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  • Find the shortest distance between lines r vector =6 i caret + 2 j caret + 2 k caret + lambda i caret - 2 j caret - 2 k caret and r vector = -4 i caret -k caret plus mu 3 i caret- 2 j caret- 2 k caret.

9.  Find the shortest distance between lines \overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda (\widehat{i}-2\widehat{j}-2\widehat{k}) and \overrightarrow{r}=-4\widehat{i}-\widehat{k}+\mu(3\widehat{i}-2\widehat{j}-2\widehat{k}).

Answers (1)

best_answer

Given lines are; 

\overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda (\widehat{i}-2\widehat{j}-2\widehat{k})  and

\overrightarrow{r}=-4\widehat{i}-\widehat{k}+\mu(3\widehat{i}-2\widehat{j}-2\widehat{k})

So, we can find the shortest distance between two lines \vec{r} = \vec{a_{1}}+\lambda \vec{b_{1}} and \vec{r} = \vec{a_{1}}+\lambda \vec{b_{1}} by the formula,

d = \left | \frac{(\vec{b_{1}}\times\vec{b_{2}}).(\vec{a_{2}}-\vec{a_{1}})}{\left | \vec{b_{1}}\times\vec{b_{2}} \right |} \right |                                               ...........................(1)

Now, we have from the comparisons of the given equations of lines.

\vec{a_{1}} = 6\widehat{i}+2\widehat{j}+2\widehat{k}     \vec{b_{1}} = \widehat{i}-2\widehat{j}+2\widehat{k}

\vec{a_{2}} = -4\widehat{i}-\widehat{k}             \vec{b_{2}} = 3\widehat{i}-2\widehat{j}-2\widehat{k}

So, \vec{a_{2}} -\vec{a_{1}} = (-4\widehat{i}-\widehat{k}) -(6\widehat{i}+2\widehat{j}+2\widehat{k}) = -10\widehat{i}-2\widehat{j}-3\widehat{k}

and \Rightarrow \vec{b_{1}}\times \vec{b_{2}} = \begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 1 &-2 &2 \\ 3& -2 &-2 \end{vmatrix} = (4+4)\widehat{i}-(-2-6)\widehat{j}+(-2+6)\widehat{k}

=8\widehat{i}+8\widehat{j}+4\widehat{k}

\therefore \left | \vec{b_{1}}\times \vec{b_{2}} \right | = \sqrt{8^2+8^2+4^2} =12

(\vec{b_{1}}\times \vec{b_{2}}).(\vec{a_{2}}-\vec{a_{1}}) = (8\widehat{i}+8\widehat{j}+4\widehat{k}).(-10\widehat{i}-2\widehat{j}-3\widehat{k}) = -80-16-12 =-108Now, substituting all values in equation (3) we get,

d = | \frac{-108}{12}| = 9

Hence the shortest distance between the two given lines is 9 units.

 

Posted by

Divya Prakash Singh

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