Question
Find the foot of perpendicular from the point (0, 2, 3) on the line
Solution
Let the given line AB be
Any point M on this line is (5 r – 3, 2 r + 1, 3 r – 4)
Let this point M be the first of perpendicular from P (0, 2, 3) on AB.
Direction-ratios of PM are
5 r – 3 – 0, 2 r + 1 – 2, 3 r – 4 – 3
i.e., 5 r – 3, 2 r – 1, 3 r – 7
Direction -ratios of AB are 5, 2, 3.
Since PM ⊥ AB
∴ (5 r – 3) (5) + (2 r – 1) (2) + (3 r – 7) (3) = 0
∴ 25 r – 15 + 4 r – 2 + 9 r – 21 = 0
∴ 38 r – 38 = 0 ⇒ r – 1 = 0 ⇒ r = 1
∴ M is (5 – 3, 2 + 1, 3 – 4) i.e.. (2, 3, – 1),
which is required foot of perpendicular.
