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Class 12 Three Dimensional Geometry · Long Answer3 Markstransversals

Class 12 Maths · Three Dimensional Geometry · Long Answer #44

Find a line through (1,1,1) meeting r=(−2,3,−1)+s(1,2,4) and r=(1,2,3)+u(2,3,4).

✅ Step-by-Step Solution

Final Answer: r=(1,1,1)+t(3,10,17).
  1. Let Q=(−2+s,3+2s,−1+4s), R=(1+2u,2+3u,3+4u), P=(1,1,1).
  2. Require Q−P and R−P proportional. The cross-product equations give s=9,u=3/11.
  3. Q=(7,21,35), R=(17/11,31/11,45/11); Q−P=2(3,10,17), R−P=(2/11)(3,10,17).
  4. Thus r=P+t(3,10,17); its parameters t=2 and 2/11 give the required intersections.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Let Q=(−2+s,3+2s,−1+4s), R=(1+2u,2+3u,3+4u), P=(1,1,1).
  • [1 mark] Require Q−P and R−P proportional. The cross-product equations give s=9,u=3/11.
  • [1 mark] Q=(7,21,35), R=(17/11,31/11,45/11); Q−P=2(3,10,17), R−P=(2/11)(3,10,17).
  • [0.5 marks] Thus r=P+t(3,10,17); its parameters t=2 and 2/11 give the required intersections.

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