Chapter 9.5, Problem 23E

To determine

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

Answer to Problem 23E

The area of parallelogram formed by the given vectorsis $9.92$ .

Explanation of Solution

Given information:

The vectorsare $u=2i-j+4k,v=\frac{1}{2}i+2j-\frac{3}{2}k$ .

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.

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}i& j& k\\ 2& -1& 4\\ \frac{1}{2}& 2& -\frac{3}{2}\end{array}\right|\\ =-\frac{13}{2}i+6j+\frac{9}{2}\end{array}$

The magnitude of cross product gives the area of parallelogram.

  $\begin{array}{c}\left|u\times v\right|=\sqrt{{\left(-\frac{13}{2}\right)}^2+{\left(6\right)}^2+{\left(\frac{9}{2}\right)}^2}\\ =9.92\end{array}$