Answer
$14i+4j+2k$
$a$ and $b$ are orthogonal
Work Step by Step
$a = <0, 2, -4>$, $b = <-1, 3, 1>$
$a \times b$ is equal to the determinant of the matrix:
\begin{vmatrix}
i & j & k \\
0 & 2 & -4 \\
-1 & 3 & 1 \\
\end{vmatrix}
This can be found with the formula $i(bz-yc)-j(az-xc)+k(ay-xb)$
given the general $3\times 3$ matrix:
\begin{vmatrix}
i & j & k \\
a & b & c \\
x & y & z \\
\end{vmatrix}
Plugging the values into this formula gives:
$i((2*1)-(3*(-4)))-j((0*1)-((-1)*(-4)))+k((0*3)-(-1*2))$
Which further simplifies into the expression:
$i(2+12)-j(-4)+k(2)$
Which can be rewritten as the vector $14i+4j+2k$
The cross product of two vectors, by definition, results in a vector that is orthogonal to both of the input vectors. We can check that this is true using the fact that the dot product of $a$ and $b$ with $a \times b$ should both equal zero.
$a \cdot (a \times b) = <0, 2, -4> \cdot <14, 4, 2> = 0*14 + 2*4 +-4*2$ = 0
$b \cdot (a \times b) = <-1, 3, 1> \cdot <14, 4, 2> = -1*14 + 3*4 +1*2$ = 0
Because both of these equations equal zero, it confirms that $a \times b$ is orthogonal to both $a$ and $b$.
