Chapter 5 - Inner Product Spaces - 5.2 Orthogonal Sets of Vectors and Orthogonal Projections - Problems - Page 361: 32

$\frac{10}{\sqrt 17}$

From exercise 28, we have the formula: $$d=\frac{|x_0.a+y_0.b+z_0.c+d}{\sqrt a^2+b^2+c^2}$$ Since $P(8,8,-1)$ and plan $P:z=2x$ we have: $$distance=\frac{|8.0+8.1+(-1).(-4)+(-2)|}{\sqrt 0^2+1^2+(-4)^2}=\frac{|0+8+4+2|}{\sqrt 17}=\frac{10}{\sqrt 17}$$