
Let P be a plane $lx + my + nz = 0$ containing the line $\frac{{1 - x}}{1} = \frac{{y + 4}}{2} = \frac{{z + 2}}{3}$. If plane P divides the segment AB joining points $A( - 3, - 6,1)$ and $B(2,4, - 3)$ in ratio $k:1$ then the value of $k$ is equal to:
a) $1.5$
b) $2$
c) $4$
d) $3$
Answer
218.7k+ views
Hint: Given P be a plane $lx + my + nz = 0$ containing the line $\frac{{1 - x}}{1} = \frac{{y + 4}}{2} = \frac{{z + 2}}{3}$. And plane P divides the segment AB joining points $A( - 3, - 6,1)$ and $B(2,4, - 3)$ in ratio $k:1$.
Firstly, we will find the equation of the plane using the equation of the given line then, we will find the value of $k$.
Complete step by step Solution:
In the question we are given the equation of line $\frac{{1 - x}}{1} = \frac{{y + 4}}{2} = \frac{{z + 2}}{3}$
The line lies on the plane
$ - l + 2m + 3n = 0$ (1)
Point $(1, - 4, - 2)$ lies on the plane
$l - 4m - 2n = 0$ (2)
now we will add equation (1) and equation (2)
$ - l + 2m + 3n + l - 4m - 4n = 0$
after adding the equation l gets eliminated and the equation is formed in terms of m and n
$ - 2m + n = 0$
taking n on the other side
$2m = n$
hence the value of m in terms of n
$m = \frac{n}{2}$
now we will put this value of m in equation (1)
$ - l + n + 3n = 0$
and we get the value of l in terms of m
$ \Rightarrow l = 4m$
forming the ratio using all the equations formed above
$l:m:n::4n:\frac{n}{2}:n$
now we will simplify the equation by removing all the fractions from the equations
$ \Rightarrow l:m:n::8n:n:2n$
from the above equation, we will remove the term n from right-hand side ratio to make it simple
$ \Rightarrow l:m:n::8:1:2$
Hence, $8x + y + 2z = 0$ be the equation of the plane.
Given, that P divides AB in ratio $k:1$
$8(\frac{{ - 3 + 2k}}{{k + 1}}) + (\frac{{ - 6 + 4k}}{{k + 1}}) + 2(\frac{{1 - 3k}}{{k + 1}}) = 0$
By solving the above equation we get
$16k - 24 + 4k - 6 - 6k + 2 = 0$
taking all the similar terms together and constant at one side we get
$ \Rightarrow 14k = 28$
value of k by taking 14 on the other side
$k = 2$
Hence, the correct option is (b).
Note:Students mostly make mistakes while finding the equation of the plane. We must be careful while finding equations. And also concentrate while calculating the value of k to avoid calculation errors.
Firstly, we will find the equation of the plane using the equation of the given line then, we will find the value of $k$.
Complete step by step Solution:
In the question we are given the equation of line $\frac{{1 - x}}{1} = \frac{{y + 4}}{2} = \frac{{z + 2}}{3}$
The line lies on the plane
$ - l + 2m + 3n = 0$ (1)
Point $(1, - 4, - 2)$ lies on the plane
$l - 4m - 2n = 0$ (2)
now we will add equation (1) and equation (2)
$ - l + 2m + 3n + l - 4m - 4n = 0$
after adding the equation l gets eliminated and the equation is formed in terms of m and n
$ - 2m + n = 0$
taking n on the other side
$2m = n$
hence the value of m in terms of n
$m = \frac{n}{2}$
now we will put this value of m in equation (1)
$ - l + n + 3n = 0$
and we get the value of l in terms of m
$ \Rightarrow l = 4m$
forming the ratio using all the equations formed above
$l:m:n::4n:\frac{n}{2}:n$
now we will simplify the equation by removing all the fractions from the equations
$ \Rightarrow l:m:n::8n:n:2n$
from the above equation, we will remove the term n from right-hand side ratio to make it simple
$ \Rightarrow l:m:n::8:1:2$
Hence, $8x + y + 2z = 0$ be the equation of the plane.
Given, that P divides AB in ratio $k:1$
$8(\frac{{ - 3 + 2k}}{{k + 1}}) + (\frac{{ - 6 + 4k}}{{k + 1}}) + 2(\frac{{1 - 3k}}{{k + 1}}) = 0$
By solving the above equation we get
$16k - 24 + 4k - 6 - 6k + 2 = 0$
taking all the similar terms together and constant at one side we get
$ \Rightarrow 14k = 28$
value of k by taking 14 on the other side
$k = 2$
Hence, the correct option is (b).
Note:Students mostly make mistakes while finding the equation of the plane. We must be careful while finding equations. And also concentrate while calculating the value of k to avoid calculation errors.
Recently Updated Pages
Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE General Topics in Chemistry Important Concepts and Tips

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

Algebra Made Easy: Step-by-Step Guide for Students

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Understanding Atomic Structure for Beginners

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

NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections

NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series

How to Convert a Galvanometer into an Ammeter or Voltmeter

NCERT Solutions For Class 11 Maths Chapter 12 Limits And Derivatives

