JEE Main20199 Apr 2019Evening ShiftMathematicsThree Dimensional GeometryActual
The vertices B and C of a Δ A B C lie on the line, x + 2 3 = y - 1 0 = z 4 such that B C = 5 units. Then the area (in sq. units) of this triangle, given the point A 1 ,   - 1 ,   2 , is
Options
- A6
- B2 34
- C34
- D5 17
Correct answer
C. 34
Step-by-step solution
The points B and C lie on the line x + 2 3 = y - 1 0 = z 4 . Draw perpendicular A D on the line B C . Clearly area of Δ A B C = 1 2 ⋅ A D ⋅ B C To find a point on the line, let x + 2 3 = y - 1 0 = z 4 = r ⇒ x + 2 = 3 r ,   y - 1 = 0 ,   z = 4 r ⇒ x = 3 r - 2 ,   y = 1 ,   z = 4 r Thus, the point D ≡ 3 r - 2 ,   1 ,   4 r The direction ratios of a line joining two points x 1 ,   y 1 ,   z 1 and x 2 ,   y 2 ,   z 2 are < x 2 - x