Answer
$\frac{-3}{2}i+\frac{7}{4}j+\frac{2}{3}k$
Yes, $a \times b$ is orthogonal to both $a$ and $b$
Work Step by Step
in this question:
$a\times b= \begin{vmatrix} i&j&k \\ \frac{1}{2} &\frac{1}{3}&\frac{1}{4}\\1&2&-3\end{vmatrix}$
expand along the first row:
$a\times b=\ i\begin {vmatrix} \frac{1}{3}&\frac{1}{4} \\ 2&-3\end{vmatrix}-\ j \begin {vmatrix} \frac{1}{2}&\frac{1}{4} \\ 1&-3\end{vmatrix}+\ k\begin {vmatrix} \frac{1}{2}&\frac{1}{3} \\ 1&2\end{vmatrix}$
$a \times b=\frac{-3}{2}i+\frac{7}{4}j+\frac{2}{3}k$
To verify that it is orthogonal to $a$, we will compute:
$(a\times b).a=(\frac{-3}{2},\frac{7}{4},\frac{2}{3}) \cdot (\frac{1}{2},\frac{1}{3}.\frac{1}{4})=0$
To verify that it is orthogonal to $b$, we will compute:
$(a\times b).b=(\frac{-3}{2},\frac{7}{4},\frac{2}{3}) \cdot (1,2,-3)=0$
Yes, $a \times b$ is orthogonal to both $a$ and $b$
