When getting the cross product of two vectors, is the i value initially positive or negative? [ERROR ON MY END]

Question

[upon further discussion it appears this was just a sign error when performing operations with the matrix (comments in first answer)]

given $$ \vec{u} = \frac{1}{2} i -1j+\frac{2} {3} k ,\ \vec{v} = 6 i -12 j -6 k , $$

We can cross multiply these vectors to find a vector that is perpendicular or orthogonal to both of them. Two of these vectors exist and you can get them both by swapping the order you multiply them in. ($u\times v$ and $v\times u$) Apparently when we cross multiply these vectors we get $-2 i -1 j +0 k $ or $14 i +7 i +0 k $ depending on which order you multiply them.

I understand $-2 i -1 j +0 k $ but why is it we get $14 i +7 i +0 k $ instead of $-14 i +7 i +0 k $ if the determinant of the i matrix when we cross multiply is $-14$ not $+14$. Is the value of i being switched here?

There is also the possibility that $14 i +7 i +0 k $ is incorrect. Do let me know if that's the case.

Answer 1

If you are using a determinant to find the cross product use the following pattern $$\begin {bmatrix} +&-&+\\-&+&-\\+&-&+\end{bmatrix}$$

For example (3i+2j-3k)\time (2i+5j+k) = det $$\begin {bmatrix} i&j&k\\3&2&-3\\2&5&1\end{bmatrix} =$$

$$i(17)-j(9)+k(11)=17i-9j+11k$$

Answer 2

$$\vec{u} \times \vec{v} = \begin{vmatrix} i & j & k \\ \frac12 & -1 & \frac23 \\ 6 & -12 & -6\end{vmatrix}= 14i + 7j$$

Recall that when we switch rows, the determinant gets multiplied by $-1$, $$\vec{v} \times \vec{u} = -(14i + 7j)$$

$-2i-j$ is not the right answer.

Edit:

notice that $$\begin{vmatrix} -12 & -6 \\-1 & \frac23\end{vmatrix}=-8-(-1)(-6)=-14$$