Let the line L pass through (1, 1, 1) and intersect the lines x-12 = y+13 = z-14 and x-31 = y-42 = z1. Then, which of the following points lies on the line L?
- (7, 15, 13)
- (4, 22, 7)
- (10, -29, -50)
- (5, 4, 3)
Answer
(1) (7, 15, 13)
Solution
Let P(1, 1, 1). Line L1 passes through A(1, -1, 1) with direction (2, 3, 4), and line L2 through B(3, 4, 0) with direction (1, 2, 1).
Since L passes through P and meets L1, it lies in the plane containing P and L1. Here AP = (0, 2, 0) and the normal is (2, 3, 4) (0, 2, 0) = (-8, 0, 4) (2, 0, -1), so this plane 1 is 2(x-1) - (z-1) = 0, i.e. 2x - z = 1.
Similarly L lies in the plane containing P and L2: BP = (-2, -3, 1) and the normal is (1, 2, 1) (-2, -3, 1) = (5, -3, 1), so 2 is 5(x-1) - 3(y-1) + (z-1) = 0, i.e. 5x - 3y + z = 3.
Hence the direction of L is (2, 0, -1) (5, -3, 1) = (-3, -7, -6) (3, 7, 6), and
L: x-13 = y-17 = z-16 = , i.e. (1+3 , 1+7 , 1+6 )
= 2 gives (7, 15, 13), which therefore lies on L.
Check: 2(7) - 13 = 1 and 5(7) - 3(15) + 13 = 3 ✓. The other points do not satisfy both plane equations.
Answer: (1)