CBSE Class 12 Mathematics 2026 Supplementary Paper 65/7/2 · Q384 Markscase_study
CBSE Class 12 Mathematics 2026 Supplementary Paper 65/7/2 · Question 38 (Three Dimensional Geometry)
A cuboid swimming pool has breadth line \((-x+2)/(-5)=y=z-3\). Students A and B swim along its length. Student A follows \((x+5)/1=(y+3)/4=(z-6)/(-3p)\). The breadth and length are perpendicular.
Part (1) 1 Marks
Find p if the length direction (1,4,−3p) is perpendicular to the breadth direction (5,1,1).
Answer: p=3
- Perpendicularity requires 5+4−3p=0.
- Hence p=3.
Part (2) 1 Marks
Write student B’s line parallel to student A’s path and passing through (2,4,−1).
Answer: r=(2,4,−1)+t(1,4,−9).
- At p=3, student A’s direction is (1,4,−9).
- Use that direction and the point (2,4,−1) to obtain the line.
Part (3) 2 Marks
Find the foot from (2,4,−1) to student A’s path.
Answer: (−4,1,−3)
- Let H=(−5,−3,6)+t(1,4,−9). The perpendicular condition is (P−H)·(1,4,−9)=0.
- Since P−(−5,−3,6)=(7,7,−7), both the projection numerator and direction squared norm equal 98, so t=1.
- Substitution gives H=(−4,1,−3).
Part (4) 2 Marks
Find the shortest distance from student A’s path to the bird line \((x+4)/2=(y+3)/8=(z-5)/(-18)\).
Answer: 4√3/7
- The bird’s direction (2,8,−18) is twice (1,4,−9), so the paths are parallel.
- The base displacement is (1,0,−1), whose cross product with (1,4,−9) is (4,8,4), of length √96. The shared direction length is √98.
- The distance is √96/√98=4√3/7.