
The shortest distance between the line ${\vec{r}}_1=4\vec{i}-3\vec{j}-\vec{k}+\lambda(\vec{i}-4\vec{j}+7\vec{k})$ and ${\vec{r}}_2=\vec{i}-\vec{j}-10\vec{k}+\lambda(2\vec{i}-3\vec{j}+8\vec{k})$ is –
A. 3
B. 1
C. 2
D. 0
Answer
162.9k+ views
Hint: To determine the shortest distance between any two lines we must determine the shortest distance between any two points on both lines. There are different types of lines: parallel lines, intersecting lines, and skew lines.
1. Parallel lines

All distance between parallel lines will be the same/constant.
2. Intersecting lines

The shortest distance between the intersecting lines will be zero.
3. Skew lines

Formula used: The formula to determine the distance between the lines $r_1=a_1+\lambda b_1$ and $r_2=a_2+\lambda b_2$ are –
$d=\left\bracevert\dfrac{(a_2-a_1).(b_1\times b_2)}{\left|b_1\times b_2\right|}\right\bracevert$
Complete step by step solution: Given, the equations of the lines are –
${\vec{r}}_1=4\vec{i}-3\vec{j}-\vec{k}+\lambda(\vec{i}-4\vec{j}+7\vec{k})$ ---------- (i)
${\vec{r}}_2=\vec{i}-\vec{j}-10\vec{k}+\lambda(2\vec{i}-3\vec{j}+8\vec{k})$ ---------- (ii)
We know that to determine the shortest distance between the lines-
$r_1=a_1+\lambda b_1$ and $r_2=a_2+\lambda b_2$
$d=\left\bracevert\dfrac{(a_2-a_1).(b_1\times b_2)}{\left|b_1\times b_2\right|}\right\bracevert$
On comparing the equations (i) and (ii) with equations of lines $[ r_1=a_1+\lambda b_1 ]$ and $[ r_2=a_2+\lambda b_2 ]$ respectively
$a_1=4\vec{i}-3\vec{j}-\vec{k}$ and $a_2=\vec{i}-\vec{j}-10\vec{k}$
$b_1=\vec{i}-4\vec{j}+7\vec{k}$ and $b_2=2\vec{i}-3\vec{j}+8\vec{k}$
$a_2-a_1=\vec{i}-\vec{j}-10\vec{k}-(4\vec{i}-3\vec{j}-\vec{k})$
$a_2-a_1=\vec{i}-\vec{j}-10\vec{k}-4\vec{i}+3\vec{j}+\vec{k}$
$a_2-a_1=-3\vec{i}+2\vec{j}-9\vec{k}$
${b}_{1}\times{b}_{2}=\vec{{i}}[(-{4})\times{8}-{7}\times(-{3})]-j[(7×2)-(8×1)]-k[(-3)×1-2×(-4)]$
${b}_{1}\times{b}_{2}=\vec{{i}}[(-{32})-(-{21})]+j[(14)-(8)]-k[(-3)-(-8)]$
${b}_{1}\times{b}_{2}=\vec{{i}}[(-{32})+({21})]+j[(14)-(8)]-k[(-3)-(-8)]$
${b}_{1}\times{b}_{2}=-{11}\vec{{i}}+{6}\vec{{j}}+{5}\vec{{k}}$
$(a_2-a_1).(b_1\times b_2)=(-3\vec{i}+2\vec{j}-9\vec{k}).(-{11}\vec{{i}}+{6}\vec{{j}}+{5}\vec{{k}})$
$(a_2-a_1).(b_1\times b_2)=(-3).(-{11})(\vec{{i}}.\vec{i})+(2.{6})(\vec{{j}}.\vec{j})+(-9).(-{5})(\vec{{k}}.\vec{k})$
As we know that $(\vec{{i}}.\vec{i})=(\vec{{j}}.\vec{j})=(\vec{{k}}.\vec{k})=1$
$(a_2-a_1).(b_1\times b_2)={33}+12-45$
$a_2-a_1).(b_1\times b_2)={45}-45$
$(a_2-a_1).(b_1\times b_2)={0}$
$\left|b_1\times b_2\right|=\sqrt{{(-11)}^2+{(6)}^2+{(5)}^2}$
$\left|b_1\times b_2\right|=\sqrt{121+36+25}$
$\left|b_1\times b_2\right|=\sqrt{182}$
$d=\left|\frac{(a_2-a_1).(b_1\times b_2)}{\left|b_1\times b_2\right|}\right|=\frac{0}{\sqrt{182}}=0$
Hence the shortest distance between the line [ ${\vec{r}}_1=4\vec{i}-3\vec{j}-\vec{k}+\lambda(\vec{i}-4\vec{j}+7\vec{k}) ]$ and $[ {\vec{r}}_2=\vec{i}-\vec{j}-10\vec{k}+\lambda(2\vec{i}-3\vec{j}+8\vec{k}) ]$ is 0.
Thus, Option (D) is correct.
Note: The direction ratios (a, b, c) are components of vector with respect to x-axis, y-axis, and z-axis respectively and direction cosines (l, m, n) are the angle subtended by the line with x-axis, y-axis, and z-axis respectively.
1. Parallel lines

All distance between parallel lines will be the same/constant.
2. Intersecting lines

The shortest distance between the intersecting lines will be zero.
3. Skew lines

Formula used: The formula to determine the distance between the lines $r_1=a_1+\lambda b_1$ and $r_2=a_2+\lambda b_2$ are –
$d=\left\bracevert\dfrac{(a_2-a_1).(b_1\times b_2)}{\left|b_1\times b_2\right|}\right\bracevert$
Complete step by step solution: Given, the equations of the lines are –
${\vec{r}}_1=4\vec{i}-3\vec{j}-\vec{k}+\lambda(\vec{i}-4\vec{j}+7\vec{k})$ ---------- (i)
${\vec{r}}_2=\vec{i}-\vec{j}-10\vec{k}+\lambda(2\vec{i}-3\vec{j}+8\vec{k})$ ---------- (ii)
We know that to determine the shortest distance between the lines-
$r_1=a_1+\lambda b_1$ and $r_2=a_2+\lambda b_2$
$d=\left\bracevert\dfrac{(a_2-a_1).(b_1\times b_2)}{\left|b_1\times b_2\right|}\right\bracevert$
On comparing the equations (i) and (ii) with equations of lines $[ r_1=a_1+\lambda b_1 ]$ and $[ r_2=a_2+\lambda b_2 ]$ respectively
$a_1=4\vec{i}-3\vec{j}-\vec{k}$ and $a_2=\vec{i}-\vec{j}-10\vec{k}$
$b_1=\vec{i}-4\vec{j}+7\vec{k}$ and $b_2=2\vec{i}-3\vec{j}+8\vec{k}$
$a_2-a_1=\vec{i}-\vec{j}-10\vec{k}-(4\vec{i}-3\vec{j}-\vec{k})$
$a_2-a_1=\vec{i}-\vec{j}-10\vec{k}-4\vec{i}+3\vec{j}+\vec{k}$
$a_2-a_1=-3\vec{i}+2\vec{j}-9\vec{k}$
${b}_{1}\times{b}_{2}=\vec{{i}}[(-{4})\times{8}-{7}\times(-{3})]-j[(7×2)-(8×1)]-k[(-3)×1-2×(-4)]$
${b}_{1}\times{b}_{2}=\vec{{i}}[(-{32})-(-{21})]+j[(14)-(8)]-k[(-3)-(-8)]$
${b}_{1}\times{b}_{2}=\vec{{i}}[(-{32})+({21})]+j[(14)-(8)]-k[(-3)-(-8)]$
${b}_{1}\times{b}_{2}=-{11}\vec{{i}}+{6}\vec{{j}}+{5}\vec{{k}}$
$(a_2-a_1).(b_1\times b_2)=(-3\vec{i}+2\vec{j}-9\vec{k}).(-{11}\vec{{i}}+{6}\vec{{j}}+{5}\vec{{k}})$
$(a_2-a_1).(b_1\times b_2)=(-3).(-{11})(\vec{{i}}.\vec{i})+(2.{6})(\vec{{j}}.\vec{j})+(-9).(-{5})(\vec{{k}}.\vec{k})$
As we know that $(\vec{{i}}.\vec{i})=(\vec{{j}}.\vec{j})=(\vec{{k}}.\vec{k})=1$
$(a_2-a_1).(b_1\times b_2)={33}+12-45$
$a_2-a_1).(b_1\times b_2)={45}-45$
$(a_2-a_1).(b_1\times b_2)={0}$
$\left|b_1\times b_2\right|=\sqrt{{(-11)}^2+{(6)}^2+{(5)}^2}$
$\left|b_1\times b_2\right|=\sqrt{121+36+25}$
$\left|b_1\times b_2\right|=\sqrt{182}$
$d=\left|\frac{(a_2-a_1).(b_1\times b_2)}{\left|b_1\times b_2\right|}\right|=\frac{0}{\sqrt{182}}=0$
Hence the shortest distance between the line [ ${\vec{r}}_1=4\vec{i}-3\vec{j}-\vec{k}+\lambda(\vec{i}-4\vec{j}+7\vec{k}) ]$ and $[ {\vec{r}}_2=\vec{i}-\vec{j}-10\vec{k}+\lambda(2\vec{i}-3\vec{j}+8\vec{k}) ]$ is 0.
Thus, Option (D) is correct.
Note: The direction ratios (a, b, c) are components of vector with respect to x-axis, y-axis, and z-axis respectively and direction cosines (l, m, n) are the angle subtended by the line with x-axis, y-axis, and z-axis respectively.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NEET 2025 – Every New Update You Need to Know

Verb Forms Guide: V1, V2, V3, V4, V5 Explained

NEET Total Marks 2025

1 Billion in Rupees
