Question
Find the length and the foot of the perpendicular drawn from the point (2, – 1, 5) to the line 
Solution
Let the given line AB be
Any point M on this line is
(10 r + 11, – 4 r – 2, – 11 r – 8).
Let this point be foot of perpendicular from P(2, – 1, 5) on AB.
Any point M on this line is
(10 r + 11, – 4 r – 2, – 11 r – 8).
Let this point be foot of perpendicular from P(2, – 1, 5) on AB.
Direction ratios of PM are
10 r + 11 – 2, – 4 r – 2 + 1, – 11 r – 8 – 5 i.e. 10 r + 9, – 4 r – 1, – 11 r – 13
Direction ratios of AB are 10, – 4, – 11.
Since PM ⊥ AB
∴ (10) (10 r + 9) + (–4) (–4 r – 1) + (–11) (– 11 r – 13) = 0
∴ 100 r + 90 + 16 r + 4 + 121 r + 143 = 0
∴ 237 r + 237 = 0 or 237 r = – 237 ⇒ r = – 1
∴ M is (– 10 + 11, 4 – 2, 11 – 8) i.e. (1, 2, 3)
which is foot of perpendicular.
Length of perpendicular = PM
