Quite a simple thing, but can't work out one step.

Charge √3Q is at cartesian point (3a, a, 2a). Calculate the electric field at the point (2a, 2a, 3a) due to this charge.

using the formula;

E(r) = q / 4 ∏ ε_{0}|r - r_{0}|^{3}(r - r_{0})

Here

r = (2a e_{x}+ 2a e_{y}+ 3a e_{z})

r_{0}= (3a e_{x}+ a e_{y}+ 2a e_{z})

r - r_{0}= (-a e_{x}+ a e_{y}+ a e_{z})

|r - r_{0}| = √-1a^{2}+1a^{2}+1a^{2 so E(r) = }√3 Q / 4 ∏ ε_{0 (}√3a)^{3}* (-a e_{x}+ a e_{y}+ a e_{z}) ------1

which goes to

^{E(r) = }Q / 4 ∏ ε_{0 }a^{2}* (-0.33 e_{x}+ 0.33 e_{y}+ 0.33 e_{z}) -------2

I need to know how to get to the last part.

i.e. the algebra bit from 1 to 2

where does the √3 bits get transformed into the 0.33 in the vector at the end.

Just feeling dumb when I REALLY shouldn't.

Thank you