Using the determinant method, the cross product is
[tex]\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\3&8&-6\\-4&-2&-3\end{vmatrix}=\begin{vmatrix}8&-6\\-2&-3end{vmatrix}\,\vec\imath-\begin{vmatrix}3&-6\\-4&-3\end{vmatrix}\,\vec\jmath+\begin{vmatrix}3&8\\-4&-2\end{vmatrix}\,\vec k=-36\,\vec\imath+33\,\vec\jmath+26\,\vec k[/tex]
so the answer is B.
Or you can apply the properties of the cross product. By distributivity, we have
(3i + 8j - 6k) x (-4i - 2j - 3k)
= -12(i x i) - 32(j x i) + 24(k x i) - 6(i x j) - 16(j x j) + 12(k x j) - 9(i x k) - 24(j x k) + 18(k x k)
Now recall that
- (i x i) = (j x j) = (k x k) = 0 (the zero vector)
- (i x j) = k
- (j x k) = i
- (k x i) = j
- (a x b) = -(b x a) for any two vectors a and b
Putting these rules together, we get
(3i + 8j - 6k) x (-4i - 2j - 3k)
= -32(-k) + 24j - 6k + 12(-i) - 9(-j) - 24i
= (-12 - 24)i + (24 + 9)j + (32 - 6)k
= -36i + 33j + 26k
