Answer
$15i-10j-3k$
Yes, $a \times b$ is orthogonal to both $a$ and $b$.
Work Step by Step
In this question:
$a\times b= \begin{vmatrix} i&j&k \\ 2& 3&0\\1&0&5\end{vmatrix}$
Expand along the first row:
$a\times b=\ i\begin {vmatrix} 3&0 \\ 0&5\end{vmatrix}-\ j \begin {vmatrix} 2&0 \\ 1&5\end{vmatrix}+\ k\begin {vmatrix} 2&3 \\ 1&0\end{vmatrix}$
$a \times b=15i-10j-3k$
To verify that it is orthogonal to $a$, we will compute:
$(a\times b).a=(15,-10,-3) \cdot (2,3,0)=0$
To verify that it is orthogonal to $b$, we will compute:
$(a\times b).b=(15,-10,-3) \cdot (1,0,5)=0$
Yes, $a \times b$ is orthogonal to both $a$ and $b$.
