CBSE Class 12 Mathematics 2026 Supplementary Paper 65/7/3 · Q263 Marksline_distance
CBSE Class 12 Mathematics 2026 Supplementary Paper 65/7/3 · Question 26 (Three Dimensional Geometry)
Find the line through (1,−1,2) parallel to the x-axis and its shortest distance to r=(2,1,−1)+t(3,−5,2). OR For A=(3,−4,2), B=(−1,−2,0), C=(4,1,−1), D=(4,0,5), find lines AC and BD and show their directions are perpendicular.
Alternative 1 3 Marks
Find the line through (1,−1,2) parallel to the x-axis and its shortest distance to r=(2,1,−1)+t(3,−5,2).
Answer: r=(1,−1,2)+s(1,0,0); distance 11/√29.
- The x-parallel direction is u=(1,0,0), so the line has fixed y=−1,z=2.
- For v=(3,−5,2), u×v=(0,−2,−5). The base displacement is (1,2,−3), with dot product 11 against that normal.
- Divide abs(11) by √29 for the distance.
Alternative 2 3 Marks
For A=(3,−4,2), B=(−1,−2,0), C=(4,1,−1), D=(4,0,5), find lines AC and BD and show their directions are perpendicular.
Answer: AC=(3,−4,2)+t(1,5,−3); BD=(−1,−2,0)+s(5,2,5).
- Subtract endpoints to get AC=(1,5,−3) and BD=(5,2,5).
- The two line equations follow by adding real multiples of these directions to A and B. Their dot product is 5+10−15=0.
- Their nonzero directions therefore make a right angle.