# Cross Product of two vectors

I have these two vectors $$\vec{a}_1=\frac{a}{2}\hat{x}+\frac{\sqrt{3}}{2}a\hat{y}$$ $$\vec{a}_2=c\hat{z}$$

I know $\vec{a}_1\times\vec{a}_2$ is equal to: $$ac\frac{\sqrt{3}}{2}\hat{x}-ac\frac{1}{2}\hat{y}$$

I would like to know the procedure followed to solve this vector operation. I am familiar with the definition of cross product but when vector $\vec{a}_1$ has components in $\hat{x}$ and $\hat{y}$, I get confused.Thanks

To take a cross product, arrange $\vec{a}_1=\frac{a}{2}\hat{x}+\frac{\sqrt{3}}{2}a\hat{y}$ and $\vec{a}_2=c\hat{z}$ in two rows and three columns:- $$\left( \begin{array}{ccc} \frac{a}{2} & \frac{\sqrt{3}}{2}a & 0 \\ 0 & 0 & c \end{array} \right)$$ Then you take the determinants of the square matrices (formed by removing the 1st, 2nd and 3rd columns respectively, of the above $3\times2$ matrix) as follows - note the minus sign before the $\hat{y}$ component:- $$\left| \begin{array}{cc} \ \frac{\sqrt{3}}{2}a & 0 \\ 0 & c \end{array} \right|\hat{x}-\left| \begin{array}{cc} \ \frac{a}{2} & 0 \\ 0 & c \end{array} \right|\hat{y}+\left| \begin{array}{cc} \ \frac{a}{2} & 0 \\ \frac{\sqrt{3}}{2}a & 0 \end{array} \right|\hat{z}=ac\frac{\sqrt{3}}{2}\hat{x}-\frac{ac}{2}\hat{y}+0\hat{z}=ac\frac{\sqrt{3}}{2}\hat{x}-\frac{ac}{2}\hat{y}$$
Use the multiplication table $$\hat{x}\times\hat{y}=\hat{z},\quad \hat{y}\times\hat{z}=\hat{x},\quad \hat{z}\times\hat{x}=\hat{y}.$$ Remember that the cross product is anti-commutative, so that $\hat{y}\times\hat{x}=-\hat{x}\times\hat{y}=-\hat{z},$ and so on.