A vector of length $l$ is turned through the angle $\theta $ about its tail. What is the change in the position vector of its head?
A) $l \cos(\dfrac{\theta}{2})$
B) $2l \sin(\dfrac{\theta}{2})$
C) $2l \cos(\dfrac{\theta}{2})$
D) $l \sin(\dfrac{\theta}{2})$

Hint: When a vector is turned through the angle $\theta $ , it will not change the magnitude of the vector. The only thing we need to calculate is the direction of the vector. Since the position of the tail will be unchanged we can calculate the change in position of the head of the vector by subtracting the initial vector from the final vector.

Complete step by step solution:
Let us denote the initial vector by $\overrightarrow {{v_1}} $ and the vector after turning $\theta $ by $\overrightarrow {{v_2}}$. The length of the vector is given which is $l$ . Now give an expression for the change in the position vector
\[\therefore \Delta \overrightarrow v = \overrightarrow {{v_2}} - \overrightarrow {{v_{_1}}} \]……….equation (1)
Squaring on both sides in equation (1)
\[\therefore {(\Delta \overrightarrow v )^2} = {(\overrightarrow {{v_2}} - \overrightarrow {{v_{_1}}} )^2}\]
We know that on squaring a vector we get its magnitude
\[\therefore \Delta v = {(\overrightarrow {{v_2}} - \overrightarrow {{v_{_1}}} )^2}\]
We know that
${\left( {\overrightarrow {{v_2}} - \overrightarrow {{v_1}} } \right)^2} = \sqrt {{{\left| {\overrightarrow {{v_2}} } \right|}^2} + {{\left| {\overrightarrow {{v_1}} } \right|}^2} - 2\left| {{v_2}} \right|\left| {{v_1}} \right|\cos \theta } $
$\therefore \Delta v = \sqrt {{{\left| {\overrightarrow {{v_2}} } \right|}^2} + {{\left| {\overrightarrow {{v_1}} } \right|}^2} - 2\left| {{v_2}} \right|\left| {{v_1}} \right|\cos \theta } $
Since the length of the vector will not change, therefore $\left| {\overrightarrow {{v_1}} } \right| = \left| {\overrightarrow {{v_2}} } \right| = l$ . Now put this value in the above equation
$\therefore \Delta v = \sqrt {{l^2} + {l^2} - 2{l^2}\cos \theta } $
$ \Rightarrow \Delta v = \sqrt {2{l^2} - 2{l^2}\cos \theta } $
Take $2{l^2}$ common
$\therefore \Delta v = \sqrt {2{l^2}(1 - \cos \theta )} $
We can write $\cos \theta $ as \[1 - 2{\sin ^2}(\dfrac{\theta}{2})\]
$\therefore \Delta v = \sqrt {2{l^2}(1 - 1 + 2{\sin ^2}(\dfrac{\theta}{2}))}$
$ \Rightarrow \Delta v = \sqrt {2{l^2} \times 2 {\sin^2}(\dfrac{\theta}{2})}$
$ \Rightarrow \Delta v = \sqrt {4{l^2}{{\sin }^2}(\dfrac{\theta}{2})}$
Solve the square root on the right-hand side and you will get the desired result
$\therefore \Delta v = 2l \sin(\dfrac{\theta}{2})$

Therefore the correct option is B.

Note: Always remember that when we turn a vector through the angle, the magnitude of the vector will remain the same since we are not altering the length of the vector. The second thing to remember, when we calculate the change in the position vector, we subtract the initial vector from the final vector.