A vector is represented by 3$\overset{\wedge }{\mathop{i}}\,$+$\overset{\wedge }{\mathop{j}}\,$ +2$\overset{\wedge }{\mathop{k}}\,$.Its length in X-Y plane is
A. 2
B. $\sqrt{14}$
C. $\sqrt{10}$
D. $\sqrt{5}$

Hint: The coefficients of $\overset{\wedge }{\mathop{i}}\,$ and $\overset{\wedge }{\mathop{j}}\,$ give the value of x coordinate and y coordinate in the plane .To find its length in X-Y plane, use the distance formula taking the coefficients of $\overset{\wedge }{\mathop{i}}\,$ and $\overset{\wedge }{\mathop{j}}\,$ as one pair of coordinates and (0,0) as other pair of coordinates.

Formula Used: Here, the formula used will be-
Distance formula=$\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}$

Complete step by step solution: It is given that vector is 3$\overset{\wedge }{\mathop{i}}\,$+$\overset{\wedge }{\mathop{j}}\,$ +2$\overset{\wedge }{\mathop{k}}\,$. We can say that coefficient of $\overset{\wedge }{\mathop{i}}\,$is 3 and coefficient of $\overset{\wedge }{\mathop{j}}\,$is 1. Hence, coordinates are (3,1).Now, to calculate distance in the X-Y plane, the coordinates of origin is (0,0).
Now, use distance formula=$\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}$
                                                 =$\sqrt{{{(3-0)}^{2}}+{{(1-0)}^{2}}}$
                                                  =$\sqrt{9+1}$
                                                  = $\sqrt{10}$ units
Hence, it is concluded that in the X-Y plane, the length of the vector is $\sqrt{10}$ units.

So, Option ‘C’ is correct.

Note: It is to be understood that if we are asked to find the length in X-Y plane, it will be coordinated in X-Y plane and there is nothing to do with Z coordinates. Conclusively, the length of the vector in the X-Y plane will be independent of the Z coordinate.