Shortest Distance Between Skew Lines

The required machinery is the shortest-distance formula for skew lines: d = |(a_2 - a_1) \cdot (b_1 \times b_2)| / |b_1 \times b_2|, where a_1, a_2 are points and b_1, b_2 are direction vectors. This scalar-triple-product method is a hallmark JEE Advanced technique. First compute b_1 \times b_2 with b_1 = (1, 0, -1) and b_2 = (0, 1, 1): the cross product is (0\cdot1 - (-1)\cdot1,\ (-1)\cdot0 - 1\cdot1,\ 1\cdot1 - 0\cdot0) = (1, -1, 1). The connecting vector a_2 - a_1 = (1, 2, 2). Its dot with (1, -1, 1) is 1 - 2 + 2 = 1, so the numerator is |1| = 1. The denominator is |b_1 \times b_2| = \sqrt{1 + 1 + 1} = \sqrt{3}, giving d = 1/\sqrt{3}. Option 3/\sqrt{3} overstates the numerator as 3. Option 2/\sqrt{3} uses a partial dot product. Option \sqrt{3} squares the denominator into the numerator. This invokes the scalar-triple-product distance theorem, whose geometric meaning is that the numerator is the volume of the parallelepiped built on the connecting vector and the two directions, while the denominator is the area of its base spanned by the two directions, so their ratio is the perpendicular height between the lines. That height is exactly the common perpendicular, the unique shortest segment joining the skew pair. Plausibility check: skew lines have a unique positive shortest distance, and 1/\sqrt{3} is positive and finite as expected, and a nonzero cross product confirms the directions are not parallel, which is a necessary precondition for the lines to be genuinely skew rather than parallel.