What is the cross product of << -1, -1, 2 >> and << 4,3,6 >> ?

Nov 9, 2016

Well, you have at least two ways to do it.

The first way:

Let $\vec{u} = \left\langle{u}_{1} , {u}_{2} , {u}_{3}\right\rangle$ and $\vec{v} = \left\langle{v}_{1} , {v}_{2} , {v}_{3}\right\rangle$. Then:

$\textcolor{b l u e}{\vec{u} \times \vec{v}} = \left\langle{u}_{2} {v}_{3} - {u}_{3} {v}_{2} , {u}_{3} {v}_{1} - {u}_{1} {v}_{3} , {u}_{1} {v}_{2} - {u}_{2} {v}_{1}\right\rangle$

$= \left\langle- 1 \cdot 6 - 2 \cdot 3 , 2 \cdot 4 - \left(- 1 \cdot 6\right) , - 1 \cdot 3 - \left(- 1 \cdot 4\right)\right\rangle$

$= \textcolor{b l u e}{\left\langle- 12 , 14 , 1\right\rangle}$

Assuming you didn't know that formula, the second way (which is a little more foolproof) is recognizing that:

$\hat{i} \times \hat{j} = \hat{k}$
$\hat{j} \times \hat{k} = \hat{i}$
$\hat{k} \times \hat{i} = \hat{j}$
$\hat{A} \times \hat{A} = \vec{0}$
$\hat{A} \times \hat{B} = - \hat{B} \times \hat{A}$

where $\hat{i} = \left\langle1 , 0 , 0\right\rangle$, $\hat{j} = \left\langle0 , 1 , 0\right\rangle$, and $\hat{k} = \left\langle0 , 0 , 1\right\rangle$.

Thus, rewriting the vectors in unit vector form:

$\left(- \hat{i} - \hat{j} + 2 \hat{k}\right) \times \left(4 \hat{i} + 3 \hat{j} + 6 \hat{k}\right)$

$= {\cancel{- 4 \left(\hat{i} \times \hat{i}\right)}}^{0} - 3 \left(\hat{i} \times \hat{j}\right) - 6 \left(\hat{i} \times \hat{k}\right) - 4 \left(\hat{j} \times \hat{i}\right) - {\cancel{3 \left(\hat{j} \times \hat{j}\right)}}^{0} - 6 \left(\hat{j} \times \hat{k}\right) + 8 \left(\hat{k} \times \hat{i}\right) + 6 \left(\hat{k} \times \hat{j}\right) + {\cancel{12 \left(\hat{k} \times \hat{k}\right)}}^{0}$

$= - 3 \hat{k} + 6 \hat{j} + 4 \hat{k} - 6 \hat{i} + 8 \hat{j} - 6 \hat{i}$

$= - 12 \hat{i} + 14 \hat{j} + \hat{k}$

$= \textcolor{b l u e}{\left\langle- 12 , 14 , 1\right\rangle}$

as expected.