#### Question

Find the length of the perpendicular from the point (3, 2, 1) to the line `(x-7)/2=(y-7)/2=(z-6)/3=lambda (say)`

#### Solution

Let M be the foot of the perpendicular drawn from point A (3, 2, 1) to the given line

`(x-7)/2=(y-7)/2=(z-6)/3=lambda (say)`

therefore x=7-2λ,y=7+2λ,z=6+3λ

are the coordinates of any point on the given line

M=(7-2λ,y=7+2λ,6+3λ)

The d.r.s of AM are 7 - 2λ - 3, 7 + 2λ - 2, 6 + 3λ-1

i.e. 4 - 2λ, 5 + 2λ, 5 + 3λ

The d.r.s of the given line are -2, 2, 3

Since, AM is perpendicular to the given line

(4 - 2λ) (-2) + (5 + 2λ) (2) + (5 + 3λ) (3) = 0

-8 + 4λ + 10 + 4λ + 15 + 9λ = 0

17λ+ 17 = 0

λ = -1 [1]

The coordinates of the foot of the perpendicular

i.e. M = (9, 5, 3)

The length of the perpendicular distance

i.e. AM = ` sqrt((9-3)^2+(5-2)^2+(3-1)^2)`

= `sqrt(49)`

l(AM) = 7 units