Question

How do we find the sine of the angle between two vectors?

Answer 1

Hint: The angle between two vectors will be deferred by a single point, which is called the shortest angle at which we have to turn around one of the vectors to the position of co-directional with another vector. We will find the angle of a vector with respect to the positive x-axis. A vector is to be in the standard position of its initial point is the origin $\left( {0,0} \right)$. Let us assume two vectors are $\overrightarrow u $ and $\overrightarrow v $. The angle between the x-axis and $\overrightarrow u$ is ‘A’, and the angle between the x-axis and $\overrightarrow v$ is ‘B’. Write these two vectors in polar form.
Here, we will use the trigonometry formula as below.
$\sin \left( {B - A} \right) = \sin B\cos A - \sin A\cos B$

Complete step-by-step answer:
In this question, we want to find the sine of the angle between two vectors.
Therefore, let us assume two vectors are $\overrightarrow u $ and $\overrightarrow v $. These are the two-dimensional vectors. And two angles are A and B.
Let us write these vectors in the polar form.
$\overrightarrow u = \left| u \right|\left( {\left( {\cos A} \right)\widehat i + \left( {\sin A} \right)\widehat j} \right)$
$\overrightarrow v = \left| v \right|\left( {\left( {\cos B} \right)\widehat i + \left( {\sin B} \right)\widehat j} \right)$
Here, A is the angle between the x-axis and $\overrightarrow u $ and B is the angle between the x-axis and$\overrightarrow v $.
Now, apply the cross-product of two vectors.
$ \Rightarrow \;\overrightarrow u \times \overrightarrow v = \left( {\left| u \right|\left( {\left( {\cos A} \right)\widehat i + \left( {\sin A} \right)\widehat j} \right)} \right) \times \left( {\left| v \right|\left( {\left( {\cos B} \right)\widehat i + \left( {\sin B} \right)\widehat j} \right)} \right)$
 That is equal to,
$ \Rightarrow \;\overrightarrow u \times \overrightarrow v = \left| u \right|\left( {\cos A} \right)\left| v \right|\left( {\sin B} \right) - \left| u \right|\left( {\sin A} \right)\left| v \right|\left( {\cos B} \right)$
$ \Rightarrow \;\overrightarrow u \times \overrightarrow v = \left| u \right|\left| v \right|\left( {\cos A\sin B - \sin A\cos B} \right)$
Let us apply the trigonometric formula$\sin \left( {B - A} \right) = \sin B\cos A - \sin A\cos B$
$ \Rightarrow \;\overrightarrow u \times \overrightarrow v = \left| u \right|\left| v \right|\sin \left( {B - A} \right)$
Let us divide by $\left| u \right|\left| v \right|$ into both sides.
$ \Rightarrow \;\dfrac{{\overrightarrow u \times \overrightarrow v }}{{\left| u \right|\left| v \right|}} = \dfrac{{\left| u \right|\left| v \right|\sin \left( {B - A} \right)}}{{\left| u \right|\left| v \right|}}$
Therefore, we can write:
$ \Rightarrow \;\sin \left( {B - A} \right) = \dfrac{{\overrightarrow u \times \overrightarrow v }}{{\left| u \right|\left| v \right|}}$

Hence, this is the sine of the angle between the two vectors.

Note:
Polar Vector: In elementary math, the term polar vector is used to refer to a representation of a vector as a vector magnitude and angle, which is equivalent to specifying its endpoints in polar coordinates. Here, the magnitude of the vector means length.