Question

If we have two unit vectors as $e_1$ and $e_2$ and theta is the angle between them ,then prove that $\sin \dfrac{\theta }{2}$ is $\dfrac{1}{2}\left| {e_1 - e_2} \right|$ .

Answer 1

We have to prove that the relation for the angle between two unit vectors is as given in the question . We solve this question using the concept of vectors . We should also have the knowledge of the various formulas of the trigonometric functions . We should also have the knowledge of the dot product of two vectors and the concept of magnitude of vectors . First we will find the value of the magnitude of the difference of two unit vectors . Then using the properties of the dot product of two vectors and the value of magnitude of unit vectors we will simplify the expression and then using the formula of the double angle of cosine function , we will get to the required expression and hence prove the required statement.

Complete step-by-step solution:
Given :
$e_1$ and $e_2$ are two unit vectors and theta is the angle between the two vectors .
To prove :
$\sin \dfrac{\theta }{2} = \dfrac{1}{2}\left| {e_1 - e_2} \right|$
Proof :
We know that the formula for magnitude of difference of two terms is given as :
$\left| {a - b} \right| = \sqrt {{a^2} + {b^2} - 2\left| a \right|.\left| b \right|}$
Using the formula , we get the value of the magnitude of difference of the two unit vectors as :
$\left| {e_1 - e_2} \right| = \sqrt {{{\left( {e_1} \right)}^2} + {{\left( {e_2} \right)}^2} - 2\left| {e_1} \right|.\left| {e_2} \right|}$
Now , we also know that the magnitude of unit vector is given as :
$\left| {e_1} \right| = \left| {e_2} \right| = 1$
We also know that the formula for dot product of two vectors is give as :
$\left| a \right|.\left| b \right| = \left| a \right|\left| b \right|\cos \theta$
Using the formula and the value of magnitude of unit vectors , we can write the expression as :
$\left| {e_1 - e_2} \right| = \sqrt {{1^2} + {1^2} - 2 \times 1 \times 1\cos \theta }$
$\left| {e_1 - e_2} \right| = \sqrt {1 + 1 - 2\cos \theta }$
Further simplifying the expression , we get the expression as :
$\left| {e_1 - e_2} \right| = \sqrt {2 - 2\cos \theta }$
Taking 2 common we get the expression as :
$\left| {e_1 - e_2} \right| = \sqrt {2\left( {1 - \cos \theta } \right)}$
We also know that the formula for double angle of cosine function in terms of sine function is given as :
$\cos 2x = 1 - 2{\sin ^2}x$
Using the formula of double angle of cosine , we can write the expression as :
$\left| {e_1 - e_2} \right| = \sqrt {2\left( {1 - \left( {1 - 2{{\sin }^2}\dfrac{\theta }{2}} \right)} \right)}$
$\left| {e_1 - e_2} \right| = \sqrt {2\left( {1 - 1 + 2{{\sin }^2}\dfrac{\theta }{2}} \right)}$
On further simplifying , we get the expression as :
$\left| {e_1 - e_2} \right| = \sqrt {2 \times 2{{\sin }^2}\dfrac{\theta }{2}}$
On solving the square root we get the expression as :
$\left| {e_1 - e_2} \right| = 2\sin \dfrac{\theta }{2}$
$\sin \dfrac{\theta }{2} = \dfrac{1}{2}\left| {e_1 - e_2} \right|$
Hence proved that $\sin \dfrac{\theta }{2}$ is $\dfrac{1}{2}\left| {e_1 - e_2} \right|$ for the given condition of vectors .

Note: While writing the formula of the dot product of two vectors we can mistakenly write the formula with sin function instead of cosine . As we may get confused about which type of product . Sine function would be used in cross vectors to two terms and cos function in the dot product .
Formula for cross product is given as :
$\left| a \right| \times \left| b \right| = \left| a \right|\left| b \right|\sin \theta$