Given the vectors a = (-2i + 6j + 2k) and b = (1i + 10k), the cross product, $$atimesb$$, is most nearly:

The cross product of two vectors will provide us a 3rd vector that is perpendicular to both our two original vectors and equal in magnitude to the area both original vectors span. The easiest and most compact way to find the cross product of two (i, j, k) vectors is to set up a matrix as follows:

$\mathit{a} x \mathit{b} = \left.open\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|$

By plugging our values from the problem into this matrix form, we get:

$\mathit{a} x \mathit{b} = \left.open\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -2& 6& 2\\ 1& 0& 10\end{array}\right|$

We now solve this matrix as we would to find the determinant of a 3x3 matrix, according to the following formulas:

$$\begin{gathered} A = \left.open\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right] \\ \left.openA\right| = a·\left.open\begin{array}{cc}e& f\\ h& i\end{array}\right|-b·\left.open\begin{array}{cc}d& f\\ g& i\end{array}\right|+c·\left.open\begin{array}{cc}d& e\\ g& h\end{array}\right| \end{gathered}$$

To find the determinant of the 2x2 matrices formed:

$A= \left.open\begin{array}{cc}a& b\\ c& d\end{array}\right], \left.openA\right|=ad-bc$

Using our original matrix and formulas:

$$\begin{gathered} \mathit{a} x \mathit{b} = \left.open\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -2& 6& 2\\ 1& 0& 10\end{array}\right| \\ \mathit{a} x \mathit{b} = \mathit{i}·\left.open\begin{array}{cc}6& 2\\ 0& 10\end{array}\right|-\mathit{j}·\left.open\begin{array}{cc}-2& 2\\ 1& 10\end{array}\right|+\mathit{k}·\left.open\begin{array}{cc}-2& 6\\ 1& 0\end{array}\right| \\ \mathit{a} x\mathbf{ }\mathit{b} = \mathit{i}\left(\right(6*10)-(2*0\left)\right)-\mathit{j}\left(\right(-2*10)-(2*1\left)\right)+\mathit{k}\left(\right(-2*0)-(6*1\left)\right) \\ \mathit{a} x \mathit{b} = \mathit{i}(60-0)-\mathit{j}(-20-2)+\mathit{k}(0-6) \\ \mathit{a} x \mathit{b} = 60\mathit{i}+22\mathit{j}-6\mathit{k} \end{gathered}$$

The cross product a x b is therefore $\mathbf{60}\mathit{i}\mathbf{+}\mathbf{22}\mathit{j}\mathbf{-}\mathbf{6}\mathit{k}$.