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Chapter 10: Q.23 (page 848)

Compute the areas of the six faces of the parallelepiped

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

Short Answer

Expert verified

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

Step by step solution

01

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 \(|w\times v|\).

02

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

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