# If a and b are unit vectors and theta is the angle between them, express abs(a-b) in terms of theta ?

Aug 24, 2017

$| \vec{a} - \vec{b} | = 2 \sin \left(\frac{\theta}{2}\right) .$

#### Explanation:

Suppose that, $\vec{a} , \mathmr{and} \vec{b}$ are such unit vectors, that,

$\hat{\left(\left(\vec{a} , \vec{b}\right)\right)} = \theta , \theta \in \left[0 , \pi\right] .$

because veca, &, vecb" are unit vectors, ":. |veca|=|vecb|=1....(0).

We know, $\left(1\right) : \vec{a} \cdot \vec{b} = | \vec{a} | \cdot | \vec{b} | \cdot \cos \theta ,$ and,

$\left(2\right) : | \vec{x} {|}^{2} = \vec{x} \cdot \vec{x} .$

$\therefore , | \vec{a} - \vec{b} {|}^{2} = \left(\vec{a} - \vec{b}\right) \cdot \left(\vec{a} - \vec{b}\right) \ldots \ldots \ldots \ldots \left[\because , \left(2\right)\right] ,$

$= \vec{a} \cdot \left(\vec{a} - \vec{b}\right) - \vec{b} \cdot \left(\vec{a} - \vec{b}\right) ,$

$= \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} ,$

$= | \vec{a} {|}^{2} - 2 \vec{a} \cdot \vec{b} + | \vec{b} {|}^{2.} . . \left[\because , \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\right] ,$

$= 1 - 2 | \vec{a} | \cdot | \vec{b} | \cdot \cos \theta + 1. \ldots . . \left[\because , \left(0\right) , \mathmr{and} , \left(1\right)\right] ,$

$= 2 - 2 \cdot 1 \cdot 1 \cos \theta \ldots \ldots \ldots . . \left[\because , \left(0\right)\right] ,$

$= 2 \left(1 - \cos \theta\right) = 2 \left(2 {\sin}^{2} \left(\frac{\theta}{2}\right)\right) .$

$\Rightarrow | \vec{a} - \vec{b} | = 2 \sin \left(\frac{\theta}{2}\right) .$

Enjoy Maths.!