# Position vector...

• Apr 22nd 2009, 07:10 AM
Mush
Position vector...
My lecturer occasionally writes his position vectors as a matrix, as follows:

$\vec{r} = \left( \begin{array}{ccc}
0 & -r_z & r_y \\
r_z & 0 & -r_x \\
-r_y & r_x & 0 \end{array} \right)$

Where does this come from? Why does he use it? Is it equivalent to $\vec{r} = \left( \begin{array}{ccc}
r_x \\
r_y \\
r_z \end{array} \right)$

For example, he uses it in deriving Newton's 2nd Law for Rotational Motion, as I have attached as a JPEG.
• Apr 22nd 2009, 07:38 PM
The Second Solution
Quote:

Originally Posted by Mush
My lecturer occasionally writes his position vectors as a matrix, as follows:

$\vec{r} = \left( \begin{array}{ccc}
0 & -r_z & r_y \\
r_z & 0 & -r_x \\
-r_y & r_x & 0 \end{array} \right)$

Where does this come from? Why does he use it? Is it equivalent to $\vec{r} = \left( \begin{array}{ccc}
r_x \\
r_y \\
r_z \end{array} \right)$

For example, he uses it in deriving Newton's 2nd Law for Rotational Motion, as I have attached as a JPEG.

Your lecturer is giving a matrix representation of the vector resulting from a cross product.
• Apr 22nd 2009, 11:59 PM
Mush
Quote:

Originally Posted by The Second Solution
Your lecturer is giving a matrix representation of the vector resulting from a cross product.

How does one derive this?