Find the shortest distance between the lines

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Question 5:- Find the shortest distance between the lines :

\dfrac{x + 1}{2} = \dfrac{y - 1}{1} = \dfrac{z - 9}{-3}

\dfrac{x - 3}{2} = \dfrac{y + 15}{-7} = \dfrac{z - 9}{5}.

Solution:- Shortest distance between two skew line

= \left|\dfrac{(\vec{a_2} - \vec{a_1}). (\vec{b_1}\times \vec{b_2})}{|\vec{b_1}\times \vec{b_2}|}\right|

Point on Ist line = (-1, 1, 9) and dr’s of Ist line = (2, 1, -3)

Point on IInd line = (3, -15, 9) and dr’s of IInd line = (2, -7, 5)

Since, \vec{a_1} = -\hat{i} + \hat{j} + 9\hat{k} \text{ and } \vec{b_1} = 2\hat{i} + \hat{j} - 3\hat{k}

\vec{a_2} = 3\hat{i} - 15\hat{j} + 9\hat{k} \text{ and } \vec{b_1} = 2\hat{i} - 7\hat{j} + 5\hat{k}

\vec{a_2} - \vec{a_1} = (3\hat{i} - 15\hat{j} + 9\hat{k}) - (-\hat{i} + \hat{j} + 9\hat{k} )

⇒ \vec{a_2} - \vec{a_1} = 4\hat{i} - 16\hat{j} + 0\hat{k}

\vec{b_1}\times \vec{b_2} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 1 & -3 \\ 2 & -7 & 5 \end{vmatrix}

⇒ \vec{b_1}\times \vec{b_2} = \hat{i}(5 - 21) - \hat{j}(10 + 6) + \hat{k}(-14-2)

⇒ \vec{b_1}\times \vec{b_2} = -16\hat{i} - 16\hat{j} -16 \hat{k}

(\vec{a_2} - \vec{a_1}).(\vec{b_1}\times \vec{b_2}) = 4.(-16) - 16(-16) + 0(-16)

⇒ (\vec{a_2} - \vec{a_1}).(\vec{b_1}\times \vec{b_2}) = -64 + 256 = 192

|\vec{b_1}\times \vec{b_2}| = \sqrt{(-16)^2 + (-16)^2 + (-16)^2}

⇒ |\vec{b_1}\times \vec{b_2}| = 16\sqrt{3}

Shortest distance between two skew line

= \left|\dfrac{(\vec{a_2} - \vec{a_1}). (\vec{b_1}\times \vec{b_2})}{|\vec{b_1}\times \vec{b_2}|}\right|

= \dfrac{192}{16\sqrt{3}}

= \dfrac{12}{\sqrt{3}}.\dfrac{\sqrt{3}}{\sqrt{3}}

= \dfrac{12\sqrt{3}}{3} = 4\sqrt{3}

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