JEE MainMathematicsThree Dimensional Geometry
Let a plane pass through the points (1, 2, 0) , (0, -5, 0) and (2, 4, 1) . The distance of the point (2, 7, 8) from the plane , measured parallel to the line x 2 = y 3 = z 6 , is :
Options
- A2
- B28
- C38
- D14
Correct answer
D. 14
Step-by-step solution
Let the given points be A(1, 2, 0) , B(0, -5, 0) and C(2, 4, 1) . Two vectors on the plane are: AB = (0 - 1) i + (-5 - 2) j + (0 - 0) k = - i - 7 j AC = (2 - 1) i + (4 - 2) j + (1 - 0) k = i + 2 j + k A normal vector n to the plane is: n = AB AC = vmatrix i & j & k -1 & -7 & 0 1 & 2 & 1 vmatrix = -7 i + j + 5 k The equation of the plane passing through A(1, 2, 0) is: -7(x - 1) + 1(y - 2) + 5(z - 0) = 0 -7x + 7 + y - 2 + 5z = 0 7x - y - 5z = 5 Now, the equation of the line passing through P(2, 7, 8) and parallel to