Question

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

Answer 1

Answer:

$\frac{7}{\sqrt{3}}$

Step-by-step explanation:

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

$d = |\frac{(ma + nb + tc - r)}{\sqrt{m^2 + n^2 + t^2}} |$                   --------------------(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;

$d = |\frac{((1*5) + (1*0) + (1 * (-6)) - 6)}{\sqrt{1^2 + 1^2 + 1^2}} |$

$d = |\frac{((5) + (0) + (-6) - 6)}{\sqrt{1 + 1 + 1}} |$

$d = |\frac{(-7)}{\sqrt{3}} |$

$d = \frac{7}{\sqrt{3}}$

Therefore, the shortest distance from the point to the plane is  $d = \frac{7}{\sqrt{3}}$