Answer
$-\frac{3}{2}i+\frac{7}{4}j+\frac{2}{3}k$
$a$ and $b$ are orthogonal.
Work Step by Step
$a = <\frac{1}{2}, \frac{1}{3}, \frac{1}{4}>$, $b = <1, 2, -3>$
$a \times b$ is equal to the determinant of the matrix:
\begin{vmatrix}
i & j & k \\
\frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\
1 & 2 & -3 \\
\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:
$i((\frac{1}{3}*(-3))-(2*\frac{1}{4}))-j((\frac{1}{2}*(-3)
)-(1*\frac{1}{4}))+k((\frac{1}{2}*\frac{1}{3})-(1*\frac{1}{3}))$
Which further simplifies into the expression:
$i(-1-\frac{1}{2})-j(-\frac{3}{2}-\frac{1}{4})+k(\frac{1}{6}-\frac{1}{3})$
Which can be rewritten as the vector $-\frac{3}{2}i+\frac{7}{4}j+\frac{2}{3}k$
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.
$a \cdot (a \times b) = <\frac{1}{2}, \frac{1}{3}, \frac{1}{4}> \cdot <-\frac{3}{2}, \frac{7}{4}, \frac{2}{3}> = \frac{1}{2}*\left(-\frac{3}{2}\right) + \frac{1}{3}*\frac{7}{4} +\frac{1}{4}*\frac{2}{3} = 0$
$b \cdot (a \times b) = <1, 2, -3> \cdot <-\frac{3}{2}, \frac{7}{4}, \frac{2}{3}> = 1*\left(-\frac{3}{2}\right) + 2*\frac{7}{4} +(-3)*\frac{2}{3}
= 0$
Because both of these equations equal zero, it confirms that $a \times b$ is orthogonal to both $a$ and $b$.
