Chapter 11.3, Problem 19E

To determine

ToFind:The value of $u\times v$ for the vectors $u=6i+2j+k$ and $v=i+3j-2k$ . Show that it is orthogonal to both $u$ and $v$

Answer to Problem 19E

  $u\times v=-7i+13j+16k$

  $u\times v$ is orthogonal to $u$

  $u\times v$ is orthogonal to $v$ .

Explanation of Solution

Given:

The vectors $u=6i+2j+k$ and $v=i+3j-2k$

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 $u=6i+2j+k$ and $v=i+3j-2k$

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}i& j& k\\ 6& 2& 1\\ 1& 3& -2\end{array}\right|\\ =i\left|\begin{array}{cc}2& 1\\ 3& -2\end{array}\right|-j\left|\begin{array}{cc}6& 1\\ 1& -2\end{array}\right|+k\left|\begin{array}{cc}6& 2\\ 1& 3\end{array}\right|\\ =i\left(\left(2\right)\left(-2\right)-\left(3\right)\left(1\right)\right)-j\left(\left(6\right)\left(-2\right)-\left(1\right)\left(1\right)\right)+k\left(\left(6\right)\left(3\right)-\left(1\right)\left(2\right)\right)\\ =i\left(-4-3\right)-j\left(-12-1\right)+k\left(18-2\right)\end{array}$

  $=-7i+13j+16k$

Therefore, $u\times v=-7i+13j+16k$

Consider

  $\begin{array}{c}\left(u\times v\right)\cdot u=\left(-7i+13j+16k\right)\cdot \left(6i+2j+k\right)\\ =\left(-7\right)\left(6\right)+\left(13\right)\left(2\right)+\left(16\right)\left(1\right)\\ =-42+26+16\\ =0\end{array}$

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

Consider

  $\begin{array}{c}\left(u\times v\right)\cdot v=\left(-7i+13j+16k\right)\cdot \left(i+3j-2k\right)\\ =\left(-7\right)\left(1\right)+\left(13\right)\left(3\right)+\left(16\right)\left(-2\right)\\ =-7+39-32\\ =0\end{array}$

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

Conclusion:

  $u\times v=-7i+13j+16k$

  $u\times v$ is orthogonal to $u$

  $u\times v$ is orthogonal to $v$ .