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