
Using direction ratio shows that the points $A(2,3, - 4), B(1, - 2,3)$ and $C(3,8, - 11)$ are collinear.
Answer
232.8k+ views
Hint: First subtract 2 from 1, 3 from -2 and -4 from 3 to obtain the direction ratio $\left( {{a_1},{a_2},{a_3}} \right)$ of the line AB. Then subtract 1 from 3, -2 from 8 and 3 from -11 to obtain the direction ratio $\left( {{b_1},{b_2},{b_3}} \right)$ of the line BC. Then divide ${a_1}$ by ${b_1}$, ${a_2}$ by ${b_2}$ and ${a_3}$ by ${b_3}$, if all the ratios are same then the given points are collinear.
Formula Used:
If $\left( {{a_1},{a_2},{a_3}} \right)$are the direction cosines of a line AB and $\left( {{b_1},{b_2},{b_3}} \right)$ are the direction cosines of a line BC, then the points A, B and C are collinear if
$\dfrac{{{a_1}}}{{{b_1}}} = \dfrac{{{a_2}}}{{{b_2}}} = \dfrac{{{a_3}}}{{{b_3}}}$.
Complete step by step solution:
The direction ratios of the line formed by A, B are
$\left( {1 - 2, - 2 - 3,3 - ( - 4)} \right)$
=$\left( { - 1, - 5,7} \right)$
The direction ratios of the line formed by B, C are
$\left( {3 - 1,8 - ( - 2), - 11 - 3} \right)$
=$\left( {2,10, - 14} \right)$
Now, divide -1 by 2, -5 by 10 and 7 by -14 to observe whether the ratios are the same or not.
$ - \dfrac{1}{2}, - \dfrac{5}{{10}}, - \dfrac{7}{{14}}$
That is, $ - \dfrac{1}{2} = - \dfrac{1}{2} = - \dfrac{1}{2}$
As the ratios of the direction cosines of the line formed by A, B and B, C are the same,
Hence, A, B and C are collinear.
Note: We commonly obtain the equation by two points and then the direction cosines; this process is time-consuming but not incorrect. To answer the question quickly, obtain the direction cosines from the given points and their ratio.
Formula Used:
If $\left( {{a_1},{a_2},{a_3}} \right)$are the direction cosines of a line AB and $\left( {{b_1},{b_2},{b_3}} \right)$ are the direction cosines of a line BC, then the points A, B and C are collinear if
$\dfrac{{{a_1}}}{{{b_1}}} = \dfrac{{{a_2}}}{{{b_2}}} = \dfrac{{{a_3}}}{{{b_3}}}$.
Complete step by step solution:
The direction ratios of the line formed by A, B are
$\left( {1 - 2, - 2 - 3,3 - ( - 4)} \right)$
=$\left( { - 1, - 5,7} \right)$
The direction ratios of the line formed by B, C are
$\left( {3 - 1,8 - ( - 2), - 11 - 3} \right)$
=$\left( {2,10, - 14} \right)$
Now, divide -1 by 2, -5 by 10 and 7 by -14 to observe whether the ratios are the same or not.
$ - \dfrac{1}{2}, - \dfrac{5}{{10}}, - \dfrac{7}{{14}}$
That is, $ - \dfrac{1}{2} = - \dfrac{1}{2} = - \dfrac{1}{2}$
As the ratios of the direction cosines of the line formed by A, B and B, C are the same,
Hence, A, B and C are collinear.
Note: We commonly obtain the equation by two points and then the direction cosines; this process is time-consuming but not incorrect. To answer the question quickly, obtain the direction cosines from the given points and their ratio.
Recently Updated Pages
Geometry of Complex Numbers Explained

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

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

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

Electricity and Magnetism Explained: Key Concepts & Applications

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Understanding the Electric Field of a Uniformly Charged Ring

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

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

Understanding How a Current Loop Acts as a Magnetic Dipole

