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  • Answer please! - Three Dimensional Geometry - JEE Main-5

If the line\small \frac{x-3}{1} =\frac{y+2}{-1}=\frac{z+\lambda }{-2}  lies in the plane,

2x−4y+3z=2,

then the shortest distance between this line and the line

\small \frac{x-1}{12} =\frac{y}{9}=\frac{z}{4} is

 

  • Option 1)

    2

  • Option 2)

    1

  • Option 3)

    0

  • Option 4)

    3

 

Answers (2)

best_answer

As we have learned

Shortest distance between two skew lines (vector form) -

Shortest distance between \vec{r}=\vec{a}+\lambda \vec{b}\, \: and \,\: \vec{r}=\vec{a_{1}}+\mu\vec{b}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 |
 

- wherein

shortest distance is among the line which is perpendicular to both

LM= \vec{b}\times \vec{b_{1}}

shortest distance will be projection of PQ = \vec{a}- \vec{a_{1}} on LM

 

 

(3 , -2 , -\lambda ) \: \: lies \: \: on \: \: plane

 

So,  6+8 -3 \lambda = 2 \\ \Rightarrow 3 \lambda = 12 \\ \lambda = 4

Where \vec b \times \vec c = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -1 &-2 \\ 12 & 9 & 4 \end{vmatrix} = 14 \hat i -28 \hat j +21 \hat k

So distance  = 0 

 

 

 


Option 1)

2

Option 2)

1

Option 3)

0

Option 4)

3

Posted by

Himanshu

View full answer

JEE Main high-scoring chapters and topics

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

As we have learned

Shortest distance between two skew lines (vector form) -

Shortest distance between \vec{r}=\vec{a}+\lambda \vec{b}\, \: and \,\: \vec{r}=\vec{a_{1}}+\mu\vec{b}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 |
 

- wherein

shortest distance is among the line which is perpendicular to both

LM= \vec{b}\times \vec{b_{1}}

shortest distance will be projection of PQ = \vec{a}- \vec{a_{1}} on LM

 

 

(3 , -2 , -\lambda ) \: \: lies \: \: on \: \: plane

 

So,  6+8 -3 \lambda = 2 \\ \Rightarrow 3 \lambda = 12 \\ \lambda = 4

Where \vec b \times \vec c = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -1 &-2 \\ 12 & 9 & 4 \end{vmatrix} = 14 \hat i -28 \hat j +21 \hat k

So distance  = 0 

 

 

 


Option 1)

2

Option 2)

1

Option 3)

0

Option 4)

3

Posted by

gaurav

View full answer

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