Chapter 11.3, Problem 28E

To determine

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

Answer to Problem 28E

  $u\times v=-2i-1j-1k$

  $u\times v$ is orthogonal to $u$

  $u\times v$ is orthogonal to $v$ .

Explanation of Solution

Given:

The vectors $u=i-2k$ and $v=-j+k$

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 $u=i-2k$ and $v=-j+k$

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

  $=-2i-1j-1k$

Therefore, $u\times v=-2i-1j-1k$

Consider

  $\begin{array}{c}\left(u\times v\right)\cdot u=\left(-2i-1j-1k\right)\cdot \left(i-2k\right)\\ =\left(-2\right)\left(1\right)+\left(-1\right)\left(0\right)+\left(-1\right)\left(-2\right)\\ =-2+0+2\\ =0\end{array}$

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

Consider

  $\begin{array}{c}\left(u\times v\right)\cdot v=\left(-2i-1j-1k\right)\cdot \left(-j+k\right)\\ =\left(-2\right)\left(0\right)+\left(-1\right)\left(-1\right)+\left(-1\right)\left(1\right)\\ =0+1-1\\ =0\end{array}$

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

Conclusion:

  $u\times v=-2i-1j-1k$

  $u\times v$ is orthogonal to $u$

  $u\times v$ is orthogonal to $v$ .