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?.
You can solve this very easily using some basic vectors. Observe- $$x'=\frac {\hat i+\hat j}{\sqrt 2}\;,\;\;\; y'=\frac {\hat i-\hat j}{\sqrt 2}$$ Solving for $\hat i$ and $\hat j$ gives us the following- $$\hat i=\frac {x'+y'}{\sqrt 2}\;,\;\;\; \hat j=\frac {x'-y'}{\sqrt 2}$$ Substituting this in $\vec A=5\hat i+4\hat j$ gives us the following - $$\vec A=\frac 9{\sqrt2} x'+\frac 1{\sqrt2} y'$$ $$\sqrt 2\vec A=9x'+y'$$