Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

On which of the following lines lies the point of intersection of the line, $\frac{x-4}{2} = \frac{y-5}{2} = \frac{z-3}{1}$ and the plane $x + y + z = 2?$

Option 1)
$\frac{x+3}{3} = \frac{4-y}{3} = \frac{z+1}{-2}$
Option 2)
$\frac{x-4}{1} = \frac{y-5}{1} = \frac{z-5}{-1}$
Option 3)
$\frac{x-2}{2} = \frac{y-3}{2} = \frac{z+3}{3}$
Option 4)
$\frac{x-1}{1} = \frac{y-3}{2} = \frac{z+4}{-5}$

Answers (1)

best_answer

Intersection of line and plane -

Let the line

$\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$ plane

$a_{1}x+b_{1}y+c_{1}z+d=0$ intersect at P

to find P assume general point on line as $\left ( x_{1}+\lambda a_{1}y_{1} +\lambda b_{1}z_{1}+\lambda c_{1}\right )$

now put it in plane to find $\lambda$ ,

$a_{1}\left ( x_{1}+\lambda a \right )+b_{1}\left ( y_{1}+\lambda b \right )+c_{1}\left ( z_{1}+\lambda c \right )+d=0$

-

General point on the given line

$\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{2}=\lambda$

$x=2\lambda +4$

$y=2\lambda +5$

$z=\lambda +3$

This point lies on the plane $x+y+z=2$

So,

$2\lambda +4+2\lambda +5+\lambda +3=2$

$5\lambda =-10$

$\lambda =-2$

Option 1)

$\frac{x+3}{3} = \frac{4-y}{3} = \frac{z+1}{-2}$

Option 2)

$\frac{x-4}{1} = \frac{y-5}{1} = \frac{z-5}{-1}$

Option 3)

$\frac{x-2}{2} = \frac{y-3}{2} = \frac{z+3}{3}$

Option 4)
$\frac{x-1}{1} = \frac{y-3}{2} = \frac{z+4}{-5}$

Posted by

admin

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main Result 2024 Live: JEE Mains session 2 results at jeemain.nta.ac.in soon; rank predictor

anam-khan

JEE Main 2024 session 2 final answer key out at jeemain.nta.ac.in; steps to check probable scores

anam-khan

JEE Main Result 2024 Live: Download session 2 final answer key at jeemain.nta.ac.in; rank predictor, cutoff

Latest Question