# How do you find the cross product and verify that the resulting vectors are perpendicular to the given vectors <3,2,0>times<1,4,0>?

Dec 21, 2016

$\left\langle0 , 0 , 10\right\rangle$

#### Explanation:

We can use the determinant notation to compute the cross product:
$\setminus \setminus \setminus \setminus \setminus \left(\begin{matrix}3 \\ 2 \\ 0\end{matrix}\right) \times \left(\begin{matrix}1 \\ 4 \\ 0\end{matrix}\right) = | \left(\underline{\hat{i}} , \underline{\hat{j}} , \underline{\hat{k}}\right) , \left(3 , 2 , 0\right) , \left(1 , 4 , 0\right) |$

$\therefore \left(\begin{matrix}3 \\ 2 \\ 0\end{matrix}\right) \times \left(\begin{matrix}1 \\ 4 \\ 0\end{matrix}\right) = | \left(2 , 0\right) , \left(4 , 0\right) | \underline{\hat{i}} - | \left(3 , 0\right) , \left(1 , 0\right) | \underline{\hat{j}} + | \left(3 , 2\right) , \left(1 , 4\right) | \underline{\hat{k}}$

$\therefore \left(\begin{matrix}3 \\ 2 \\ 0\end{matrix}\right) \times \left(\begin{matrix}1 \\ 4 \\ 0\end{matrix}\right) = \left(0 - 0\right) \underline{\hat{i}} - \left(0 - 0\right) \underline{\hat{j}} + \left(12 - 2\right) \underline{\hat{k}}$

$\therefore \left(\begin{matrix}3 \\ 2 \\ 0\end{matrix}\right) \times \left(\begin{matrix}1 \\ 4 \\ 0\end{matrix}\right) = 0 \underline{\hat{i}} + 0 \underline{\hat{j}} + 10 \underline{\hat{k}}$
$\therefore \left(\begin{matrix}3 \\ 2 \\ 0\end{matrix}\right) \times \left(\begin{matrix}1 \\ 4 \\ 0\end{matrix}\right) = \left(\begin{matrix}0 \\ 0 \\ 10\end{matrix}\right)$, or $\left\langle0 , 0 , 10\right\rangle$

To confirm that this vector is perpendicular we can check the dot product is zero:

$\left\langle3 , 2 , 0\right\rangle \cdot \left\langle0 , 0 , 10\right\rangle = 0 + 0 + 0 = 0$
$\left\langle1 , 4 , 0\right\rangle \cdot \left\langle0 , 0 , 10\right\rangle = 0 + 0 + 0 = 0$