Chapter 11.3, Problem 6E

To determine

ToEvaluate:The expression $v\times \left(u\times u\right)$ for the vectors $u=3i-j+2k$ and $v=2i+j-k$

Answer to Problem 6E

  $v\times \left(u\times u\right)=0i-0j+0k=0$

Explanation of Solution

Given:

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

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

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

Calculation:

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

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

  $=0i-0j+0k$

Then,

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

  $=0i-0j+0k$

Therefore, $v\times \left(u\times u\right)=0i-0j+0k=0$

Conclusion:

  $v\times \left(u\times u\right)=0i-0j+0k=0$