Answer:
[tex]\frac{7}{\sqrt{3}}[/tex]
Step-by-step explanation:
The shortest distance d, of a point (a, b, c) from a plane mx + ny + tz = r is given by:
[tex]d = |\frac{(ma + nb + tc - r)}{\sqrt{m^2 + n^2 + t^2}} |[/tex] --------------------(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;
[tex]d = |\frac{((1*5) + (1*0) + (1 * (-6)) - 6)}{\sqrt{1^2 + 1^2 + 1^2}} |[/tex]
[tex]d = |\frac{((5) + (0) + (-6) - 6)}{\sqrt{1 + 1 + 1}} |[/tex]
[tex]d = |\frac{(-7)}{\sqrt{3}} |[/tex]
[tex]d = \frac{7}{\sqrt{3}}[/tex]
Therefore, the shortest distance from the point to the plane is [tex]d = \frac{7}{\sqrt{3}}[/tex]
