Applications of Dot (Scalar) Product
Finding Angle between Two Vectors
If \(\overrightarrow{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}\) and \(\overrightarrow{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}\) are non-zero vectors, then the angle between them is given by \(\cos \theta =\frac{\overrightarrow{a.}\overrightarrow{b}}{\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|}=\frac{{{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}}{\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}\sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}}\).
Also
\(\frac{{{\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}} \right)}^{2}}}{\left( a_{1}^{2}+a_{2}^{2}+a_{3}^{2} \right)\left( b_{1}^{2}+b_{2}^{2}+b_{3}^{2} \right)}={{\cos }^{2}}\theta \le 1\).
Or \({{\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}} \right)}^{2}}\le \left( a_{1}^{2}+a_{2}^{2}+a_{3}^{2} \right)\left( b_{1}^{2}+b_{2}^{2}+b_{3}^{2} \right)\).
Cosine Rule using Dot Product:
Using vector method, prove that in a triangle a² = b² + c² – 2bc cos A (Cosine law).
In ΔABC,
Let \(\overrightarrow{AB}=\overrightarrow{c}\), \(\overrightarrow{BC}=\overrightarrow{a}\), \(\overrightarrow{CA}=\overrightarrow{b}\).
Since \(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0\), we have \(\overrightarrow{a}=-\left( \overrightarrow{b}+\overrightarrow{c} \right)\),
∴ \(\left| \overrightarrow{a} \right|=\left| -\left( \overrightarrow{b}+\overrightarrow{c} \right) \right|\),
or \({{\left| \overrightarrow{a} \right|}^{2}}={{\left| \overrightarrow{b}+\overrightarrow{c} \right|}^{2}}\),
or \({{\left| \overrightarrow{a} \right|}^{2}}={{\left| \overrightarrow{b} \right|}^{2}}+{{\left| \overrightarrow{c} \right|}^{2}}+2\,\overrightarrow{b}.\overrightarrow{c}\),
or \({{\left| \overrightarrow{a} \right|}^{2}}={{\left| \overrightarrow{b} \right|}^{2}}+{{\left| \overrightarrow{c} \right|}^{2}}+2\left| \overrightarrow{b} \right|\left| \overrightarrow{c} \right|\cos \left( \pi -A \right)\) (∵ angle between \(\overrightarrow{b}\) and \(\overrightarrow{c}\) = angle between CA produced and AB)
or a² = b² + c² – 2bc cos A.