0
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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:- $$\left| \begin{array}{cc} \ \frac{\sqrt{3}}{2}a & 0 \\ 0 & c \end{array} \right|\hat{x}-\left| \begin{array}{cc} \ \frac{a}{2} & 0 \\ 0 & c \end{array} \right|\hat{y}+\left| \begin{array}{cc} \ \frac{a}{2} & 0 \\ \frac{\sqrt{3}}{2}a & 0 \end{array} \right|\hat{z}=ac\frac{\sqrt{3}}{2}\hat{x}-\frac{ac}{2}\hat{y}+0\hat{z}=ac\frac{\sqrt{3}}{2}\hat{x}-\frac{ac}{2}\hat{y}$$

$\endgroup$
0
$\begingroup$

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.

With this table you can easily compute your cross product.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.