Chapter 11.3, Problem 10E

To determine

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

Answer to Problem 10E

  $\left(-2u\right)\times v=2i-14j-10k$

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 $v=2i+j-k$

That is

  $\begin{array}{c}-2u=-2\left(3i-j+2k\right)\\ =-6i+2j-4k\end{array}$

Then,

  $\begin{array}{c}\left(-2u\right)\times v=\left|\begin{array}{ccc}i& j& k\\ -6& 2& -4\\ 2& 1& -1\end{array}\right|\\ =i\left|\begin{array}{cc}2& -4\\ 1& -1\end{array}\right|-j\left|\begin{array}{cc}-6& -4\\ 2& -1\end{array}\right|+k\left|\begin{array}{cc}-6& 2\\ 2& 1\end{array}\right|\\ =i\left(\left(2\right)\left(-1\right)-\left(1\right)\left(-4\right)\right)-j\left(\left(-6\right)\left(-1\right)-\left(2\right)\left(-4\right)\right)+k\left(\left(-6\right)\left(1\right)-\left(2\right)\left(2\right)\right)\\ =i\left(-2+4\right)-j\left(6+8\right)+k\left(-6-4\right)\end{array}$

  $=2i-14j-10k$

Therefore, $\left(-2u\right)\times v=2i-14j-10k$

Conclusion:

  $\left(-2u\right)\times v=2i-14j-10k$