Chapter 11.3, Problem 24E

To determine

ToFind:The value of $u\times v$ for the vectors $u=1i+\frac{3}{2}j-\frac{5}{2}k$ and $v=\frac{1}{2}i-\frac{3}{4}j+\frac{1}{4}k$ . Show that it is orthogonal to both $u$ and $v$

Answer to Problem 24E

  $u\times v=-\frac{3}{2}i-\frac{3}{2}j-\frac{3}{2}k$

  $u\times v$ is orthogonal to $u$

  $u\times v$ is orthogonal to $v$ .

Explanation of Solution

Given:

The vectors $u=1i+\frac{3}{2}j-\frac{5}{2}k$ and $v=\frac{1}{2}i-\frac{3}{4}j+\frac{1}{4}k$

Concept Used:

For the two vectors $u=\langle u_1,u_2,u_3\rangle$ and $v=\langle v_1,v_2,v_3\rangle$ , the cross product of them can be calculated as

  $u\times v=\left|\begin{array}{ccc}i& j& k\\ u_1& u_2& u_3\\ v_1& v_2& v_3\end{array}\right|$

Two vectors $u=\langle u_1,u_2,u_3\rangle$ and $v=\langle v_1,v_2,v_3\rangle$ are orthogonal, if their dot product is zero. That is $u\cdot v=0$

Calculation:

Forthe vectors $u=1i+\frac{3}{2}j-\frac{5}{2}k$ and $v=\frac{1}{2}i-\frac{3}{4}j+\frac{1}{4}k$

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}i& j& k\\ 1& \frac{3}{2}& -\frac{5}{2}\\ \frac{1}{2}& -\frac{3}{4}& \frac{1}{4}\end{array}\right|\\ =i\left|\begin{array}{cc}\frac{3}{2}& -\frac{5}{2}\\ -\frac{3}{4}& \frac{1}{4}\end{array}\right|-j\left|\begin{array}{cc}1& -\frac{5}{2}\\ \frac{1}{2}& \frac{1}{4}\end{array}\right|+k\left|\begin{array}{cc}1& \frac{3}{2}\\ \frac{1}{2}& -\frac{3}{4}\end{array}\right|\\ =i\left(\left(\frac{3}{2}\right)\left(\frac{1}{4}\right)-\left(-\frac{3}{4}\right)\left(-\frac{5}{2}\right)\right)-j\left(\left(1\right)\left(\frac{1}{4}\right)-\left(\frac{1}{2}\right)\left(-\frac{5}{2}\right)\right)+k\left(\left(1\right)\left(-\frac{3}{4}\right)-\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\right)\\ =i\left(\frac{3}{8}-\frac{15}{8}\right)-j\left(\frac{1}{4}+\frac{5}{4}\right)+k\left(-\frac{3}{4}-\frac{3}{4}\right)\end{array}$

  $\begin{array}{l}=-\frac{12}{8}i-\frac{6}{4}j-\frac{6}{4}k\\ =-\frac{3}{2}i-\frac{3}{2}j-\frac{3}{2}k\end{array}$

Therefore, $u\times v=-\frac{3}{2}i-\frac{3}{2}j-\frac{3}{2}k$

Consider

  $\begin{array}{c}\left(u\times v\right)\cdot u=\left(-\frac{3}{2}i-\frac{3}{2}j-\frac{3}{2}k\right)\cdot \left(1i+\frac{3}{2}j-\frac{5}{2}k\right)\\ =\left(-\frac{3}{2}\right)\left(1\right)+\left(-\frac{3}{2}\right)\left(\frac{3}{2}\right)+\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\\ =-\frac{3}{2}-\frac{9}{4}+\frac{15}{4}\\ =-\frac{3}{2}+\frac{6}{4}\end{array}$

  $\begin{array}{l}=-\frac{3}{2}+\frac{3}{2}\\ =0\end{array}$

Since, $\left(u\times v\right)\cdot u=0$ . Therefore, $u\times v$ is orthogonal to $u$ .

Consider

  $\begin{array}{c}\left(u\times v\right)\cdot v=\left(-\frac{3}{2}i-\frac{3}{2}j-\frac{3}{2}k\right)\cdot \left(\frac{1}{2}i-\frac{3}{4}j+\frac{1}{4}k\right)\\ =\left(-\frac{3}{2}\right)\left(\frac{1}{2}\right)+\left(-\frac{3}{2}\right)\left(-\frac{3}{4}\right)+\left(-\frac{3}{2}\right)\left(\frac{1}{4}\right)\\ =-\frac{3}{4}+\frac{9}{8}-\frac{3}{8}\\ =-\frac{3}{4}+\frac{6}{8}\end{array}$

  $\begin{array}{l}=-\frac{3}{4}+\frac{3}{4}\\ =0\end{array}$

Since, $\left(u\times v\right)\cdot v=0$ . Therefore, $u\times v$ is orthogonal to $v$ .

Conclusion:

  $u\times v=-\frac{3}{2}i-\frac{3}{2}j-\frac{3}{2}k$

  $u\times v$ is orthogonal to $u$

  $u\times v$ is orthogonal to $v$ .