Chapter 10.3, Problem 21E

To determine

To find: The value of $u\times v$ for the vectors $u=\langle 2,4,3\rangle$ and $v=\langle 0,-2,1\rangle$ with the help of graphing utility.

Answer to Problem 21E

The value of $u\times v$ for the vectors $u=\langle 2,4,3\rangle$ and $v=\langle 0,-2,1\rangle$ is $10\text{i}-2\text{j}-4\text{k}$ .

Explanation of Solution

Given information:

The vectors are $u=\langle 2,4,3\rangle$ and $v=\langle 0,-2,1\rangle$ .

Calculation:

Use the formula for cross product.

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 2& 4& 3\\ 0& -2& 1\end{array}\right|\\ =\text{i}\left|\begin{array}{cc}4& 3\\ -2& 1\end{array}\right|-\text{j}\left|\begin{array}{cc}2& 3\\ 0& 1\end{array}\right|+\text{k}\left|\begin{array}{cc}2& 4\\ 0& -2\end{array}\right|\end{array}$

To find the value of determinants with the help of graphing utility follow the steps below:

Press “ON” button on graphical calculator. First enter the matrix to the calculator. Enter the keystrokes.

  $\boxed{2\text{nd}}\to \boxed{\text{MATRX}}\to \boxed{\text{EDIT}}\to \boxed{\left[\text{A}\right]}$

Enter the dimensions and elements of the matrix $2\times 2$ as shown below in the figure.

  

Now, return to the home screen by using the keystrokes.

  $\boxed{2\text{nd}}\to \boxed{\text{MODE}}$

To find the determinant press the keystrokes as:

  $\boxed{2\text{nd}}\to \boxed{\text{MATRX}}\to \boxed{\text{MATH}}\to \boxed{\det (}\to \boxed{\text{ENTER}}$

Select the matrix in which to find the determinant by keystrokes.

  $\boxed{2\text{nd}}\to \boxed{\text{MATRX}}\to \boxed{\text{NAMES}}\to \boxed{\left[\text{A}\right]}\to \boxed{)}\to \boxed{\text{ENTER}}$

The determinant of the given matrix is as shown below on the screen.

  

So, the determinant of the matrix $\left[\begin{array}{cc}4& 3\\ -2& 1\end{array}\right]$ is $10$ .

Use the similar procedure for the determinant of matrix $\left[\begin{array}{cc}2& 3\\ 0& 1\end{array}\right]$ and $\left[\begin{array}{cc}2& 4\\ 0& -2\end{array}\right]$ .

The determinant of matrix $\left[\begin{array}{cc}2& 3\\ 0& 1\end{array}\right]$ and $\left[\begin{array}{cc}2& 4\\ 0& -2\end{array}\right]$ is $2$ and $-4$ respectively.

Substitute the values in the formula.

  $\begin{array}{c}u\times v=\text{i}\left|\begin{array}{cc}4& 3\\ -2& 1\end{array}\right|-\text{j}\left|\begin{array}{cc}2& 3\\ 0& 1\end{array}\right|+\text{k}\left|\begin{array}{cc}2& 4\\ 0& -2\end{array}\right|\\ =10\text{i}-2\text{j}+\left(-4\right)\text{k}\\ =10\text{i}-2\text{j}-4\text{k}\end{array}$

Therefore, the value of $u\times v$ for the vectors $u=\langle 2,4,3\rangle$ and $v=\langle 0,-2,1\rangle$ is $10\text{i}-2\text{j}-4\text{k}$ .