Question

Compute the areas of the six faces of the parallelepiped

determined by \(u=i\), \(v=2j\), and \(w=2k\).

Short answer

The area of the faces are \(2\) sq units, \(2\) sq units, and \(1\) sq unit.

Step-by-step solution

Given Information

The three vectors are \(u=i\), \(v=2j\), and \(w=2k\).

The opposite faces of the parallelepiped are identical.

The area of the first face of the parallelepiped is \(|u\times v|\).

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

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

Find the area of the faces

The area of 1st face: \(|u\times v|=\sqrt{2^2}\)

\(=2\) sq units

The area of the 2nd face: \(|u\times w|=\sqrt{2^2}\)

\(=2\) sq units

The area of the 3rd face: \(|w\times v|=\sqrt{1^2}\).
\(=1\) sq unit