}) + μ (4 \hat{i} + 4 \hat{j} - 7 \hat{k})$ ek doosre se kis angle par jhuki hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 8NCERT Example 8 · p. 385

Cartesian equations se angle

Do lines Cartesian form mein di gayi hain, $\frac{x - 1}{2} = \frac{y + 3}{1} = \frac{z - 4}{- 1}$ aur $\frac{x}{1} = \frac{y - 2}{- 2} = \frac{z + 5}{- 3}$ . Unke beech ka angle kitna bada hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

k

ki kis value ke liye direction ratios

k, 2, - 1

aur

3, k, 4

wali lines perpendicular hain?

Answer dekho

Answer 0.8

3 k + 2 k - 4 = 0

, isliye

k = \frac{4}{5} = 0.8

11.5 Do Lines ke beech Shortest Distance

Plane mein do lines ya to milti hain ya parallel hoti hain. Space mein ek teesri possibility hai, woh jo metro viaduct ke neeche dikhti hai: neeche ki road aur upar ka track parallel nahi hain, phir bhi kabhi milte nahi (Fig. 11.6). Aisi lines skew lines hain. Koi ek plane dono ko contain nahi karta, isliye woh coplanar nahi hain.

3/3Road ke upar ek flyover: lines na milti hain na parallel chalti hain. Pillar dono ke perpendicular hai.

Fig. 11.6Space mein do lines ya to milti hain, ya parallel chalti hain, ya skew hoti hain.NCERT Fig. 11.5 · p. 385

Tirchhe se dikhti ek flat green zameen par ek seedhi road solid blue line se bani hai. Pehle step mein ek dashed orange road ise ek marked point par cross karti hai, label hai mil gayi, distance 0. Doosre step mein orange road blue road ke saath-saath bina cross kiye chalti hai, label hai parallel road, har jagah same gap. Teesre step mein orange line zameen se upar uthe flyover deck ka edge hai jo road ke aar-paar jaata hai; ek purple vertical pillar deck ko road se jodta hai, dono ends par right-angle marks hain, aur note par na milein, na parallel likha hai.

Definition

Skew lines

Space mein lines ka aisa pair jinka na koi common point hai na common direction. Koi ek plane dono ko nahi rakh sakta.

ExampleEk seedha flyover deck aur woh road jise woh cross karta hai.

Do lines ke beech ka shortest distance us sabse chhote segment ki length hai jiska ek end har line par ho. Milne wali lines ke liye yeh 0 hai. Parallel lines ke liye yeh unke beech ka perpendicular gap hai, jo kahin bhi naapo same rehta hai. Skew lines ke liye theek ek shortest segment hota hai, aur woh dono lines ko right angles par milta hai. Ise unka common perpendicular kehte hain; flyover wali picture mein yeh pillar hai.

Try karke dekho

Do pencils aur ek ruler

Tumhe chahiye

do lambi pencils
clay ka ek tukda ya ek eraser
ek ruler

Pehle guess karo

Tumhe kya lagta hai, pencils ke along sabse chhota gap kahan hoga?

Pencils ke ends parJahan unke beech rakha ruler dono pencils ke saath square hoKahin bhi: gap poori length mein same haiJahan pencils cross karti hui lagti hain

Ek pencil ko table par flat rakho.
Doosri pencil ko clay se upar fix karo taaki woh pehli ke upar se cross kare, bina use chhue aur bina uske parallel hue.
Ruler ko pencils ke beech kai jagah rakho aur har baar gap note karo.
Woh jagah dhoondo jahan gap sabse chhota hai, aur set-square se check karo ki wahan ruler dono pencils ke saath square hai ya nahi.

11.5.1 Do skew lines ke beech distance

DerivationSkew lines ke beech shortest distance

Maan lo lines $l_{1} : \vec{r} = {\vec{a}}_{1} + λ {\vec{b}}_{1}$ aur $l_{2} : \vec{r} = {\vec{a}}_{2} + μ {\vec{b}}_{2}$ hain, jo $A_{1}$ aur $A_{2}$ se guzarti hain. Maan lo $M N$ unka common perpendicular hai, $M$ , $l_{1}$ par aur $N$ , $l_{2}$ par, aur $d = M N$ (Fig. 11.7).

$M N$ , ${\vec{b}}_{1}$ aur ${\vec{b}}_{2}$ dono ke perpendicular hai, isliye yeh unke cross product ke along hai. Unit vector $\hat{n} = \frac{{\vec{b}}_{1} \times {\vec{b}}_{2}}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}$ ke saath humein $\vec{M N} = \pm d \hat{n}$ milta hai.

Ab $A_{1}$ se $A_{2}$ tak lambe raaste se jao: $l_{1}$ ke along $M$ tak, aar-paar $N$ tak, phir $l_{2}$ ke along $A_{2}$ tak. Vectors mein, kuch numbers $s$ aur $t$ ke liye ${\vec{a}}_{2} - {\vec{a}}_{1} = s {\vec{b}}_{1} \pm d \hat{n} + t {\vec{b}}_{2}$ .

Ab dono sides ka $\hat{n}$ ke saath dot product lo aur yaad rakho ki $\hat{n}$ , ${\vec{b}}_{1}$ aur ${\vec{b}}_{2}$ ke bhi perpendicular hai, isliye line ke along wale do hisse hat jaate hain aur sirf $\pm d$ bachta hai. To $d$ , $\vec{A_{1} A_{2}}$ ke $\hat{n}$ par projection ki length hai. $\hat{n}$ ko poora likhne par neeche wala formula milta hai.

Skew lines

d = \frac{∣ \vec{N} \cdot ({\vec{a}}_{2} - {\vec{a}}_{1}) ∣}{∣ \vec{N} ∣}

(11.11)

jahan

$\vec{N}$: dono directions ka cross product ${\vec{b}}_{1} \times {\vec{b}}_{2}$
${\vec{a}}_{1}, {\vec{a}}_{2}$: $l_{1}$ par point $A_{1}$ aur $l_{2}$ par point $A_{2}$ ke position vectors

Poora likhein to $d$ , $({\vec{b}}_{1} \times {\vec{b}}_{2}) \cdot ({\vec{a}}_{2} - {\vec{a}}_{1})$ ki absolute value ko cross product ki length $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣$ se divide karke milta hai.

4/4 $l_{1}$ ke along, $M N$ ke aar-paar, $l_{2}$ ke along: $\hat{n}$ ke saath dot product mein sirf beech wala hissa bachta hai.

Fig. 11.7Common perpendicular

M N

\vec{A_{1} A_{2}}

{\vec{b}}_{1} \times {\vec{b}}_{2}

par projection hai.NCERT Fig. 11.6 · p. 386

Neeche left mein chhote axes hain. Ek solid blue line l1 picture mein neeche se point A1 se guzarti hai, uske along arrow b1 hai, aur ek dashed orange line l2 upar ek alag direction mein A2 se guzarti hai, arrow b2 ke saath. Doosre step mein ek purple arrow A1 ko A2 se jodta hai aur us par a2 minus a1 likha hai. Teesre step mein lines ke beech common perpendicular MN banaya gaya hai, l1 par M aur l2 par N par right-angle marks ke saath, label d hai, aur note kehta hai ki n hat, b1 cross b2 ke parallel hai. Aakhri step mein ek chauda green raasta A1 se l1 ke along M tak, upar N tak aur l2 ke along A2 tak jaata hai, aur label par d barabar (a2 minus a1) dot n hat ki absolute value likha hai.

Formula (11.11) pair ko classify bhi karta hai. Agar ${\vec{b}}_{1} \times {\vec{b}}_{2} = \vec{0}$ hai to lines parallel hain aur formula lagu nahi hota (Section 11.5.2 dekho). Warna zero numerator ka matlab $d = 0$ hai: lines milti hain. Non-zero numerator ka matlab hai ki woh skew hain.

Cartesian form. $(x_{1}, y_{1}, z_{1})$ se guzarti direction ratios $a_{1}, b_{1}, c_{1}$ wali $l_{1}$ , aur $(x_{2}, y_{2}, z_{2})$ se guzarti $a_{2}, b_{2}, c_{2}$ wali $l_{2}$ ke liye, (11.11) ka numerator ek scalar triple product hai, jo ek determinant hai. $Δ x = x_{2} - x_{1}$ aur isi tarah baaki likhkar:

D = ∣ \begin{array}{ccc} Δ x & Δ y & Δ z \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} ∣

(11.12)

d = \frac{∣ D ∣}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}

(11.13)

jahan

$∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣$: $b_{1} c_{2} - b_{2} c_{1}$ , $c_{1} a_{2} - c_{2} a_{1}$ aur $a_{1} b_{2} - a_{2} b_{1}$ ke squares ke sum ka square root

Example 9NCERT Example 9 · p. 388

Skew lines ke beech shortest distance

Lines $\vec{r} = (2 \hat{i} - \hat{j}) + λ (\hat{i} + 2 \hat{j} + 2 \hat{k})$ aur $\vec{r} = (5 \hat{i} - 2 \hat{j} + 4 \hat{k}) + μ (2 \hat{i} + \hat{j} - 2 \hat{k})$ ek doosre ke kitne paas aati hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Zero distance: lines milti hain

Dikhao ki lines $\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - \hat{j} + 2 \hat{k})$ aur $\vec{r} = (- \hat{j} + 7 \hat{k}) + μ (\hat{i} + \hat{j} - \hat{k})$ intersect karti hain, aur common point nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

11.5.2 Parallel lines ke beech distance

Jab do lines parallel hon to hum dono ke liye ek hi direction vector $\vec{b}$ use kar sakte hain, zaroorat ho to ek equation ko rescale karke: $l_{1} : \vec{r} = {\vec{a}}_{1} + λ \vec{b}$ aur $l_{2} : \vec{r} = {\vec{a}}_{2} + μ \vec{b}$ . Woh ek common plane mein hoti hain, aur unke beech ka gap unki length ke along nahi badalta.

DerivationParallel lines ke beech distance

Woh parallelogram banao jiski ek side $\vec{b}$ ho, jo $A_{1}$ se $l_{1}$ ke along rakhi ho, aur doosri side $\vec{A_{1} A_{2}} = {\vec{a}}_{2} - {\vec{a}}_{1}$ ho (Fig. 11.8). Chapter 10 se, iska area $∣ \vec{b} \times ({\vec{a}}_{2} - {\vec{a}}_{1}) ∣$ hai.

Iska base $∣ \vec{b} ∣$ hai, aur iski height dono lines ke beech ka perpendicular gap $d$ hai, kyunki opposite side $l_{2}$ ke along hai. Isliye iska area $∣ \vec{b} ∣ d$ bhi hai. Dono expressions ko barabar rakhne par distance milta hai.

Parallel lines

d = \frac{∣ \vec{b} \times ({\vec{a}}_{2} - {\vec{a}}_{1}) ∣}{∣ \vec{b} ∣}

(11.14)

3/3Iski height gap $d$ hai, isliye area $∣ \vec{b} ∣ d$ bhi hai.

Fig. 11.8Ek parallelogram, uska area nikaalne ke do tareeke.NCERT Fig. 11.7 · p. 387

Do parallel tirchhi lines, neeche solid blue l1 aur upar dashed orange l2. l1 par point A1 se l1 ke along arrow b hai, aur ek purple arrow A1 se l2 par point A2 tak jaata hai. Doosre step mein sides b aur A1A2 wala parallelogram shade kiya gaya hai aur label par area barabar b cross (a2 minus a1) ki absolute value likha hai. Teesre step mein A2 se l1 par uske foot F tak ek dashed perpendicular girta hai, right-angle mark ke saath, label d hai, aur label badalkar area barabar b ki absolute value times d ho jaata hai.

Example 10NCERT Example 10 · p. 388

Parallel lines ke beech distance

Dikhao ki lines $\vec{r} = (\hat{i} + \hat{j}) + λ (\hat{i} - 2 \hat{j} + 2 \hat{k})$ aur $\vec{r} = (4 \hat{i} + 2 \hat{j} - 2 \hat{k}) + μ (2 \hat{i} - 4 \hat{j} + 4 \hat{k})$ parallel hain, aur unke beech ka gap nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

InteractiveShortest distance explorer

Do lines ka three-dimensional view, ek solid blue aur ek dashed orange, chaar ready-made pairs ke saath: ek road aur flyover, Example 9 ki skew lines, doosre example ki milne wali lines, aur Example 10 ki parallel lines. Pehli line par point P aur doosri par point Q do sliders, lambda aur mu, se set hote hain, aur ek purple segment unhe jodta hai. Ek panel P aur Q ke coordinates, length PQ, aur dono direction vectors ke saath PQ ke dot products dikhata hai, saath mein chapter ke formula se mili distance bhi. Closest points button P aur Q ko common perpendicular ke ends par le jaata hai, jahan dono dot products zero hain. Picture ko drag karne ya do view sliders use karne se woh ghoomti aur jhukti hai. Jab tak learner kuch nahi karta, kuch nahi hilta.

Fig. 11.9Lines ka ek pair chuno, phir har line par ek point khiskao aur gap ko jitna ho sake utna chhota karne ki koshish karo. Picture ko ghumane ke liye drag karo.

explorer mein Example 9 chuno aur $λ$ aur $μ$ ko tab tak khiskao jab tak $P Q$ jitna ho sake utna chhota na ho jaaye. Sabse chhoti value jahan tak tum pahunch sakte ho woh 4 hai, aur wahan dono dot products $\vec{P Q} \cdot {\vec{b}}_{1}$ aur $\vec{P Q} \cdot {\vec{b}}_{2}$ 0 hain: shortest segment dono lines ke perpendicular hai. Yahi parallel pair ke saath try karo, jahan har $λ$ ka apna closest $μ$ hota hai.

Quick check

4 mein se 1

Lines

\vec{r} = λ (\hat{i} + \hat{j})

aur

\vec{r} = 3 \hat{k} + μ (\hat{i} - \hat{j})

hain

skew, 3 units doorintersectingparallelskew,

\sqrt{2}

units door

Answer dekho

Answer (A) skew, 3 units door

Directions proportional nahi hain, aur lines parallel planes

z = 0

aur

z = 3

mein hain, isliye woh kabhi nahi milti: woh skew hain. (11.11) se,

{\vec{b}}_{1} \times {\vec{b}}_{2} = - 2 \hat{k}

aur

3 \hat{k}

ke saath iska dot product

- 6

hai, isliye

d = \frac{6}{2} = 3

Practice · 17 questions

Practice 11.2: answers ke saath

ProofEasyNCERT Exercise 11.2, Q1
Lines $L_{1}$ , $L_{2}$ , $L_{3}$ ke direction cosines $\frac{2}{3}, \frac{2}{3}, \frac{1}{3}$ ; $- \frac{2}{3}, \frac{1}{3}, \frac{2}{3}$ aur $\frac{1}{3}, - \frac{2}{3}, \frac{2}{3}$ hain. Prove karo ki har ek baaki do ke perpendicular hai.
Solution dikhao
Pehli aur doosri: $\frac{- 4 + 2 + 2}{9} = 0$ .
Pehli aur teesri: $\frac{2 - 4 + 2}{9} = 0$ .
Doosri aur teesri: $\frac{- 2 - 2 + 4}{9} = 0$ .
Har pair perpendicular hai. (Har set ke squares ka sum bhi 1 hai, isliye yeh sach mein direction cosines hain.)
ProofEasyNCERT Exercise 11.2, Q2
Line $A B$ , $A (2, 0, - 1)$ ko $B (4, 3, 1)$ se jodti hai, aur line $C D$ , $C (1, 1, 5)$ ko $D (5, - 1, 4)$ se. Prove karo ki $A B ⊥ C D$ .
Solution dikhao
Pehli line ke direction ratios $2, 3, 2$ hain aur doosri ke $4, - 2, - 1$ .
Unke products ka sum $8 - 6 - 2 = 0$ hai, isliye lines perpendicular hain.
ProofEasyNCERT Exercise 11.2, Q3
Yahan diye gaye points $E (1, 5, - 2)$ , $F (4, 3, 2)$ , $G (0, - 1, 6)$ aur $H (- 6, 3, - 2)$ ke liye kya $E F$ , $G H$ ke parallel hai? Apna answer prove karo.
Solution dikhao
Pehli line ke direction ratios $3, - 2, 4$ hain aur doosri ke $- 6, 4, - 8$ .
Doosra set pehle ka $- 2$ guna hai, isliye lines parallel hain.
Short answerEasyNCERT Exercise 11.2, Q4
Ek laser beam $(- 2, 0, 5)$ se shuru hoti hai aur direction $\hat{i} - 4 \hat{j} + 3 \hat{k}$ mein chalti hai. Jis line ke along woh chalti hai uski vector equation likho.
Solution dikhao
Answer $\vec{r} = (- 2 \hat{i} + 5 \hat{k}) + λ (\hat{i} - 4 \hat{j} + 3 \hat{k})$
Yahan point ka position vector $\vec{a} = - 2 \hat{i} + 5 \hat{k}$ hai aur direction $\vec{b} = \hat{i} - 4 \hat{j} + 3 \hat{k}$ di gayi hai.
To line hai $\vec{r} = (- 2 \hat{i} + 5 \hat{k}) + λ (\hat{i} - 4 \hat{j} + 3 \hat{k})$ .
Short answerEasyNCERT Exercise 11.2, Q5
Us line ke dono forms, vector aur Cartesian, do jo point $3 \hat{i} + \hat{j} - 2 \hat{k}$ (ek position vector) se shuru hoti hai aur $2 \hat{i} - \hat{j} + 3 \hat{k}$ ke along chalti hai.
Solution dikhao
Answer $\vec{r} = (3 \hat{i} + \hat{j} - 2 \hat{k}) + λ (2 \hat{i} - \hat{j} + 3 \hat{k})$ ; $\frac{x - 3}{2} = \frac{y - 1}{- 1} = \frac{z + 2}{3}$
Vector form yeh hai: $\vec{r} = (3 \hat{i} + \hat{j} - 2 \hat{k}) + λ (2 \hat{i} - \hat{j} + 3 \hat{k})$ .
Cartesian form, point $(3, 1, - 2)$ aur ratios $2, - 1, 3$ se: $\frac{x - 3}{2} = \frac{y - 1}{- 1} = \frac{z + 2}{3}$
Short answerMediumNCERT Exercise 11.2, Q6
Line $\frac{2 - x}{4} = \frac{y + 1}{3} = \frac{z - 7}{2}$ ko bina ghumaye tab tak khiskaya jaata hai jab tak woh $(1, - 3, 4)$ se na guzre. Nayi line ki Cartesian equation likho.
Solution dikhao
Answer $\frac{x - 1}{- 4} = \frac{y + 3}{3} = \frac{z - 4}{2}$
$\frac{2 - x}{4} = \frac{x - 2}{- 4}$ , isliye di gayi line ke direction ratios $- 4, 3, 2$ hain.
Parallel line ke ratios wahi hote hain, isliye $(1, - 3, 4)$ se guzarti line hai $\frac{x - 1}{- 4} = \frac{y + 3}{3} = \frac{z - 4}{2}$
Short answerEasyNCERT Exercise 11.2, Q7
Line $\frac{x + 2}{5} = \frac{y - 7}{- 2} = \frac{z + 1}{4}$ ko vector form mein likho.
Solution dikhao
Answer $\vec{r} = (- 2 \hat{i} + 7 \hat{j} - \hat{k}) + λ (5 \hat{i} - 2 \hat{j} + 4 \hat{k})$
Line $(- 2, 7, - 1)$ se guzarti hai aur iske direction ratios $5, - 2, 4$ hain.
To $\vec{r} = (- 2 \hat{i} + 7 \hat{j} - \hat{k}) + λ (5 \hat{i} - 2 \hat{j} + 4 \hat{k})$ .
8
Short answerMediumNCERT Exercise 11.2, Q8
Lines ka har pair kitna angle banata hai?
1. $\vec{r} = (\hat{i} + 2 \hat{j}) + λ (\hat{i} + 4 \hat{j} + 8 \hat{k})$ aur $\vec{r} = (3 \hat{j} - \hat{k}) + μ (2 \hat{i} - \hat{j} + 2 \hat{k})$
  Solution dikhao
  Answer $\cos^{- 1} \frac{14}{27} \approx {58.8}^{\circ}$
  ${\vec{b}}_{1} \cdot {\vec{b}}_{2} = 2 - 4 + 16 = 14$ , $∣ {\vec{b}}_{1} ∣ = 9$ aur $∣ {\vec{b}}_{2} ∣ = 3$ .
  To $\cos θ = \frac{14}{27}$ aur $θ \approx {58.8}^{\circ}$ .
2. $\vec{r} = (\hat{i} - \hat{j} + \hat{k}) + λ (2 \hat{i} + \hat{j} - 2 \hat{k})$ aur $\vec{r} = (2 \hat{i} + 3 \hat{k}) + μ (6 \hat{i} - 2 \hat{j} + 3 \hat{k})$
  Solution dikhao
  Answer $\cos^{- 1} \frac{4}{21} \approx {79.0}^{\circ}$
  ${\vec{b}}_{1} \cdot {\vec{b}}_{2} = 12 - 2 - 6 = 4$ , $∣ {\vec{b}}_{1} ∣ = 3$ aur $∣ {\vec{b}}_{2} ∣ = 7$ .
  To $\cos θ = \frac{4}{21}$ aur $θ \approx {79.0}^{\circ}$ .
9
Short answerMediumNCERT Exercise 11.2, Q9
Yeh pairs Cartesian form mein hain. Har case mein angle nikaalo.
1. $\frac{x - 1}{3} = \frac{y + 2}{- 1} = \frac{z}{2}$ aur $\frac{x + 3}{1} = \frac{y - 1}{2} = \frac{z - 4}{- 4}$
  Solution dikhao
  Answer $\cos^{- 1} \frac{1}{\sqrt{6}} \approx {65.9}^{\circ}$
  Direction ratios $3, - 1, 2$ aur $1, 2, - 4$ . Products ka sum $3 - 2 - 8 = - 7$ hai.
  Lengths $\sqrt{14}$ aur $\sqrt{21}$ hain, jinka product $7 \sqrt{6}$ hai.
  To $\cos θ = \frac{7}{7 \sqrt{6}} = \frac{1}{\sqrt{6}}$ aur $θ \approx {65.9}^{\circ}$ .
2. $\frac{x}{1} = \frac{y - 3}{2} = \frac{z + 1}{2}$ aur $\frac{x - 2}{4} = \frac{y + 5}{4} = \frac{z - 1}{7}$
  Solution dikhao
  Answer $\cos^{- 1} \frac{26}{27} \approx {15.6}^{\circ}$
  Direction ratios $1, 2, 2$ (length 3) aur $4, 4, 7$ (length 9).
  Products ka sum $4 + 8 + 14 = 26$ hai, isliye $\cos θ = \frac{26}{27}$ aur $θ \approx {15.6}^{\circ}$ .
Short answerHardNCERT Exercise 11.2, Q10
$p$ ki kaunsi value lines $\frac{2 - x}{2} = \frac{3 y - 6}{p} = \frac{z + 1}{4}$ aur $\frac{1 - x}{p} = \frac{y + 2}{6} = \frac{3 - z}{2}$ ko perpendicular banati hai?
Solution dikhao
Answer $p = 2$
Pehli line: $\frac{x - 2}{- 2} = \frac{y - 2}{p / 3} = \frac{z + 1}{4}$ , direction ratios $- 2, \frac{p}{3}, 4$ ke saath.
Doosri line: $\frac{x - 1}{- p} = \frac{y + 2}{6} = \frac{z - 3}{- 2}$ , direction ratios $- p, 6, - 2$ ke saath.
Perpendicular: $2 p + 2 p - 8 = 0$ , isliye $p = 2$ .
ProofEasyNCERT Exercise 11.2, Q11
Kya lines $\frac{x - 3}{4} = \frac{y + 1}{- 1} = \frac{z - 2}{- 2}$ aur $\frac{x}{2} = \frac{y - 5}{6} = \frac{z}{1}$ perpendicular hain? Justify karo.
Solution dikhao
Direction ratios $4, - 1, - 2$ aur $2, 6, 1$ hain.
Unke products ka sum $8 - 6 - 2 = 0$ hai, isliye lines perpendicular hain.
NumericalMediumNCERT Exercise 11.2, Q12
Lines $\vec{r} = (\hat{i} + 2 \hat{k}) + λ (\hat{i} + 2 \hat{j} - \hat{k})$ aur $\vec{r} = (3 \hat{j} - \hat{k}) + μ (2 \hat{i} + \hat{j} + \hat{k})$ ek doosre ke kitne paas aati hain?
Solution dikhao
Answer $\frac{1}{\sqrt{3}} \approx 0.577$ units
${\vec{a}}_{2} - {\vec{a}}_{1} = - \hat{i} + 3 \hat{j} - 3 \hat{k}$ .
${\vec{b}}_{1} \times {\vec{b}}_{2} = 3 \hat{i} - 3 \hat{j} - 3 \hat{k}$ , length $\sqrt{27} = 3 \sqrt{3}$ .
Dot product $- 3 - 9 + 9 = - 3$ hai.
To $d = \frac{3}{3 \sqrt{3}} = \frac{1}{\sqrt{3}} \approx 0.577$ .
NumericalMediumNCERT Exercise 11.2, Q13
Lines $\frac{x - 2}{3} = \frac{y + 1}{- 1} = \frac{z - 4}{2}$ aur $\frac{x + 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{- 1}$ ke beech, sabse paas hone par, gap calculate karo.
Solution dikhao
Answer $\frac{25}{\sqrt{83}} \approx 2.74$ units
Points $(2, - 1, 4)$ aur $(- 1, 2, - 3)$ hain, isliye $Δ x, Δ y, Δ z = - 3, 3, - 7$ . Direction ratios $3, - 1, 2$ aur $1, 2, - 1$ hain.
Determinant ko uski pehli row ke along expand karne par $D = - 3 (1 - 4) - 3 (- 3 - 2) - 7 (6 + 1) = 9 + 15 - 49 = - 25$ milta hai.
Directions ke cross product ke components $- 3, 5, 7$ hain, isliye uski length $\sqrt{83}$ hai.
Isliye $d = \frac{25}{\sqrt{83}} \approx 2.74$ .
NumericalMediumNCERT Exercise 11.2, Q14
$\vec{r} = (\hat{i} + \hat{j} + \hat{k}) + λ (2 \hat{i} + \hat{j} - 2 \hat{k})$ aur $\vec{r} = (4 \hat{i} - \hat{j} + 2 \hat{k}) + μ (2 \hat{i} - 2 \hat{j} + \hat{k})$ ke common perpendicular ki length nikaalo.
Solution dikhao
Answer $\frac{1}{3}$ unit
${\vec{a}}_{2} - {\vec{a}}_{1} = 3 \hat{i} - 2 \hat{j} + \hat{k}$ .
${\vec{b}}_{1} \times {\vec{b}}_{2} = - 3 \hat{i} - 6 \hat{j} - 6 \hat{k}$ , length 9.
Dot product $- 9 + 12 - 6 = - 3$ hai, isliye $d = \frac{3}{9} = \frac{1}{3}$ .
NumericalHardNCERT Exercise 11.2, Q15
Do lines parametric form mein di gayi hain: $\vec{r} = (2 + t) \hat{i} + (1 - 2 t) \hat{j} + 3 t \hat{k}$ aur $\vec{r} = (2 s - 1) \hat{i} + (s + 4) \hat{j} + (1 - s) \hat{k}$ . Sabse paas hone par woh kitni door hain?
Solution dikhao
Answer $\frac{29}{5 \sqrt{3}} = \frac{29 \sqrt{3}}{15} \approx 3.35$ units
Pehli line $(2 \hat{i} + \hat{j}) + t (\hat{i} - 2 \hat{j} + 3 \hat{k})$ hai aur doosri $(- \hat{i} + 4 \hat{j} + \hat{k}) + s (2 \hat{i} + \hat{j} - \hat{k})$ .
To ${\vec{a}}_{2} - {\vec{a}}_{1} = - 3 \hat{i} + 3 \hat{j} + \hat{k}$ aur ${\vec{b}}_{1} \times {\vec{b}}_{2} = - \hat{i} + 7 \hat{j} + 5 \hat{k}$ , length $\sqrt{75} = 5 \sqrt{3}$ .
Dot product $3 + 21 + 5 = 29$ hai, isliye $d = \frac{29}{5 \sqrt{3}} = \frac{29 \sqrt{3}}{15} \approx 3.35$ .
Quick answerMedium
Lines $\vec{r} = (\hat{i} + \hat{j}) + λ (2 \hat{i} - \hat{j} + \hat{k})$ aur $\vec{r} = (3 \hat{i} + 2 \hat{k}) + μ (- 4 \hat{i} + 2 \hat{j} - 2 \hat{k})$ hain
ek hi lineintersecting linesskew linesalag-alag parallel lines
Answer dekho
Answer (D) alag-alag parallel lines
Doosri direction pehli ka $- 2$ guna hai, isliye lines parallel hain. Join $2 \hat{i} - \hat{j} + 2 \hat{k}$ , $2 \hat{i} - \hat{j} + \hat{k}$ ke along nahi hai, isliye woh ek hi line nahi hain; (11.14) se woh $\sqrt{\frac{5}{6}} \approx 0.91$ door hain.
Quick answerMedium
Parallel lines $\vec{r} = λ (2 \hat{i} - 2 \hat{j} + \hat{k})$ aur $\vec{r} = (3 \hat{i} + 3 \hat{j}) + μ (2 \hat{i} - 2 \hat{j} + \hat{k})$ ke beech distance do decimal places tak nikaalo.
Answer dekho
Answer 4.24
$\vec{b} \times (3 \hat{i} + 3 \hat{j}) = - 3 \hat{i} + 3 \hat{j} + 12 \hat{k}$ , length $\sqrt{162} = 9 \sqrt{2}$ , isliye $d = 3 \sqrt{2} \approx 4.24$ .

Recap

Summary aur key points

Ek directed line positive axes ke saath angles $α, β, γ$ banati hai; inke cosines uske direction cosines $l, m, n$ hain aur arrow ulta karne par teeno ke sign badal jaate hain — Section 11.2
Direction cosines ka har set $l^{2} + m^{2} + n^{2} = 1$ satisfy karta hai — (11.1)
Direction cosines ko kisi bhi non-zero number se scale karne par direction ratios milte hain; wapas jaane ke liye $a, b, c$ ko $\sqrt{a^{2} + b^{2} + c^{2}}$ se divide karo — (11.2)
Do points se guzarti line ke liye coordinates ke differences direction ratios hain, aur distance se divide karne par cosines milte hain — Section 11.2.1
Point $\vec{a}$ se guzarti $\vec{b}$ ke parallel line $\vec{r} = \vec{a} + λ \vec{b}$ hai; points $\vec{a}$ aur $\vec{b}$ se guzarti line $\vec{r} = \vec{a} + λ (\vec{b} - \vec{a})$ hai — (11.4)
Coordinates mein wahi line $\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c}$ hai — (11.6)
Do lines ke beech ka angle unki directions ke beech ka acute angle hai, $\cos θ = \frac{∣ {\vec{b}}_{1} \cdot {\vec{b}}_{2} ∣}{∣ {\vec{b}}_{1} ∣ ∣ {\vec{b}}_{2} ∣}$ — Section 11.4
Lines perpendicular hoti hain jab $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$ ho, aur parallel jab unke direction ratios proportional hon — test
Space mein do lines ya to milti hain, ya parallel chalti hain, ya skew hoti hain (dono mein se koi nahi); skew lines ke liye shortest gap common perpendicular ke along hota hai — Section 11.5
Skew lines ke liye: $d = \frac{∣ ({\vec{b}}_{1} \times {\vec{b}}_{2}) \cdot ({\vec{a}}_{2} - {\vec{a}}_{1}) ∣}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}$ — (11.11)
Parallel lines ke liye: $d = \frac{∣ \vec{b} \times ({\vec{a}}_{2} - {\vec{a}}_{1}) ∣}{∣ \vec{b} ∣}$ — (11.14)

Flash cards · 6

Yaad rakhne wale formulas

1 / 6

Chapter 11 par Miscellaneous Exercise

Practice · 6 questions

Practice questions answers ke saath (NCERT “Miscellaneous exercise”)

Short answerMediumNCERT Miscellaneous Exercise, Q1
Real numbers $p, q, r$ ke liye do lines ke direction ratios $p + q, q + r, r + p$ aur $p - q, q - r, r - p$ hain (koi bhi set poora zero nahi). Unke beech ka angle nikaalo.
Solution dikhao
Answer $90^{\circ}$
Matching ratios multiply karo: har product do squares ka difference hai, $(p + q) (p - q) = p^{2} - q^{2}$ , aur baaki do ke liye bhi aisa hi.
Sum $(p^{2} - q^{2}) + (q^{2} - r^{2}) + (r^{2} - p^{2}) = 0$ hai, isliye $\cos θ = 0$ aur lines perpendicular hain, $p$ , $q$ , $r$ chahe kuch bhi hon.
Short answerEasyNCERT Miscellaneous Exercise, Q2
$(0, 2, - 3)$ se guzarti $z$ -axis ke parallel line ki vector aur Cartesian equations likho.
Solution dikhao
Answer $\vec{r} = (2 \hat{j} - 3 \hat{k}) + λ \hat{k}$ ; $\frac{x}{0} = \frac{y - 2}{0} = \frac{z + 3}{1}$ , yaani $x = 0$ aur $y = 2$
Direction $\hat{k}$ hai, isliye $\vec{r} = (2 \hat{j} - 3 \hat{k}) + λ \hat{k}$ .
Cartesian form mein $\frac{x}{0} = \frac{y - 2}{0} = \frac{z + 3}{1}$ , jahan zero denominators ka matlab hai ki har point ke liye $x = 0$ aur $y = 2$ , jabki $z$ free hai.
Short answerMediumNCERT Miscellaneous Exercise, Q3
Pata hai ki lines $\frac{x + 2}{k} = \frac{y - 1}{3} = \frac{z + 4}{- 2}$ aur $\frac{x - 5}{2} = \frac{y + 3}{k} = \frac{z - 1}{4}$ perpendicular hain. $k$ kya hai?
Solution dikhao
Answer $k = \frac{8}{5}$
Perpendicular: $2 k + 3 k - 8 = 0$ , isliye $5 k = 8$ aur $k = \frac{8}{5}$ .
NumericalMediumNCERT Miscellaneous Exercise, Q4
Nikaalo ki yeh lines sabse paas hone par kitni door hain: $\vec{r} = (3 \hat{i} - \hat{k}) + λ (2 \hat{i} + 2 \hat{j} + \hat{k})$ aur $\vec{r} = (4 \hat{i} - 2 \hat{j} + \hat{k}) + μ (- 2 \hat{i} + \hat{j} + 2 \hat{k})$ .
Solution dikhao
Answer3 units
${\vec{a}}_{2} - {\vec{a}}_{1} = \hat{i} - 2 \hat{j} + 2 \hat{k}$ .
${\vec{b}}_{1} \times {\vec{b}}_{2} = 3 \hat{i} - 6 \hat{j} + 6 \hat{k}$ , length 9.
Dot product $3 + 12 + 12 = 27$ hai, isliye $d = \frac{27}{9} = 3$ .
Short answerHardNCERT Miscellaneous Exercise, Q5
$(2, - 1, 3)$ se guzarti ek line ko do di gayi lines, $\frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 4}{- 2}$ aur $\vec{r} = (\hat{i} + 4 \hat{j}) + μ (3 \hat{i} + \hat{k})$ , ke right angles par hona hai. Uski vector equation likho.
Solution dikhao
Answer $\vec{r} = (2 \hat{i} - \hat{j} + 3 \hat{k}) + λ (2 \hat{i} - 7 \hat{j} - 6 \hat{k})$
Directions $\hat{i} + 2 \hat{j} - 2 \hat{k}$ aur $3 \hat{i} + \hat{k}$ hain.
Unka cross product woh determinant hai jiski rows $\hat{i}, \hat{j}, \hat{k}$ ; $1, 2, - 2$ ; $3, 0, 1$ hain.
Yeh cross product $2 \hat{i} - 7 \hat{j} - 6 \hat{k}$ hai. Check: dono directions ke saath iske dot products $2 - 14 + 12 = 0$ aur $6 - 6 = 0$ hain.
To line hai $\vec{r} = (2 \hat{i} - \hat{j} + 3 \hat{k}) + λ (2 \hat{i} - 7 \hat{j} - 6 \hat{k})$ .
Quick answerMedium
Agar ${\vec{b}}_{1} \times {\vec{b}}_{2} \neq \vec{0}$ aur $({\vec{b}}_{1} \times {\vec{b}}_{2}) \cdot ({\vec{a}}_{2} - {\vec{a}}_{1}) = 0$ ho, to lines $\vec{r} = {\vec{a}}_{1} + λ {\vec{b}}_{1}$ aur $\vec{r} = {\vec{a}}_{2} + μ {\vec{b}}_{2}$ intersect karti hain.
SahiGalat
Answer dekho
Answer Sahi
Directions parallel nahi hain, aur (11.11) se $d = 0$ milta hai, isliye lines ka ek common point hai.

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 11.1, Q1	Practice 11.1 · Q1
NCERT Exercise 11.1, Q2	Practice 11.1 · Q2
NCERT Exercise 11.1, Q3	Practice 11.1 · Q3
NCERT Exercise 11.1, Q4	Practice 11.1 · Q4
NCERT Exercise 11.1, Q5	Practice 11.1 · Q5
NCERT Exercise 11.2, Q1	Practice 11.2 · Q1
NCERT Exercise 11.2, Q2	Practice 11.2 · Q2
NCERT Exercise 11.2, Q3	Practice 11.2 · Q3
NCERT Exercise 11.2, Q4	Practice 11.2 · Q4
NCERT Exercise 11.2, Q5	Practice 11.2 · Q5
NCERT Exercise 11.2, Q6	Practice 11.2 · Q6
NCERT Exercise 11.2, Q7	Practice 11.2 · Q7
NCERT Exercise 11.2, Q8	Practice 11.2 · Q8
NCERT Exercise 11.2, Q9	Practice 11.2 · Q9
NCERT Exercise 11.2, Q10	Practice 11.2 · Q10
NCERT Exercise 11.2, Q11	Practice 11.2 · Q11
NCERT Exercise 11.2, Q12	Practice 11.2 · Q12
NCERT Exercise 11.2, Q13	Practice 11.2 · Q13
NCERT Exercise 11.2, Q14	Practice 11.2 · Q14
NCERT Exercise 11.2, Q15	Practice 11.2 · Q15
NCERT Miscellaneous Exercise, Q1	Miscellaneous exercise · Q1
NCERT Miscellaneous Exercise, Q2	Miscellaneous exercise · Q2
NCERT Miscellaneous Exercise, Q3	Miscellaneous exercise · Q3
NCERT Miscellaneous Exercise, Q4	Miscellaneous exercise · Q4
NCERT Miscellaneous Exercise, Q5	Miscellaneous exercise · Q5

Quick check: MCQs

11.1 Introduction3 questions
Direction ratios3 questions
11.2.1 Do points se guzarne wali line ke direction cosines2 questions
11.3.1 Diye gaye point se guzarne wali aur diye gaye vector ke parallel line ki equation3 questions
11.4 Do Lines ke beech Angle3 questions
11.5.2 Parallel lines ke beech distance4 questions

Glossary · 11

Key terms (English + Hindi)

Cartesian equation of a lineरेखा का कार्तीय समीकरणrekha ka kaarteey sameekaran: Coordinates mein likhi line, $\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c}$ , jo $(x_{1}, y_{1}, z_{1})$ se guzarti hai aur jiske direction ratios $a, b, c$ hain.Chapter mein pehli baar yahan aaya →
collinearसंरेखsanrekh: Ek hi seedhi line par hona.Chapter mein pehli baar yahan aaya →
common perpendicularउभयनिष्ठ लंबubhayanishth lamb: Woh ek segment jo do skew lines ko milta hai aur dono ke perpendicular hai; iski length unka shortest distance hai.Chapter mein pehli baar yahan aaya →
coplanarसमतलीयsamtaleey: Ek hi plane mein hona. Do lines coplanar tabhi hoti hain jab woh milti hain ya parallel hoti hain.Chapter mein pehli baar yahan aaya →

directed lineदिष्ट रेखाdisht rekha: Ek line jiski do directions mein se ek chuni gayi ho, arrow se dikhayi gayi.Chapter mein pehli baar yahan aaya →
direction anglesदिक्-कोणdik-kon: Woh angles $α, β, γ$ (har ek $0^{\circ}$ se $180^{\circ}$ tak) jo directed line positive $x$ -, $y$ - aur $z$ -axes ke saath banati hai.Chapter mein pehli baar yahan aaya →
direction cosinesदिक्-कोसाइनdik-cosine: Direction angles ke cosines $l, m, n$ ; yeh $l^{2} + m^{2} + n^{2} = 1$ satisfy karte hain.Chapter mein pehli baar yahan aaya →
direction ratiosदिक्-अनुपातdik-anupaat: Koi bhi teen numbers, sab zero nahi, jo line ke direction cosines ke same ratio mein hon.Chapter mein pehli baar yahan aaya →

parameterप्राचलpraachal: $\vec{r} = \vec{a} + λ \vec{b}$ mein real number $λ$ ; har value line ka ek point chunti hai.Chapter mein pehli baar yahan aaya →

shortest distanceन्यूनतम दूरीnyoontam doori: Ek line ke point ko doosri line ke point se jodne wale sabse chhote segment ki length.Chapter mein pehli baar yahan aaya →
skew linesविषमतलीय रेखाएँvishamtaleey rekhaayen: Space mein lines ka aisa pair jo kabhi milta nahi aur parallel bhi nahi hai; koi plane dono ko nahi rakhta.Chapter mein pehli baar yahan aaya →

11.1 Introduction

11.2 Line ke Direction Cosines aur Direction Ratios

Direction ratios

11.2.1 Do points se guzarne wali line ke direction cosines

11.3 Space mein Line ki Equation

11.3.1 Diye gaye point se guzarne wali aur diye gaye vector ke parallel line ki equation

11.4 Do Lines ke beech Angle

11.5 Do Lines ke beech Shortest Distance

11.5.1 Do skew lines ke beech distance

11.5.2 Parallel lines ke beech distance

Chapter 11 par Miscellaneous Exercise

Apna NCERT exercise question dhoondo

Quick check: MCQs

Tum chapter ke end tak pahunch gaye