## Thomas' Calculus 13th Edition

$7 \sqrt 3$
The distance formula for two vectors can be calculated as: $d=\dfrac{|p \times q|}{|q|}$ Thus, we have $p \times q=\lt -28,-56,28\gt$ and $|p \times q|=\sqrt{(-28)^2+(-56)^2+(28)^2}=\sqrt{56+112+56}=28 \sqrt {6}$ Thus, $d=\dfrac{28 \sqrt {6}}{ \sqrt {(4)^2+(0)^2+(4)^2}}=\dfrac{28 \sqrt {6}}{ 4 \sqrt {2}}=7 \sqrt 3$