
The coordinate of point in which the line joining the point $(2,5, - 7)$ and $( - 3, - 1,8)$ are intersected by the y-z planes are
A. $(0,\dfrac{{13}}{5}, - 1)$
B. $(0,\dfrac{{ - 13}}{5}, - 2)$
C. $(0,\dfrac{{ - 13}}{5},\dfrac{2}{5})$
D. $(0,\dfrac{{13}}{5},\dfrac{2}{5})$
Answer
217.8k+ views
Hint: Given, the line joining the point $(2,5, - 7)$ and $( - 3, - 1,8)$. We have to find the coordinate of the point which is the intersection point of the line joining the point $(2,5, - 7)$ and $( - 3, - 1,8)$ and the y-z plane. First, we will find the equation of the line and then use the concept that when the line intersects the y-z plane x-coordinate is zero to find the coordinates points.
Formula used:
Equation of line joining two points $\dfrac{{x - {x_1}}}{{{x_2} - {x_1}}} = \dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{z - {z_1}}}{{{z_2} - {z_1}}}$
Complete step by step solution:
Given, the line joining the point $(2,5, - 7)$ and $( - 3, - 1,8)$.
Equation of line joining two points $\dfrac{{x - {x_1}}}{{{x_2} - {x_1}}} = \dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{z - {z_1}}}{{{z_2} - {z_1}}}$
So, using the above formula
$\dfrac{{x - 2}}{-5} = \dfrac{{y - 5}}{-6} = \dfrac{{z + 7}}{{15}}$
Multiplying by -1,
$\dfrac{{x - 2}}{5} = \dfrac{{y - 5}}{6} = \dfrac{{z + 7}}{{ - 15}}$
When the line meets y-z plane
$ \Rightarrow x = 0$
$ \Rightarrow \dfrac{{ - 2}}{5} = \dfrac{{y - 5}}{6} = \dfrac{{z + 7}}{{ - 15}}$
Comparing first two
$\dfrac{{ - 2}}{5} = \dfrac{{y - 5}}{6}$
Cross multiplying
$ - 2 \times 6 = 5(y - 5)$
After solving
$ - 12 = 5y - 25$
Shifting variable on one side and constant on other
$5y = - 12 + 25$
$5y = 13$
Dividing both sides by 5
$y = \dfrac{{13}}{5}$
Comparing last two
$\dfrac{{ - 2}}{5} = \dfrac{{z + 7}}{{ - 15}}$
Cross multiplying
$30 = 5z + 35$
Shifting variable on one side and constant on other
$5z = - 35 + 30$
$5z = - 5$
Dividing both sides by 5
$z = - 1$
Hence, $(x,y,z) = (0,\dfrac{{13}}{5}, - 1)$
So, option (A) is the correct answer.
Note: Students should find the equation of the line correctly so that they can do further calculations correctly. And should understand the concept that when a line intersects the y-z plane means x-coordinate is zero. This will help to find the value of y and z. They should pay attention to every detail while solving question.
Formula used:
Equation of line joining two points $\dfrac{{x - {x_1}}}{{{x_2} - {x_1}}} = \dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{z - {z_1}}}{{{z_2} - {z_1}}}$
Complete step by step solution:
Given, the line joining the point $(2,5, - 7)$ and $( - 3, - 1,8)$.
Equation of line joining two points $\dfrac{{x - {x_1}}}{{{x_2} - {x_1}}} = \dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{z - {z_1}}}{{{z_2} - {z_1}}}$
So, using the above formula
$\dfrac{{x - 2}}{-5} = \dfrac{{y - 5}}{-6} = \dfrac{{z + 7}}{{15}}$
Multiplying by -1,
$\dfrac{{x - 2}}{5} = \dfrac{{y - 5}}{6} = \dfrac{{z + 7}}{{ - 15}}$
When the line meets y-z plane
$ \Rightarrow x = 0$
$ \Rightarrow \dfrac{{ - 2}}{5} = \dfrac{{y - 5}}{6} = \dfrac{{z + 7}}{{ - 15}}$
Comparing first two
$\dfrac{{ - 2}}{5} = \dfrac{{y - 5}}{6}$
Cross multiplying
$ - 2 \times 6 = 5(y - 5)$
After solving
$ - 12 = 5y - 25$
Shifting variable on one side and constant on other
$5y = - 12 + 25$
$5y = 13$
Dividing both sides by 5
$y = \dfrac{{13}}{5}$
Comparing last two
$\dfrac{{ - 2}}{5} = \dfrac{{z + 7}}{{ - 15}}$
Cross multiplying
$30 = 5z + 35$
Shifting variable on one side and constant on other
$5z = - 35 + 30$
$5z = - 5$
Dividing both sides by 5
$z = - 1$
Hence, $(x,y,z) = (0,\dfrac{{13}}{5}, - 1)$
So, option (A) is the correct answer.
Note: Students should find the equation of the line correctly so that they can do further calculations correctly. And should understand the concept that when a line intersects the y-z plane means x-coordinate is zero. This will help to find the value of y and z. They should pay attention to every detail while solving question.
Recently Updated Pages
Elastic Collision in Two Dimensions Explained Simply

Elastic Collisions in One Dimension Explained

Electric Field of Infinite Line Charge and Cylinders Explained

Electric Flux and Area Vector Explained Simply

Electric Field of a Charged Spherical Shell Explained

Electricity and Magnetism Explained: Key Concepts & Applications

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

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 Atomic Structure for Beginners

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

Understanding the Electric Field of a Uniformly Charged Ring

