Chapter 9.5, Problem 21E

To determine

To find: The area of parallelogram formed by the given vectors.

Answer to Problem 21E

The area of parallelogram formed by the given vectorsis $4\sqrt{6}$ .

Explanation of Solution

Given information:

The vectorsare $u=\langle 3,2,1\rangle ,v=\langle 1,2,3\rangle$ .

Calculation:

The cross product is given as,

  $a\times b=\left|\begin{array}{ccc}i& j& k\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|$

Substitute the values in above equation.

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}i& j& k\\ 3& 2& 1\\ 1& 2& 3\end{array}\right|\\ =4i-8j+4k\end{array}$

The magnitude of cross product gives the area of parallelogram.

  $\begin{array}{c}\left|u\times v\right|=\sqrt{{\left(4\right)}^2+{\left(-8\right)}^2+{\left(4\right)}^2}\\ =4\sqrt{6}\end{array}$