If the lines (x−1)/−3 = (y−2)/2k = (z−3)/2 and (x−1)/3k = (y−1)/1 = (z−6)/−5 are at right angles, the value of k is: A. −10/7 B. 10/7 C. −7/10 D. 7/10 Answer: A) −10/7 Explanation: For perpendicular lines, a₁a₂ + b₁b₂ + c₁c₂ = 0. So, (−3)(3k) + (2k)(1) + (2)(−5) = 0 → −9k + 2k − 10 = 0 → −7k = 10 → k = −10/7.

imo class 12 three dimensional geometry

If the lines (x−1)/−3 = (y−2)/2k = (z−3)/2 and (x−1)/3k = (y−1)/1 = (z−6)/−5 are at right angles, the value of k is:

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If the lines (x−1)/−3 = (y−2)/2k = (z−3)/2 and (x−1)/3k = (y−1)/1 = (z−6)/−5 are at right angles, the value of k is:

Answer: A) −10/7

Explanation: For perpendicular lines, a₁a₂ + b₁b₂ + c₁c₂ = 0. So, (−3)(3k) + (2k)(1) + (2)(−5) = 0 → −9k + 2k − 10 = 0 → −7k = 10 → k = −10/7.