Question

Find the image of the point (2, -1, 5) in the line `(x - 11)/(10) = (y + 2)/(-4) = (z + 8)/(-11)`. Also, find the length of the perpendicular from the point (2, -1, 5) to the line. 

Answer 1

The given line `(x - 11)/(10) = (y + 2)/(-4) = (z + 8)/(-11)`

Let N be the foot of the perpendicular drawn from the point P(2, -1, 5).
Any point on line (1) is N (11 + 10t, -2 – 4t, -8 – 11t).
Now, direction ratio of NP is: < 9 + 10t, -1 – 4t, -13 -11t >
and direction ratio of line (1) is < 10, -4, -11 >
10(9 + 10t) + 4(1 + 4t) + 11(13 + 11t) = 0
⇒ 90 + 100t + 4 + 16t+ 143+ 121t = 0
⇒ 237t + 237 = 0
⇒ t = -1
N = (1, 2, 3)
Now, image of point P (2, -1, 5) is

(1, 2, 3) = `((2 + x)/2, (-1+y)/(2), (5 +z)/(2))`

⇒ `(2 + x)/(2) =1`                    ⇒ x = 0

⇒ `(-1 + y)/(2) = 2`                 ⇒ y = 5

⇒ `(5 + z)/(2) = 3`                  ⇒ z = 1

∴ Images is Q (0, 5, 1)
Required length of perpendicular from the point (2, -1, 5) to line 
= `sqrt((2 - 1)^2 + (-1-2)^2 + (5 -3)^2) = sqrt(1 +9 + 4) = sqrt14  "sq. units".`