Question

How do you find the \[sine\] of the angle between two vectors?

Answer 1

Hint: In this question, we have been given a condition of two vectors and we need to find the\[sine\] of the angle between them.
According to the standard formula, the angle $\theta $ between two vectors \[\overrightarrow u \] and \[\overrightarrow v \;\] is given by the formula
\[\theta = {\cos ^{ - 1}}(\dfrac{{\overrightarrow u \; \cdot \overrightarrow v \;}}{{|\overrightarrow u \;||\overrightarrow v \;\;|}})\]
We can also translate our two vectors so that their tails are at the origin. Then draw a line through each of the vectors. The smaller of the two angles is called the "angle between the two vectors".
Or we can use a slightly harder way (because it involves calculating a cross product). You know that
\[|\overrightarrow u \; \times \overrightarrow v \;\left| = \right|\overrightarrow u \;||\overrightarrow v \;|sin(\theta )\]
So, we can find the cross product and all of the norms and then plug in as shown in the equation shown above:

Complete step-by-step solution:
In general, for \[a,b \in {R^3}\] , we have the standard sine angle formula to calculate angle between two vectors:
\[\parallel a \times b\parallel = \parallel a\parallel \parallel b\parallel sin\theta ,\]
Where, $\theta $ is the angle between vector a and vector b.
In case we will assume mean real valued two dimensional vectors. Given vectors, \[\overrightarrow u \] and \[\overrightarrow v \] , note that they can be represented in polar form as:
\[\overrightarrow u = \left| u \right|\left( {\left( {cos\alpha } \right)\hat i + \left( {sin\alpha } \right)\hat j} \right)\]
\[\overrightarrow v = \left| v \right|\left( {\left( {cos\beta } \right)\hat i + \left( {sin\beta } \right)\hat j} \right)\]
Where α and β are the angles that \[\overrightarrow u \] and \[\overrightarrow v \] make with the x axis.
Then:
\[\overrightarrow u \times \overrightarrow v = \left| u \right|\left( {cos\alpha } \right)\left| v \right|\left( {sin\beta } \right) - \left| u \right|\left( {sin\alpha } \right)\left| v \right|\left( {cos\beta } \right)\]
\[\overrightarrow u \times \overrightarrow v = \left| u \right|\left| v \right|\left( {cos\alpha sin\beta - sin\alpha cos\beta } \right)\]
\[\overrightarrow u \times \overrightarrow v = \left| u \right|\left| v \right|sin\left( {\beta - \alpha } \right)\]
So, the final equation in angle will come out to be as shown below, which is the \[sine\] of the angle between the two vectors.:
\[sin\left( {\beta - \alpha } \right) = \dfrac{{\overrightarrow u \times \overrightarrow v }}{{\left| u \right|\left| v \right|}}\]

Note: For 3 dimensional vectors \[\overrightarrow u \] and \[\overrightarrow v \] , the cross product is a vector quantity rather than a scalar one, but the absolute value of the sine of the angle between \[\overrightarrow u \] and \[\overrightarrow v \] is expressible in terms of the length of that vector quantity as:
\[\left\| {\overrightarrow u \times \overrightarrow v } \right\| = \left| u \right|.\left| v \right|\]