# Cosine Rule Using Dot Product

## Cosine Rule Using Dot Product

Using vector method, prove that in a triangle $${{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\operatorname{Cos}A$$.  (Cosine law)

In Triangle ABC,

Let $$\overrightarrow{AB}=\overrightarrow{c}$$

$$\overrightarrow{BC}=\overrightarrow{a}$$ $$\overrightarrow{CA}=\overrightarrow{b}$$

Since $$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$$

$$\Rightarrow \overrightarrow{a}=-\left( \overrightarrow{b}+\overrightarrow{c} \right)$$ $$\Rightarrow |\overrightarrow{a}|=\left| -\left( \overrightarrow{b}+\overrightarrow{c} \right) \right|$$ $$\Rightarrow |\overrightarrow{a}|=\left| \left( \overrightarrow{b}+\overrightarrow{c} \right) \right|$$

Squaring on both sides

$$\Rightarrow |\overrightarrow{a}{{|}^{2}}={{\left| \overrightarrow{b}+\overrightarrow{c} \right|}^{2}}$$ $$\Rightarrow |\overrightarrow{a}{{|}^{2}}={{\left| \overrightarrow{b} \right|}^{2}}+{{\left| \overrightarrow{c} \right|}^{2}}+2\ \overrightarrow{b}.\overrightarrow{c}$$ $$\Rightarrow |\overrightarrow{a}{{|}^{2}}={{\left| \overrightarrow{b} \right|}^{2}}+{{\left| \overrightarrow{c} \right|}^{2}}+2\ \left| \overrightarrow{b} \right|\left| \overrightarrow{c} \right|\cos \left( \pi -A \right)$$ $$\Rightarrow |\overrightarrow{a}{{|}^{2}}={{\left| \overrightarrow{b} \right|}^{2}}+{{\left| \overrightarrow{c} \right|}^{2}}+2\ \left| \overrightarrow{b} \right|\left| \overrightarrow{c} \right|\left( -\cos \left( A \right) \right)$$ $$\Rightarrow |\overrightarrow{a}{{|}^{2}}={{\left| \overrightarrow{b} \right|}^{2}}+{{\left| \overrightarrow{c} \right|}^{2}}-2\ \left| \overrightarrow{b} \right|\left| \overrightarrow{c} \right|\cos \left( A \right)$$

(since angle between  and  = angle between CA produced and AB)

$${{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\operatorname{Cos}A$$

Example: Find the angle between the vectors   $$\hat{i}-2\hat{j}+3\hat{k}$$and $$3\hat{i}-2\hat{j}+\hat{k}$$.

Solution

Given that

$$\hat{i}-2\hat{j}+3\hat{k}$$ $$3\hat{i}-2\hat{j}+\hat{k}$$ $$\left| \overrightarrow{a} \right|=\sqrt{{{1}^{2}}+{{(-2)}^{2}}+{{3}^{2}}}=\sqrt{1+4+9}=\sqrt{14}$$ $$\left| \overrightarrow{b} \right|=\sqrt{{{3}^{2}}+{{(-2)}^{2}}+{{1}^{2}}}=\sqrt{1+4+9}=\sqrt{14}$$

.$$\overrightarrow{a}\ .\ \overrightarrow{b}=(\hat{i}-2\hat{j}+3\hat{k}).(3\hat{i}-2\hat{j}+\hat{k})$$

= 1.3 + (-2) (-2) + 3.1

= 3 + 4 + 3

=10