The length of the shortest distance between the two lines ${\bf{\vec r}} = \left( { - 3\widehat {\bf{i}} + 6\widehat {\bf{j}}} \right) + s\left( { - 4\widehat {\bf{i}} + 3\widehat {\bf{j}} + 2{\bf{\hat k}}} \right)$ and ${\bf{\vec r}} = \left( { - 2\widehat {\bf{i}} + 7{\bf{\hat k}}} \right) + t\left( { - 4\widehat {\bf{i}} + \widehat {\bf{j}} + {\bf{\hat k}}} \right){\rm{\;}}$ is

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

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Application +4 | -1

The length of the shortest distance between the two lines

\mathbf{\vec r} = \left( { - 3\widehat \mathbf{i} + 6\widehat \mathbf{j}} \right) + s\left( { - 4\widehat \mathbf{i} + 3\widehat \mathbf{j} + 2\mathbf{\hat k}} \right)

and

\mathbf{\vec r} = \left( { - 2\widehat \mathbf{i} + 7\mathbf{\hat k}} \right) + t\left( { - 4\widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k}} \right)\mathrm{\;}

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Correct option: D

9 units

Option D (9 units) is correct because the shortest distance between the two lines is calculated using the formula involving the vector difference and the cross product of the direction vectors.

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

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