Question

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

Short answer

The lengths of the diagonals are $\sqrt{6},\sqrt{42},\sqrt{41},$ and $3\sqrt{6}$.

Step-by-step solution

Given Information

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:

$\begin{array}{rl}& d_1=|u+v|\\ & d_2=|v+w|\\ & d_3=|u+w|\end{array}$

The length of the body diagonal is:

$d_4=|u+v+w|$

Find the length of the first face diagonal

$\begin{array}{rl}{d}_1& =|(2i+4j-k)+(-3j+2k)|\\ & =|2i+j+k|\\ & =\sqrt{(2{)}^2+(1{)}^2+(1{)}^2}\\ & =\sqrt{4+1+1}\\ & =\sqrt{6}\end{array}$

The length of the 2nd face diagonal

$\begin{array}{rl}{d}_2& =|(-3j+2k)+(-i+j+5k)|\\ & =|-i-2j+7k|\\ & =\sqrt{(-1{)}^2+(-2{)}^2+(7{)}^2}\\ & =\sqrt{1+4+49}\\ & =\sqrt{54}\\ & =3\sqrt{6}\end{array}$

Find the length of the 3rd face diagonal

$\begin{array}{rl}{d}_3& =|(2i+4j-k)+(-i+j+5k)|\\ & =|i+5j+4k|\\ & =\sqrt{(1{)}^2+(5{)}^2+(4{)}^2}\\ & =\sqrt{1+25+16}\\ & =\sqrt{42}\end{array}$

Find the length of the body diagonal

$\begin{array}{rl}{d}_4& =|u+v+w|\\ & =|(2i+4j-k)+(-3j+2k)+(-i+j+5k)|\\ & =|i+2j+6k|\\ & =\sqrt{(1{)}^2+(2{)}^2+(6{)}^2}\\ & =\sqrt{1+4+36}\\ & =\sqrt{41}\end{array}$