# How to find the new coordinates for a vector when there is a shift in a new system of coordinates?

#### Chemist116

The problem is as follows:

The figure from below shows two systems of cartesian coordinates. On ($x,y$) vector $\vec{A}$ is expressed as $\vec{A}=5\hat{i}+4\hat{j}$. Find the vector $\sqrt{2}\vec{A}$ on the system $x',y'$.

The alternatives given are as follows:

$\begin{array}{ll} 1.&9\hat{i'}+\hat{j'}\\ 2.&\hat{i'}+9\hat{j'}\\ 3.&-9\hat{i'}+\hat{j'}\\ 4.&9\hat{i'}-\hat{j'}\\ 5.&\hat{i'}-9\hat{j'}\\ \end{array}$

Does it exist a way to solve this problem visually or with least use of algebra? The only thing which I could spot for the vector given is that the angle for the vector is:

$\tan\omega=\frac{4}{5}$

therefore $\omega=\tan^{-1}\left(\frac{4}{5}\right)$

This angle is not very known. How exactly can I relate it with the tilt in the new system of coordinates? What exactly is what should I do?

#### skeeter

Math Team
$|\vec{A}| = \sqrt{41} \implies |\sqrt{2} \cdot \vec{A}| = \sqrt{82}$

Note $\vec{A}$ lies below the $x'$ axis, so the angle between $\vec{A}$ and $x'$ = $\theta' = -\left[45^\circ - \arctan\left(\dfrac{4}{5}\right)\right]$

$\tan(\theta') = \dfrac{\frac{4}{5}-1}{1+\frac{4}{5}} = \dfrac{-1}{9}$

$\dfrac{A_{y'}}{A_{x'}} = \dfrac{-1 \cdot k}{9 \cdot k} \implies \sqrt{(-k)^2 + (9k)^2} = \sqrt{82} \implies k=1 \implies A' = 9i' - j'$

topsquark

#### skipjack

Forum Staff
As $(x,\, 0)$ and $(0,\, y)$ become $(x\cos(45^\circ),\, -x\sin(45^\circ))$ and $(y\sin(45^\circ),\, y\cos(45^\circ))$ respectively,
$(x,\, y)$ becomes $(x\cos(45^\circ) + y\sin(45^\circ),\, -x\sin(45^\circ) + y\cos(45^\circ))$.

Hence $\sqrt2(5, \,4)$ becomes $(9,\,-1)$.

topsquark