Question

Find the perpendicular distance of the point (3, −1, 11) from the line \[\frac{x}{2} = \frac{y - 2}{- 3} = \frac{z - 3}{4} .\]

Answer 1

Given equation of line AB is

`x/2=("y"-2)/3=("z"-3)/4=lambda`[say]

or  `x/2=lambda,("y"-2)/3=lambda`

and  `("z"-3)/4=lambda`

or x = 2λ, y = 3λ + 2

and z = 4λ + 3

∴ Any point P on the given line

= (2λ, 3λ + 2, 4λ + 3) 

Let P be the foot of Perpendicular drawn from point

Q(3, -1, 11) on line AB. Now, DR's of line

QP = (2λ - 3, 3λ + 2 + 1, 4λ + 3 - 11)

= (2λ - 3, 3λ + 3, 4λ - 8)

Here, a₁ = 2λ -3, b₁ = 3λ + 3, c₁ = 4λ - 8,

and a₂ = 2, b₂ = 3, c₂ = 4

Since, QP ⊥ AB

∴ We have, a₁a₂ + b₁b₂ + c₁c₂ = 0      ...(i) 

or 2(2λ - 3) + 3 (3λ + 3) + 4 (4λ - 8) = 0

or 4λ - 6 + 9λ + 9 + 16λ - 32 = 0

or 29λ - 29 = 0 or 29λ = 29 or λ = 1

∴ Foot of Perpendicular 

 P = (2, 3 + 2, 4 + 3)

= (2, 5, 7)

Now, equation of Perpendicular QP, Where Q (3, -1, 11) and P (2, 5, 7) is

`(x-3)/(2-3)=("y"+1)/(5+1)=("z"-11)/(7-11)`

`[("using two points form of equation of line"), ("i"."e". (x-x_1)/(x_2-x_1)=("y"-"y"_1)/("y"_2-"y"_1)=("z"-"z"_1)/("z"_2-"z"_1))]`

or `(x-3)/-1=("y"+1)/6=("z"-11)/-4`

Now, length of Perpendicular QP = distance between points Q(3, -1, 11) and P (2, 5, 7)

= `sqrt((2-3)^2+(5+1)^2+(7-11)^2)`

`[(∵ "distance"), (=sqrt((x_2-x_1)^2+("y"_2-"y"_1)^2+("z"_2-"z"_1)^2))]`

= `sqrt(1+36+16)`

= `sqrt53`

Hence, length of Perpendicular is `sqrt53`.