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  • I need help with - Three Dimensional Geometry - JEE Main-3

Shortest distance between lines \vect{r}=(3\hat{i}-10\hat{j}+\hat{k})+\lambda(2\hat{i}-9\hat{j}-2\hat{k}) and \vect{r}=(-5\hat{i}-3\hat{j}+6\hat{k})+\mu(2\hat{i}-9\hat{j}-2\hat{k}) is

  • Option 1)

    5

  • Option 2)

    6

  • Option 3)

    7

  • Option 4)

    8

 

Answers (1)

best_answer

As we have learned

Shortest distance between two parallel lines (vector form) -

Shortest distance between \vec{r}=\vec{p_{1}}+\lambda \vec{\nu }\: and \: \vec{r}=\vec{p_{2}}+\mu \vec{\nu }is given by

\frac{\left |\left ( \vec{p_{2} }-\vec{p_{1} } \right )\times \vec{\nu } \right |}{\left | \vec{\nu } \right |}

 

- wherein

 

 we see lines are parallel with

\vec{p_1}=3\hat{i}-10\hat{j}+\hat{k}; \vec{p_2}=-5\hat{i}-3\hat{j}+6\hat{k}

\vec{v}=2\hat{i}-9\hat{j}-2\hat{k}

\therefore \vec{p_2}-\vec{p_1}=-8\hat{i}+7\hat{j}+5\hat{k}

(\vec{p_2}-\vec{p_1})\times\vec{v}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ -8& 7 &5 \\ 2 & -9 &-2 \end{vmatrix}=31\hat{i}-6\hat{j}+58\hat{k}

\therefore Shortest Distance =\frac{|(\vec {p}_2-\vec{p}_1)|}{|\vec{v}|}=\frac{|31\hat{i}-6\hat{j}+58\hat{k}|}{|2\hat{i}-9\hat{j}-2\hat{k}|}

\therefore \frac{\sqrt{4361}}{\sqrt{89}}=\sqrt{49}=7


Option 1)

5

Option 2)

6

Option 3)

7

Option 4)

8

Posted by

Himanshu

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