$\stackrel{\mathrm{\_}}{\mathrm{A}}=10\hat{\mathrm{i}}-10\sqrt{3}\hat{\mathrm{j}}$.  Now $\stackrel{\rightharpoonup }{A}$ is rotated about its tail in anticlockwise direction in xy plane by 105°.  Then the new vector $\left(\overline{\mathrm{B}}\right)$ formed is ____ ?

$\begin{array}{l}\text{ (a) }\overline{\mathrm{A}}=10\widehat{\mathrm{i}}-10\sqrt{3}\widehat{\mathrm{j}}\\ \mathrm{A}=\sqrt{{10}^2+(-10\sqrt{3}{)}^2}=20\end{array}$

$\begin{array}{l}\tan \mathrm{\alpha }=\frac{10\sqrt{3}}{10}=\sqrt{3}\\ \mathrm{\alpha }={60}^{\circ }\end{array}$
$\text{(b})\text{\hspace{0.17em}A\hspace{0.17em}rotated\hspace{0.17em}by\hspace{0.17em}}{105}^0\text{\hspace{0.17em}in\hspace{0.17em}A.C.W\hspace{0.17em}direction\hspace{0.17em}to\hspace{0.17em}obtain\hspace{0.17em}}$

$\overline{\mathrm{B}}{\text{=20\hspace{0.17em}cos\hspace{0.17em}45}}^0+20\sin {45}^0$
$\text{ =10 }\sqrt{2}\hat{\mathrm{i}}\text{+10}\sqrt{2}\hat{\mathrm{j}}$