Question

Prove by the vector method ,the law of sine in trigonometry:
$\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}$

Answer 1

Hint: For solving this question we will assume that \[AB = \overrightarrow c ,BC = \overrightarrow a ,AC = \overrightarrow b \] and use the following known information:
 For a triangle ABC ,
\[\overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CA} = 0\],
Then just solve the question by using the cross product/ vector product of vectors method to get the desired answer.
The magnitude of cross product of two vectors is equal to the product of magnitude of both the vectors and the sine of angle between them.
\[\left| {\overrightarrow x \times \overrightarrow y } \right| = \left| {\overrightarrow x } \right|.\left| {\overrightarrow y } \right|.\sin \theta \]
Where \[\theta \] is the angle between vectors x and y

Complete step-by-step answer:
Let us assume \[AB = \overrightarrow c ,BC = \overrightarrow a ,AC = \overrightarrow b \]
Now as we know that for a triangle
\[\overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CA} = 0\]
Since we know that:
\[\overrightarrow {CA} = - \overrightarrow {AC} \]
Hence putting the value we get:-
\[\overrightarrow {CA} = - \overrightarrow b \]
Hence putting the respective values we get:-
\[\vec a - \vec b + \vec c = 0\]
Using cross product of vectors method, multiply by $\vec a$ both side
\[\vec a \times (\vec a - \vec b + \vec c) = a \times \vec 0\]
Now as we know that:
\[\overrightarrow x \times \left( {\overrightarrow y + \overrightarrow z } \right) = \overrightarrow x \times \overrightarrow y + \overrightarrow x \times \overrightarrow z \]
Applying this formula we get:-
\[(\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{a} \times \vec a) - (\vec a \times \vec b) + (\vec a \times \vec c) = \vec 0\]
We know that the cross product of same vectors is zero
Therefore,
\[(\vec a \times \vec a) = 0\]
Putting this value we get:-
\[
  0 - (\vec a \times \vec b) + (\vec a \times \vec c) = \vec 0 \\
  (\vec a \times \vec b) = (\vec a \times \vec c) \\
 \]
Now as we know that the magnitude of cross product of two vectors is equal to the product of magnitude of both the vectors and the sine of angle between them.
\[\left| {\overrightarrow x \times \overrightarrow y } \right| = \left| {\overrightarrow x } \right|.\left| {\overrightarrow y } \right|.\sin \theta \]
Where \[\theta \] is the angle between vectors x and y.
Hence applying this formula for both LHS and RHS we get:-
Since it is given that the angle between vectors a and b is \[\pi - C\] and angle between the vectors a and c is \[\pi - B\]
Therefore substituting the values in above formula we get:-
\[
  \left| a \right|\left| b \right|\sin (\pi - C) = \left| c \right|\left| a \right|\sin (\pi - B) \\
   \Rightarrow ab\sin (\pi - C) = ca\sin (\pi - B) \\
 \]
Now since we know that
\[\sin (\pi - \theta ) = \sin \theta \]
Hence we get:-
$b\sin C = c\sin B$
\[ \Rightarrow \dfrac{{\sin C}}{c} = \dfrac{{\sin B}}{b}\]……………….(1)
Similarly, we can prove that
$\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b}$………………..(2)
Hence from equations 1 and 2 we get:-
$\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}$
Hence proved
$\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}$

Note: In this type of question students should first analyse the diagram instead of solving the equations. All the vectors should be taken according to their directions.
Also while calculating the magnitude of the vector product, students should take the correct angle between the two vectors.