CBSE Class 12 Mathematics 2026 Main Paper 65/5/2 · Q335 Markspoint_projection
CBSE Class 12 Mathematics 2026 Main Paper 65/5/2 · Question 33 (Three Dimensional Geometry)
Find the foot from P=(0,2,3) to (−x−3)/(−5)=(1−y)/(−2)=(3z+12)/9 and the perpendicular length. OR The lines r=(1,2,1)+λ(1,−1,1) and r=(p,−1,−1)+μ(2,1,2) are 3/√2 units apart. Find p.
Alternative 1 5 Marks
Find the foot from P=(0,2,3) to (−x−3)/(−5)=(1−y)/(−2)=(3z+12)/9 and the perpendicular length.
Answer: Foot (2,3,−1); length √21.
- Write the foot as H=(−3+5t,1+2t,−4+3t).
- Set (H−P)·(5,2,3)=38t−38=0, so t=1.
- Then H=(2,3,−1), and H−P=(2,1,−4) has norm √21.
Alternative 2 5 Marks
The lines r=(1,2,1)+λ(1,−1,1) and r=(p,−1,−1)+μ(2,1,2) are 3/√2 units apart. Find p.
Answer: p=−4 or p=2.
- Their direction cross product is (−3,0,3). The displacement between base points is (p−1,−3,−2).
- The shortest distance is the absolute scalar triple product divided by the normal length, giving |p+1|/√2.
- Set |p+1|=3 and solve both signs.