Question

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Answer 1

$\large\underline{\sf{Solution-}}$

The given equation of lines are

$\rm :\longmapsto\:\dfrac{x - 3}{3} = \dfrac{y - 8}{ - 1} = \dfrac{z - 3}{1} - - - (1)$

and

$\rm :\longmapsto\:\dfrac{x + 3}{ - 3} = \dfrac{y + 7}{2} = \dfrac{z - 6}{4} - - - (2)$

Now,

Line (1) passes through the point (3, 8, 3) and having direction ratios (3, - 1, 1).

In Vector form, it can be represented as

$\rm :\longmapsto\:\vec{a_1} = 3\hat{i} + 8\hat{j} + 3\hat{k}$

and

$\rm :\longmapsto\:\vec{b_1} = 3\hat{i} - \hat{j} + \hat{k}$

Also,

Line (2) passes through the point (- 3, - 7, 6) and having direction ratios (- 3, 2, 4).

In Vector form, it can be represented as

$\rm :\longmapsto\:\vec{a_2} = - 3\hat{i} - 7\hat{j} + 6\hat{k}$

and

$\rm :\longmapsto\:\vec{b_1} = - 3\hat{i} + 2\hat{j} + 4\hat{k}$

Now,

Consider,

$\red{\rm :\longmapsto\:\vec{a_2} -\vec{a_1}}$

$\rm \: = \: - 3\hat{i} - 7\hat{j} + 6\hat{k} - 3\hat{i} - 8\hat{j} - 3\hat{k}$

$\rm \: = \: - 6\hat{i} - 15\hat{j} + 3\hat{k}$

Now,

Consider,

$\red{\rm :\longmapsto\:\vec{b_1} \times \vec{b_2}}$

$\rm \: = \: \: \: \begin{gathered}\sf \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\3& - 1& 1\\ - 3&2& 4\end{array}\right | \end{gathered}$

$\rm \: = \: - 6\hat{i} - 15\hat{j} + 3\hat{k}$

Therefore,

$\red{\rm :\longmapsto\: |\vec{b_1} \times \vec{b_2}|}$

$\rm \: = \: \sqrt{ {( - 6)}^{2} + {(15)}^{2} + {(3)}^{2} }$

$\rm \: = \: \sqrt{36 + 225 + 9}$

$\rm \: = \: \sqrt{270}$

Now, we know that,

Shortest distance between two skew lines is given by

$\red{\rm :\longmapsto\:S.D. \: = \dfrac{ | \: (\vec{a_2} -\vec{a_1}).(\vec{b_1} \times \vec{b_2}) \: |}{ |\vec{b_1} \times \vec{b_2}| } }$

So, on substituting all these values, we get

$\rm \: = \: \dfrac{ |(\: - 6\hat{i} - 15\hat{j} + 3\hat{k}) \: .\: ( - 6\hat{i} - 15\hat{j} + 3\hat{k}) \: | }{ \sqrt{270} }$

$\rm \: = \: \dfrac{ |36 + 225 + 9| }{ \sqrt{270} }$

$\rm \: = \: \dfrac{ 270 }{ \sqrt{270} }$

$\rm \: = \: \sqrt{270} \: units$

Hence,

$\boxed{ \bf{ \: Shortest \: Distance \: between \: lines \: is \: \sqrt{270} \: units}}$

Additional Information :-

Let us consider two parallel lines,

$\rm :\longmapsto\:\vec{r} = \vec{a_1} + \alpha \: \vec{b}$

and

$\rm :\longmapsto\:\vec{r} = \vec{a_2} + \beta \: \vec{b}$

then shortest distance between two parallel lines is

$\rm :\longmapsto\:S.D = \dfrac{ |(\vec{a_2} -\vec{a_1}) \times \vec{b}| }{ |\vec{b}| }$