Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Solve this problem - Three Dimensional Geometry - JEE Main-4

If the length of the perpendicular from the point (\beta ,0,\beta ) 

(\beta \neq0) to the line, \frac{x}{1}=\frac{y-1}{0}=\frac{z+1}{-1} is \sqrt{\frac{3}{2}} ,

then \beta is equal to : 

  • Option 1)

    1

  • Option 2)

    2

  • Option 3)

    -1

  • Option 4)

    -2

 

Answers (1)

 

Let point P(\beta ,0,\beta ) given that length of perpendicular distance   

from P to line is \sqrt{\frac{3}{2}}.

\frac{x}{1}=\frac{y-1}{0}=\frac{z+1}{-1}

R=(\lambda ,1,-\lambda -1)

Direction ratio of PR=(\lambda-\beta ,1,-\lambda -\beta -1)

PR is perpendicular to the line

=> (\lambda -\beta )(1)+(1)0+(-1)(-\lambda -\beta -1)=0

=> \lambda -\beta +\lambda +\beta +1=0

=> \lambda =\frac{1}{2}

PR=\sqrt{(\lambda -\beta )^{2}+1^{2}+(-\lambda -1-\beta )^{2}}=\sqrt{\frac{3}{2}}

      => 2\beta ^{2}+2\beta =0

      =>\beta =0,\beta =-1

Correct option is (3)

      


Option 1)

1

Option 2)

2

Option 3)

-1

Option 4)

-2

Posted by

Vakul

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

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions
anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in
anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question