Chapter 11, Problem 11PS

To determine

To Prove: that the volume V of a parallelepiped with vector u, v and w as adjacent edges is given by $V=\left|u\cdot \left(v\times w\right)\right|.$

Explanation of Solution

Given information:

A parallelepiped with vector u, v and w as adjacent edges.

Concept used:

A polyhedron whose all faces are parallelogram is called parallelepiped.

The volume of the parallelepiped will be the product of its height and area of its base.

The area of base will be the cross product of the base vector.

Calculation:

  

This figure shows a parallelepiped with u, v and w as adjacent edges.

Let $\theta$ be the angle between u and $v\times w$ .

Here area of base will be the cross product of the base vector that is $\Vert v\times w\Vert$ .

Height of the parallelepiped is the projection of vector u on $v\times w$ . That is $\Vert {\text{proj}}_{v\times w}u\Vert$

Also $\Vert {\text{proj}}_{v\times w}u\Vert =\left|n\right|\left|u\right|\cos \theta$ where n is a unit vector.

Hence

  $\begin{array}{c}\text{Volume}=\left(\text{Area}\text{of}\text{base}\right)\left(\Vert {\text{proj}}_{v\times w}u\Vert \right)\\ =\left(\text{Area}\text{of}\text{base}\right)\left(\left|n\right|\left|u\right|\cos \theta \right)\left(\begin{array}{l}\because \Vert {\text{proj}}_{v\times w}u\Vert =\left|n\right|\left|u\right|\cos \theta \\ n\text{is}\text{a}\text{unit}\text{vector}\end{array}\right)\\ =\Vert v\times w\Vert \left(1\cdot \left|u\right|\cos \theta \right)\left(\because \left|n\right|\right)\\ =\Vert v\times w\Vert \left(\left|u\right|\cos \theta \right)\\ =\left|u\cdot \left(v\times w\right)\right|\left(\because x\cdot y=\left|x\right|\left|y\right|\cos \theta \right)\end{array}$

Therefore the volume of a parallelepiped is $\left|u\cdot \left(v\times w\right)\right|$ . Hence the theorem is proved.

Conclusion:

The volume of a parallelepiped with vector u, v and w as adjacent edges is $\left|u\cdot \left(v\times w\right)\right|$ .