Originally Posted by
Chris L T521 I would assume that those angles are with respect to the positive coordinate axes?
We would then need to use these equations:
$\displaystyle \cos\alpha=\frac{r_x}{||\bold r||}~~~~~~~~~~\cos\beta=\frac{r_y}{||\bold r||}~~~~~~~~~~\cos\gamma=\frac{r_z}{||\bold r||}$
where $\displaystyle r_x, \ r_y, \text{ and }r_z$ are the components of the vector $\displaystyle \bold r$. These equations are used to find direction cosines. They come in useful here.
In our case, $\displaystyle \bold r=-\bold j$
We want $\displaystyle \bold r$ to be rotated and have the new direction angles $\displaystyle \alpha=45^{\circ},~\beta=10^{\circ},~\text{and}~\g amma=90^{\circ}$. So let's see how we would calculate the equation of the new vector.
Let's find the x component first.
Since $\displaystyle \cos\alpha=\frac{r_x}{||\bold r||}$, and $\displaystyle \alpha=45^{\circ}~\text{and}~||\bold r||=1$, we see that
$\displaystyle r_x=\cos(45^{\circ})=\frac{\sqrt{2}}{2}\approx\col or{red}\boxed{.707}$
Let's now find the y component,
Since $\displaystyle \cos\beta=\frac{r_y}{||\bold r||}$, and $\displaystyle \beta=10^{\circ}~\text{and}~||\bold r||=1$, we see that
$\displaystyle r_y=\cos(10^{\circ})\approx\color{red}\boxed{.985}$
Let's now find the z component,
Since $\displaystyle \cos\gamma=\frac{r_z}{||\bold r||}$, and $\displaystyle \gamma=90^{\circ}~\text{and}~||\bold r||=1$, we see that
$\displaystyle r_z=\cos(90^{\circ})=\color{red}\boxed{0}$
So we now see that the "rotated" vector has the equation $\displaystyle \color{red}\boxed{\bold r_n=.707\bold i+.985\bold j}$
Does this make sense?
--Chris