The shortest distance between the lines x2=y2=z1 and x+2−1=y−48=z−54 lies in the interval

|x2−x1y2−y1z2−z1a1b1c1a2b2c2|=-245221-184=54|ijka1b1c1a2b2c2|=ijk221-184=-9j+18kshortest distance=5495=65=2.683

The shortest distance between the lines x2=y2=z1 and x+2−1=y−48=z−54 lies in the interval

|x2−x1y2−y1z2−z1a1b1c1a2b2c2|=-245221-184=54|ijka1b1c1a2b2c2|=ijk221-184=-9j+18kshortest distance=5495=65=2.683

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The shortest distance between the lines $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{x + 2}{- 1} = \frac{y - 4}{8} = \frac{z - 5}{4}$ lies in the interval

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$[1, 2)$

$(3, 4]$

$(2, 3]$

$[0, 1)$

answer is C.

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Detailed Solution

$| \begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix} | = |\begin{matrix} - 2 & 4 & 5 \\ 2 & 2 & 1 \\ - 1 & 8 & 4 \end{matrix}| = 54$

$| \begin{matrix} i & j & k \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix} | = |\begin{matrix} i & j & k \\ 2 & 2 & 1 \\ - 1 & 8 & 4 \end{matrix}| = - 9 j + 18 k$