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  • Solve this problem - Three Dimensional Geometry - JEE Main-2

lines x-4=y-5=\frac{z+1}{2}  and \frac{x-6}{2}=\frac{y-11}{4}=z+3  are

  • Option 1)

    Parallel lines 

  • Option 2)

    Coincident lines 

  • Option 3)

    skew lines 

  • Option 4)

    intersecting lines 

 

Answers (1)

best_answer

As we have learned

Condition for lines to be intersecting (vector form) -

Their shortest distance should be 0

\left ( \vec{b}\times \vec{b_{1}} \right )\cdot \left ( \vec{a}- \vec{a_{1}} \right )=0

Also the condition for coplanar lines

-

 

 lines can be written in vector form as 

\vec{r}=(4\hat{i}+5\hat{j}-\hat{k})+\lambda(\hat{i}+\hat{j}+2\hat{k})

\vec{r}=(6\hat{i}+11\hat{j}-3\hat{k})+\mu (2\hat{i}+4\hat{j}+\hat{k})

Now ,

\vec{a}=4\hat{i}+5\hat{j}-\hat{k}, \vec{b}=\hat{i}+\hat{j}+2\hat{k}

\vec{a_1}=6\hat{i}+11\hat{j}-3\hat{k}, \vec{b_1}=2\hat{i}+4\hat{j}+\hat{k}

\therefore \vec{b}\times \vec{b_1}=\begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ 1 & 1 & 2\\ 2 &4 & 1 \end{vmatrix}= \hat{i}(-7)-\hat{j}(-3)+2\hat{k}

\vec{a}- \vec{a_1}= -2\hat{i}-6\hat{j}+2\hat{k}

(\vec{b}\times \vec{b_1})\cdot (\vec{a}- \vec{a_1})=14-18+4=0

Now, \vec{b}\neq k\vec{b_1} so lines are neither parallel nor co-incident

and 

(\vec{b}\times \vec{b_1})\cdot (\vec{a}- \vec{a_1})=0

\therefore Lines are intersecting

 

 

 

 


Option 1)

Parallel lines 

Option 2)

Coincident lines 

Option 3)

skew lines 

Option 4)

intersecting lines 

Posted by

Himanshu

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