Question

Find the shortest distance, d, from the point (5, 0, −6) to the plane x + y + z = 6.

Answer 1

`(7)/(√(3))`

Explanation:

The shortest distance d, of a point (a, b, c) from a plane mx + ny + tz = r is given by:

`d = |((ma + nb + tc - r))/(√(m^2 + n^2 + t^2)) |` --------------------(i)

From the question,

the point is (5, 0, -6)

the plane is x + y + z = 6

Therefore,

a = 5

b = 0

c = -6

m = 1

n = 1

t = 1

r = 6

Substitute these values into equation (i) as follows;

`d = |(((1*5) + (1*0) + (1 * (-6)) - 6))/(√(1^2 + 1^2 + 1^2)) |`

`d = |(((5) + (0) + (-6) - 6))/(√(1 + 1 + 1)) |`

`d = |((-7))/(√(3)) |`

`d = (7)/(√(3))`

Therefore, the shortest distance from the point to the plane is
`d = (7)/(√(3))`