Question

Find the areas of the six faces of the parallelepiped determined

by \(u=<2,4,-1>\), \(v=<0,-3,2>\), and \(w=<-1,1,5>\).

Short answer

The lengths of the faces are \(\sqrt{77},\sqrt{77},\sqrt{558},\sqrt{558},\sqrt{302}\), and \(\sqrt{302}\).

Step-by-step solution

Given Information

The sides of the parallelepied are \(u=<2,4,-1>\), \(v=<0,-3,2>\), and \(w=<-1,1,5>\).

There are 6 faces but opposite faces are similar. As the area of the similar figure is equal.

To find the area of the three adjacent faces.

The area of the 1st face is \(|u\times v|\).

The area of the 2nd face is \(|u\times w|\).

The area of the 3rd face is \(|w\times v|\).

Find the area of the first face

\(\begin{align*}|u\times v|&= \left | \begin{vmatrix}i&j&k\\2&4&-1\\0&-3&2\end{vmatrix} \right |\\&=|5i-4j-6k|\\&=\sqrt{5^2+(-4)^2+(-6)^2}\\&=\sqrt{25+16+36}\\&=\sqrt{77}\end{align*}\)

Find the area of the second face

\(\begin{align*}|u\times w|&= \left | \begin{vmatrix}i&j&k\\2&4&-1\\-1&1&5\end{vmatrix} \right |\\&=|21i-9j+6k|\\&=\sqrt{(21)^2+(-9)^2+(6)^2}\\&=\sqrt{441+81+36}\\&=\sqrt{558}\end{align*}\)

Find the area of the third face

\(\begin{align*}|w\times v|&= \left | \begin{vmatrix}i&j&k\\-1&1&5\\0&-3&2\end{vmatrix} \right |\\&=|17i+2j+3k|\\&=\sqrt{(17)^2+(2)^2+(3)^2}\\&=\sqrt{289+4+9}\\&=\sqrt{302}\end{align*}\)