Let a line L1 pass through the origin and be perpendicular to the lines L2 : r=(3+t) i+(2t-1) j+(2t+4) k and L3 : r=(3+2s) i+(3+2s) j+(2+s) k, t, s R. If (a, b, c), a Z, is the point on L3 at a distance of 17 from the point of intersection of L1 and L2, then (a+b+c)2 is equal to .
Three Dimensional Geometry
Previous year questions with solutions · 2025 – 2026
Year-wise statistics questions · papers · average per paper · weightage
| Year | Questions | Papers | Avg / paper | Weightage |
|---|---|---|---|---|
| 2026 | 32 | 19 | 1.7 | 6.74% ↓3.8% |
| 2025 | 21 | 12 | 1.8 | 7.00% |
Let the image of the point P(0, -5, 0) in the line x-12= y1= z+1-2 be the point R and the image of the point Q (0, -12, 0 ) in the line x-1-1= y+94= z+11 be the point S. Then the square of the area of the parallelogram PQRS is .
If the distance of the point P(43, , ), < 0, from the line r = 4 i - k + (2 i + 3 k), R along a line with direction ratios 3, -1, 0 is 13 10, then 2 + 2 is equal to
Let a line L passing through the point P(1, 1, 1) be perpendicular to the lines x-44 = y-11 = z-11 and x-171 = y-711 = z0. Let the line L intersect the yz-plane at the point Q. Another line parallel to L and passing through the point S(1, 0, -1) intersects the yz-plane at the point R. Then the square of the area of the parallelogram PQRS is equal to .
If the image of the point P(a, 2, a) in the line x2 = y + a1 = z1 is Q and the image of Q in the line x - 2b2 = y - a1 = z + 2b-5 is P , then a + b is equal to .
Let the area of the triangle formed by the lines x + 2 = y - 1 = z, x - 35 = y-1 = z - 11 and x-3 = y - 33 = z - 21 be A. Then A2 is equal to .