Find the shortest distance, d, from the point (6, 0, -4) to the plane x + y + z = 4?

Solution:

Given the equation of the plane: x + y + z = 4

To find: the distance from the point (6, 0, -4) to the plane.

We have, distance of the point \((x_{1}, y_{1}, z_{1})\) to the plane Ax+By+Cz+D=0 is

P= \( | \frac{Ax_{1}+By_{1}+Cz_{1}+D}{\sqrt{A^{2}+B^{2}+C^{2}}}|\) --- (i)

Substituting the values of \((x_{1}, y_{1}, z_{1})\) = (6, 0, -4) in the equation(i)

P= \( | \frac{x_{1}+y_{1}+z_{1}-4}{\sqrt{A^{2}+B^{2}+C^{2}}}  |\)

P= |(6 + 0 - 4 - 4)/√3|

P = 2/√3

Find the shortest distance, d, from the point (6, 0, -4) to the plane x + y + z = 4?

Summary:

The shortest distance, d, from the point (6, 0, -4) to the plane x + y + z = 4 is 2/√3.