# Applications of Dot (Scalar) Product

## 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.