The position vector of point P (20, 10) is rotated anti-clockwise in X-Y plane by an angle 6 =30° such that point P occupies position Q, as shown in the figure. The coordinates (x, y) of Q are 

The correct answer is (12.32, 18.66).

Step-by-step Explanation:

1. Identify the given information:
From the problem description and the provided diagram (which displays the coordinate axes X and Y, origin O, initial position vector OP, final position vector OQ, and the rotation angle θ), we have:
- The coordinates of point P are: $(x,y)=(20,10)$
- The angle of anti-clockwise rotation is: $\theta ={30}^{\circ }$

2. Rotation Transformation Formula:
When a position vector OP with coordinates $(x,y)$ is rotated anti-clockwise about the origin by an angle $\theta$ in the X-Y plane to occupy a new position Q with coordinates $(x^{\prime },y^{\prime })$, the new coordinates are determined using the standard rotation matrix equations:

$x^{\prime }=xcos\theta -ysin\theta$

$y^{\prime }=xsin\theta +ycos\theta$

3. Calculate the new X-coordinate ($x^{\prime }$):
Substitute the values of $x$, $y$, and $\theta$ into the equation for $x^{\prime }$:

$x^{\prime }=20cos\left({30}^{\circ }\right)-10sin\left({30}^{\circ }\right)$

$x^{\prime }=20\left(0.866\right)-10\left(0.5\right)$

$x^{\prime }=17.32-5=12.32$

4. Calculate the new Y-coordinate ($y^{\prime }$):
Substitute the values into the equation for $y^{\prime }$:

$y^{\prime }=20sin\left({30}^{\circ }\right)+10cos\left({30}^{\circ }\right)$

$y^{\prime }=20\left(0.5\right)+10\left(0.866\right)$

$y^{\prime }=10+8.66=18.66$

Thus, the final coordinates $(x,y)$ of point Q are (12.32, 18.66).