Single Choice

The shortest distance between the lines $$x = y + 2 = 6z - 6$$ and $$x + 1 = 2y = -12z$$ is

A$$\dfrac {1}{3}$$
B$$2$$
Correct Answer
C$$1$$
D$$\dfrac {3}{2}$$

Solution

The lines are $$\dfrac {x}{6} = \dfrac {y + 2}{6} = \dfrac {z - 1}{1}$$
and $$\dfrac {x + 1}{12} = \dfrac {y}{6} = \dfrac {z}{-1}$$
Here,
$$\vec {a}_{1} = -2\hat {j} + \hat {k}, b_{1} = 6\hat {i} + 6\hat {j} + \hat {k}, \vec {a}_{2} = -\hat {i}, \vec {b}_{2} = 12\hat {i} + 6\hat {j} - \hat {k}$$
$$\vec {b}_{1} \times \vec {b}_{2} = \begin{vmatrix} \hat {i}& \hat {j} & k\\ 6 & 6 &1 \\ 12 & 6 & -1\end{vmatrix} = -12\hat {i} + 18\hat {j} - 36\hat {k}$$
Shortest distance $$= \dfrac {|(\vec {a_{2}} - \vec {a_{1}}).(\vec {b_{1}} \times \vec {b_{2}})|}{|\vec {b_{1}} \times \vec {b_{2}}|}$$
$$= \dfrac {|(-\hat {i} + 2\hat {j} - \hat {k}).(-12i + 18\hat {j} - 36\hat {k})|}{\sqrt {(-12)^{2} + (18)^{2} + (-36)^{2}}}$$
$$= \dfrac {|+12 + 36 + 36|}{\sqrt {1764}} = \dfrac {84}{42} = 2$$
Hence, the shortest distance is $$2$$.

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