Question

find the minimum distance from the point (8,0,8) to the plane x - y + z = 5. (hint: to simplify the computations, minimize the square of the distance.)

Explanation:

Step1: Recall the distance formula

The distance \(d\) from a point \((x_0,y_0,z_0)\) to a plane \(Ax + By+ Cz+D = 0\) is \(d=\frac{\vert Ax_0 + By_0 + Cz_0+D\vert}{\sqrt{A^{2}+B^{2}+C^{2}}}\). First, rewrite the plane equation \(x - y+z = 5\) as \(x - y+z-5=0\). Here \(A = 1\), \(B=-1\), \(C = 1\), \(D=-5\), and the point \((x_0,y_0,z_0)=(8,0,8)\).

Step2: Substitute values into the formula

Substitute \(x_0 = 8\), \(y_0 = 0\), \(z_0 = 8\), \(A = 1\), \(B=-1\), \(C = 1\), \(D=-5\) into the formula \(d=\frac{\vert Ax_0 + By_0 + Cz_0+D\vert}{\sqrt{A^{2}+B^{2}+C^{2}}}\). We get \(d=\frac{\vert1\times8+(-1)\times0 + 1\times8-5\vert}{\sqrt{1^{2}+(-1)^{2}+1^{2}}}\).

Step3: Calculate numerator and denominator

Calculate the numerator: \(\vert8 + 0+8 - 5\vert=\vert11\vert = 11\). Calculate the denominator: \(\sqrt{1 + 1+1}=\sqrt{3}\).

Answer:

\(\frac{11}{\sqrt{3}}\)