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  • Answer please! - Three Dimensional Geometry - JEE Main-2

The shortest distance between the lines

\frac{x}{2}= \frac{y}{2}= \frac{z}{1}  and  \frac{x+2}{-1}= \frac{y-4}{8}= \frac{z-5}{4}   

lies in the interval :

  • Option 1)

     [0, 1)

  • Option 2)

     [1, 2)

  • Option 3)

    (2, 3]

  • Option 4)

    (3, 4]

 

Answers (1)

As we learnt in 

Shortest distance between two skew lines (Cartesian form) -

Shortest distance between

\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}and \frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}is given by

\left | \frac{\left ( \vec{b} \times \vec{b_{1}}\right )\cdot \left ( \vec{a} -\vec{a_{1}}\right )}{\left | \vec{b} \times \vec{b_{1}} \right |} \right | Where 

\vec{a}= x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k}

\vec{a_{1}}= x_{2}\hat{i}+y_{2}\hat{j}+z_{2}\hat{k}

\vec{b}= a_{1}\hat{i}+b_{1}\hat{j}+c_{1}\hat{k}

\vec{b_{1}}= a_{2}\hat{i}+b_{2}\hat{j}+c_{2}\hat{k}

 

-

 

 Cross product of DRS is b_{1}^{-1}\times b_{2}^{-1}=(2\hat{i}+2\hat{j}+\hat{k})\times (\hat{-i}+8\hat{j}+4\hat{k})=9\hat{j}+18\hat{k}

So shortest distance = \frac{\left |-\hat{j}+2\hat{k} . (2\hat{i}-4\hat{j}-5\hat{k}) \right |}{\sqrt{5}}=\left | \frac{4-10}{\sqrt{5}} \right |=\frac{6}{\sqrt{5}}

It lies in interval (2,3)


Option 1)

 [0, 1)

This option is incorrect

Option 2)

 [1, 2)

This option is incorrect

Option 3)

(2, 3]

This option is correct

Option 4)

(3, 4]

This option is incorrect

Posted by

Sabhrant Ambastha

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