1. \(\overline {AB}\) = (3 – 1)i + (- 3 – 3)j + (3 – 0)k = 2i – 6j + 3k.
2. The Cartesian equation of a line passing through the point (4, 0, -1 ) and parallel to the vector \(\overline {AB}\) is
3. We take, \(\frac{x-4}{2} = \frac{y}{-6} = \frac{z+1}{3}\) = r then any point of the line can be taken as (2r + 4, -6r, 3r – 1).
Assume that this point be the foot of the perpendicular drawn from (2, 3, 4 ). The dr’s of the line is 2 : – 6 : 3 and dr’s of the perpendicular line L is
2r + 4 – 2 : -6r – 3 : 3r – 1 – 4 ⇒ 2r + 2: -6r – 3 : 3r – 5
Since perpendicular,
2(2 r + 2) – 6(-6 r -3) + 3(3r – 5) = 0
49r = -7 ⇒ r = −7/49 = –1/7.
Therefore the foot of the perpendicular is