Discussion
The shortest distance between the point (2, 3, 4) and the line \(\frac{x-4}{-2} = \frac{y-4}{2} = \frac{z-6}{1}\) is
Answer: C
The shortest distance is 3.
A general point of the line is \(Q = (4 - 2t, 4 + 2t, 6 + t)\), with direction (-2, 2, 1). For the foot of the perpendicular from P(2, 3, 4), \(\vec{PQ}\) must be perpendicular to the direction:
\[ \begin{aligned} \vec{PQ} &= (2 - 2t, 1 + 2t, 2 + t) \\ \vec{PQ} \cdot (-2, 2, 1) &= 9t = 0 \Rightarrow t = 0 \end{aligned} \]
So the foot is (4, 4, 6) and the distance is \(\sqrt{2^{2} + 1^{2} + 2^{2}} = \sqrt{9} = 3\). The other values do not come from a perpendicular: 9 is the square of the distance, and \(\sqrt{5}\) and \(\sqrt{3}\) correspond to no point of the line at right angles to P.
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