Question

The distance of the point (1,-2,3) from the plane x-y+z=5 measured along a line parallel to x/2=y/3=z/-6

Answer 1

Answer:

1 unit

Step-by-step explanation:

Hi,

Given plane x - y + z = 5,

Any line passing through (1, -2, 3) and having dr's parallel to line

x/2 = y/3 = z/-6

will be in the form

x-1/2 = y+2/3 = z-3/-6

Any point on the above line would be of the form

(1 +2λ , -2 + 3λ, 3 -6λ),

for this point to be on the given plane

1 + 2λ + 2 - 3λ + 3 - 6λ = 5

=> 7λ = 1

=> λ = 1/7

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

( 9/7, -11/7, 15/7)

The distance of (1, -2, 3 ) from the point ( 9/7, -11/7, 15/7)

is √(2/7)² + (3/7)² + (6/7)² = 1

Hope , it helped !