Question

The distance of the point (1, 5, 9) from the plane x y + z = 5 measured along the line x = y = z is

Answer 1

Answer:

10√3 units

Step-by-step explanation:

Given plane x - y + z = 5,

(Please note -y misprint in question)

Also please note was (1, -5, 9) not (1, 5, 9) again a misprint, if it was ( 1, 5, 9) it lies on the plane and distance is zero.

Any line passing through (1, -5, 9) and having dr's parallel to line

x/1 = y/1 = z/1

will be in the form

x-1/1 = y+5/1 = z-9/1

Any point on the above line would be of the form

(1 + λ , -5 + λ, 9 + λ),

for this point to be on the given plane

1 + λ + 5 - λ + 9 + λ = 5

=> λ = -10

Thus, the point on the plane measured along line x = y = z is

( 1-10, -5-10, 9-10)

(-9, -15, -1).

The distance of (1, -5, 9 ) from the point (-9, -15, -1)

is √10² + 10² + 10² = 10√3.

Hope , it helped !