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  2. How to solve this problem- - Three Dimensional Geometry - JEE Main-4
# Engineering

The length of the perpendicular from the point (2,-1,4) on the straight line, \frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1} is :

  • Option 1)

    greater than 2 but less than 3

     

  • Option 2)

    greater than 4

  • Option 3)

    less than 2

     

  • Option 4)

    greater than 3 but less than 4

 

Answers (1)
S solutionqc

Point=(2,-1,4)=P

Line :   \frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}

Length of perpendicular from point to line ?

Any point Q on line =10\lambda -3,\; -7\lambda+2,\lambda

DR's of PQ=\left ( 10\lambda-3-2,-7\lambda+2+1,\lambda-4 \right )

                       =\left ( 10\lambda-5,-7\lambda+3,\lambda-4 \right )

PQ is perpendicular to given line

\therefore 10(10\lambda-5)+(-7)(-7\lambda+3)+1(\lambda-4)=0

\\=100\lambda-50+49\lambda-21+\lambda-4=0\\\; \; \; \\150\lambda-75=0

\lambda=\frac{1}{2}

PQ=\sqrt{(10\lambda-5)^{2}+(-7\lambda+3)^{2}+(\lambda-4)^{2}}

          =\sqrt{(10\times\frac{1}{2}-5)^{2}+(\frac{-7}{2}+3)^{2}+(\frac{1}{2}-4)^{2}}

          =\sqrt{0+(\frac{-1}{2})^{2}+(\frac{1-8}{2})^{2}}

          =\sqrt{\frac{1}{4}+\frac{49}{A}}=\sqrt{\frac{50}{4}}=\sqrt{\frac{25}{2}}=\frac{5}{\sqrt{2}}

          3<\frac{5}{\sqrt{2}}<4

 

 

 

 


Option 1)

greater than 2 but less than 3

 

Option 2)

greater than 4

Option 3)

less than 2

 

Option 4)

greater than 3 but less than 4

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