Find the velocity of image of point object P $\left(\text{in cm/s}\right)$ formed by the plane mirror :

The correct option is B $2\hat{i}-10\hat{j}$
Given,

Velocity of object $\left(\text{in cm/s}\right)$,
${\overrightarrow{v}}_o=10\sqrt{2}\sin 45{}^{\circ }\hat{i}-10\sqrt{2}\cos 45{}^{\circ }\hat{j}=10\hat{i}-10\hat{j}$

Velocity of mirror $\left(\text{in cm/s}\right)$
${\overrightarrow{v}}_m=10\sin 37{}^{\circ }\hat{i}+10\cos 37{}^{\circ }\hat{j}=6\hat{i}+8\hat{j}$

In the direction normal to the mirror,

|Relative velocity of image with respect to mirror| = |Relative velocity of object with respect to mirror|

$\text{(i.e)}{\overrightarrow{v}}_{ix}-{\overrightarrow{v}}_{mx}=-({\overrightarrow{v}}_{ox}-{\overrightarrow{v}}_{mx})$

$\Rightarrow {\overrightarrow{v}}_{ix}=2{\overrightarrow{v}}_{mx}-{\overrightarrow{v}}_{ox}$

From the data given in the question,

${\overrightarrow{v}}_{mx}=6\hat{i}\text{cm/s},{\overrightarrow{v}}_{ox}=10\hat{i}\text{cm/s}$

$\Rightarrow {\overrightarrow{v}}_{ix}=2\times 6\hat{i}-10\hat{i}=2\hat{i}\text{cm/s}$

In the direction parallel to the surface of the mirror,

Along $Y-$ direction,
${\overrightarrow{v}}_{oy}={\overrightarrow{v}}_{iy}$

$\therefore {\overrightarrow{v}}_{iy}=-10\hat{j}$$\text{cm/s}$

Therefore, velocity of image $\overrightarrow{v}={\overrightarrow{v}}_{ix}+{\overrightarrow{v}}_{iy}$

$\Rightarrow \overrightarrow{v}=(2\hat{i}-10\hat{j})\text{cm/s}$