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  2. Solve this problem - Three Dimensional Geometry - JEE Main-4
# Engineering

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)
V Vakul

 

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

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