Question

The distance of the point (1, 1, 9) from the point of intersection of the line `(x - 3)/1 = ("y" - 4)/2 = ("z" - 5)/2` and the plane x + y + z = 17 is ______.

Options

• `sqrt(38)`

• `19sqrt(2)`

• `2sqrt(19)`

• 38

Answer 1

The distance of the point (1, 1, 9) from the point of intersection of the line `(x - 3)/1 = ("y" - 4)/2 = ("z" - 5)/2` and the plane x + y + z = 17 is `underline(sqrt(38))`.

Explanation:

Given: Point is P(1, 1, 9)

Line is `(x - 3)/1 = ("y" - 4)/2 = ("z" - 5)/2`

Plane is x + y + z = 17

`(x - 3)/1 = ("y" - 4)/2 = ("z" - 5)/2` = k (say)

⇒ x = k + 3, y = 2k + 4, z = 2k + 5

(k + 3, 2k + 4, 2k + 5)

Points (x, y, z) lies on plane x + y + z = 17

⇒ k + 3 + 2k + 4 + 2k + 5 = 17

⇒ 5k = 5

⇒ k = 1

Point of intersection of line and plane is

Q(k + 3, 2k + 4, 2k + 5) i.e.

Q(4, 6, 7)

Required distance = PQ

= `sqrt((4 - 1)^2 + (6 - 1)^2 + (9 - 7)^2`  ...(Using distance formula)

= `sqrt(3^2 + 5^2 + 2^2)`

= `sqrt(38)`