Find the lengths of the four diagonals of the parallelepiped determined by \(u=<2,4,-1>\), \(v=<0,-3,2>\), and \(w=<-1,1,5>\).

The given vectors are \(u=<2,4,-1>\), \(v=<0,-3,2>\), and \(w=<-1,1,5>\).

There are four diagonals three face diagonals and one body diagonal.

The lengths of the face diagonals are:

\(\begin{align*}&d_1=|u+v|\\&d_2=|v+w|\\&d_3=|u+w|\end{align*}\)

The length of the body diagonal is:

\(d_4=|u+v+w|\)

\(\begin{align*}d_1&=|(2i+4j-k)+(-3j+2k)|\\&=|2i+j+k|\\&=\sqrt{(2)^2+(1)^2+(1)^2}\\&=\sqrt{4+1+1}\\&=\sqrt6\end{align*}\)

\(\begin{align*}d_2&=|(-3j+2k)+(-i+j+5k)|\\&=|-i-2j+7k|\\&=\sqrt{(-1)^2+(-2)^2+(7)^2}\\&=\sqrt{1+4+49}\\&=\sqrt{54}\\&=3\sqrt6\end{align*}\)

\(\begin{align*}d_3&=|(2i+4j-k)+(-i+j+5k)|\\&=|i+5j+4k|\\&=\sqrt{(1)^2+(5)^2+(4)^2}\\&=\sqrt{1+25+16}\\&=\sqrt{42}\end{align*}\)

\(\begin{align*}d_4&=|u+v+w|\\&=|(2i+4j-k)+(-3j+2k)+(-i+j+5k)|\\&=|i+2j+6k|\\&=\sqrt{(1)^2+(2)^2+(6)^2}\\&=\sqrt{1+4+36}\\&=\sqrt{41}\end{align*}\)