JEE MainMathematicsThree Dimensional Geometry
The base BC of a triangle ABC lies on the line L₁: x 2 = y-1 1 = z-2 2 and has a length of 3 2 units. The vertex A lies on the line L₂: x-2 1 = y-2 2 = z-4 -1 . If the area of ABC is 10 sq. units, there are two possible positions for the vertex A , say A₁ and A₂ . The value of OA₁^2 + OA₂^2 , where O is the origin, is _______.
Correct answer
96
Step-by-step solution
Let the height of ABC be h . The area is given by: 1 2 BC h = 10 1 2 3 2 h = 10 h = 20 3 2 h^2 = 400 18 = 200 9 . The vertex A lies on L₂ , so its coordinates can be written parametrically as A(k+2, 2k+2, -k+4) . The line L₁ passes through P₀(0, 1, 2) and has a direction vector b = 2 i + j + 2 k . The magnitude is | b | = 4+1+4 = 3 . The vector P₀A = (k+2 - 0) i + (2k+2 - 1) j + (-k+4 - 2) k = (k+2) i + (2k+1) j + (-k+2) k . The perpendicular distance h from A to L₁ is given by h = | P₀A b | | b | . P₀A b = vmatrix