Question:
Let a line having direction ratios, 1, \(-4\), 2 intersect the lines \(\frac{x - 7}{3} = \frac{y - 1}{-1} = \frac{z + 2}{1}\) and \(\frac{x}{2} = \frac{y - 7}{3} = \frac{z}{1}\) at the points A and B. Then \((AB)^2\) is equal to ____________.
Let a line having direction ratios, 1, \(-4\), 2 intersect the lines \(\frac{x - 7}{3} = \frac{y - 1}{-1} = \frac{z + 2}{1}\) and \(\frac{x}{2} = \frac{y - 7}{3} = \frac{z}{1}\) at the points A and B. Then \((AB)^2\) is equal to ____________.
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Take A = (7+3s, 1-s, -2+s) and B = (2t, 7+3t, t); require that AB is proportional to (1, -4, 2) and solve for s and t.
Updated On: Nov 4, 2026
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Correct Answer: 84
Solution and Explanation
Let \(A = (7 + 3s,\ 1 - s,\ -2 + s)\) on the first line and \(B = (2t,\ 7 + 3t,\ t)\) on the second.
\(\vec{AB} = (2t - 7 - 3s,\ 6 + 3t + s,\ t + 2 - s)\) must be proportional to (1, -4, 2), i.e. \(\frac{2t - 7 - 3s}{1} = \frac{6 + 3t + s}{-4} = \frac{t + 2 - s}{2}\).
First = third: \(2(2t - 7 - 3s) = t + 2 - s \Rightarrow 3t - 5s = 16\).
First = second: \(-4(2t - 7 - 3s) = 6 + 3t + s \Rightarrow 11t - 11s = 22 \Rightarrow t - s = 2\).
Substituting \(t = s + 2\): \(3s + 6 - 5s = 16 \Rightarrow s = -5\), \(t = -3\).
\(A = (-8, 6, -7)\), \(B = (-6, -2, -3)\).
\(AB^2 = (2)^2 + (-8)^2 + (4)^2 = 4 + 64 + 16 = 84\).
Answer: 84
\(\vec{AB} = (2t - 7 - 3s,\ 6 + 3t + s,\ t + 2 - s)\) must be proportional to (1, -4, 2), i.e. \(\frac{2t - 7 - 3s}{1} = \frac{6 + 3t + s}{-4} = \frac{t + 2 - s}{2}\).
First = third: \(2(2t - 7 - 3s) = t + 2 - s \Rightarrow 3t - 5s = 16\).
First = second: \(-4(2t - 7 - 3s) = 6 + 3t + s \Rightarrow 11t - 11s = 22 \Rightarrow t - s = 2\).
Substituting \(t = s + 2\): \(3s + 6 - 5s = 16 \Rightarrow s = -5\), \(t = -3\).
\(A = (-8, 6, -7)\), \(B = (-6, -2, -3)\).
\(AB^2 = (2)^2 + (-8)^2 + (4)^2 = 4 + 64 + 16 = 84\).
Answer: 84
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