Angle Between Two Vectors

We can find the angle between two vectors using scalar product of vectors and vector product of two vectors. We will learn both the ways here.

How to find the angle between two vectors using Scalar Product formula

Let α and β be two vectors and the angles between them is θ. Then by scalar product formula of two vectors we have,

\[\overrightarrow{\alpha }.\overrightarrow{\beta }=\left| \overrightarrow{\alpha } \right|\left| \overrightarrow{\beta } \right|\cos \theta \]

\[\Rightarrow \cos \theta =\frac{\overrightarrow{\alpha }.\overrightarrow{\beta }}{\left| \overrightarrow{\alpha } \right|\left| \overrightarrow{\beta } \right|}\]

\[\therefore \theta ={{\cos }^{-1}}\left( \frac{\overrightarrow{\alpha }.\overrightarrow{\beta }}{\left| \overrightarrow{\alpha } \right|\left| \overrightarrow{\beta } \right|} \right)\]

\[Let~~\overrightarrow{\alpha }={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}~~and~~\overrightarrow{\beta }={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}\]

\[Then,~~\overrightarrow{\alpha }.\overrightarrow{\beta }={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}~~and\]

\[\left| \overrightarrow{\alpha } \right|=\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}},~~\left| \overrightarrow{\beta } \right|=\sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}\]

\[\therefore \theta ={{\cos }^{-1}}\left( \frac{{{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}}{\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}.\sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}} \right)\]

How to find the angle between two vectors using Vector Product formula

Let α and β be two vectors and the angles between them is θ. Then by vector product formula of two vectors we have,

\[\overrightarrow{\alpha }\times \overrightarrow{\beta }=\left| \overrightarrow{\alpha } \right|\left| \overrightarrow{\beta } \right|\sin \theta \left( \overrightarrow{n} \right)\]

\[\Rightarrow {{\left| \overrightarrow{\alpha }\times \overrightarrow{\beta } \right|}^{2}}={{\left| \overrightarrow{\alpha } \right|}^{2}}{{\left| \overrightarrow{\beta } \right|}^{2}}{{\sin }^{2}}\theta {{\left| \overrightarrow{n} \right|}^{2}}\]

\[\Rightarrow {{\left| \overrightarrow{\alpha }\times \overrightarrow{\beta } \right|}^{2}}={{\left| \overrightarrow{\alpha } \right|}^{2}}{{\left| \overrightarrow{\beta } \right|}^{2}}{{\sin }^{2}}\theta ~~\left[ \because \left| \overrightarrow{n} \right|=1 \right]\]

\[\therefore {{\sin }^{2}}\theta =\frac{{{\left| \overrightarrow{\alpha }\times \overrightarrow{\beta } \right|}^{2}}}{{{\left| \overrightarrow{\alpha } \right|}^{2}}{{\left| \overrightarrow{\beta } \right|}^{2}}}\]

\[Let~~\overrightarrow{\alpha }={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}~~and~~\overrightarrow{\beta }={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}\]

\[\therefore {{\sin }^{2}}\theta =\frac{{{\left( {{a}_{2}}{{b}_{3}}-{{b}_{2}}{{a}_{3}} \right)}^{2}}+{{\left( {{a}_{3}}{{b}_{1}}-{{b}_{3}}{{a}_{1}} \right)}^{2}}+{{\left( {{a}_{1}}{{b}_{2}}-{{b}_{1}}{{a}_{2}} \right)}^{2}}}{\left( a_{1}^{2}+a_{2}^{2}+a_{3}^{2} \right)\left( b_{1}^{2}+b_{2}^{2}+b_{3}^{2} \right)}\]

\[\Rightarrow \theta ={{\sin }^{-1}}\left( \sqrt{\frac{{{\left( {{a}_{2}}{{b}_{3}}-{{b}_{2}}{{a}_{3}} \right)}^{2}}+{{\left( {{a}_{3}}{{b}_{1}}-{{b}_{3}}{{a}_{1}} \right)}^{2}}+{{\left( {{a}_{1}}{{b}_{2}}-{{b}_{1}}{{a}_{2}} \right)}^{2}}}{\left( a_{1}^{2}+a_{2}^{2}+a_{3}^{2} \right)\left( b_{1}^{2}+b_{2}^{2}+b_{3}^{2} \right)}} \right)\]

Find the angle between the two vectors

\[3\hat{i}+2\hat{j}-\hat{k}~~and~~\hat{i}-3\hat{j}+2\hat{k}\]

Solution:

\[Let~~\overrightarrow{\alpha }=3\hat{i}+2\hat{j}-\hat{k}~~and~~\overrightarrow{\beta }=~\hat{i}-3\hat{j}+2\hat{k}\]

And θ be the angles between the vectors.

\[\overrightarrow{\alpha }.\overrightarrow{\beta }=\left| \overrightarrow{\alpha } \right|\left| \overrightarrow{\beta } \right|\cos \theta \]

\[\Rightarrow \cos \theta =\frac{\overrightarrow{\alpha }.\overrightarrow{\beta }}{\left| \overrightarrow{\alpha } \right|\left| \overrightarrow{\beta } \right|}\]

\[\Rightarrow \cos \theta =\frac{3-6-2}{\sqrt{{{3}^{2}}+{{2}^{2}}+{{\left( -1 \right)}^{2}}}.\sqrt{{{1}^{2}}+{{\left( -3 \right)}^{2}}+{{2}^{2}}}}=\frac{-5}{14}\]

\[\therefore \theta ={{\cos }^{-1}}\left( \frac{-5}{14} \right)\]

Therefore the angles between the two vectors is

\[{{\cos }^{-1}}\left( \frac{-5}{14} \right)\]

If θ be the angle between the unit vectors a and b then prove that

\[(i)~~\cos \frac{\theta }{2}=\frac{1}{2}\left| \hat{a}+\hat{b} \right|\]

\[(ii)~~\sin \frac{\theta }{2}=\frac{1}{2}\left| \hat{a}-\hat{b} \right|\]

\[(iii)~~\tan \frac{\theta }{2}=\frac{\left| \hat{a}-\hat{b} \right|}{\left| \hat{a}+\hat{b} \right|}\]

Solution:

\[{{\left| \hat{a}+\hat{b} \right|}^{2}}={{\left| {\hat{a}} \right|}^{2}}+{{\left| {\hat{b}} \right|}^{2}}+2\hat{a}.\hat{b}\]

\[\Rightarrow {{\left| \hat{a}+\hat{b} \right|}^{2}}=1+1+2\left| {\hat{a}} \right|\left| {\hat{b}} \right|\cos \theta ~~~~\left[ \because \left| {\hat{a}} \right|=\left| {\hat{b}} \right|=1 \right]\]

\[\Rightarrow {{\left| \hat{a}+\hat{b} \right|}^{2}}=2+2\cos \theta =2\left( 1+\cos \theta  \right)\]

\[\Rightarrow {{\left| \hat{a}+\hat{b} \right|}^{2}}=2\left( 2{{\cos }^{2}}\frac{\theta }{2} \right)=4{{\cos }^{2}}\frac{\theta }{2}\]

\[\Rightarrow \cos \frac{\theta }{2}=\frac{1}{2}\left| \hat{a}+\hat{b} \right|~~~~\left[ \because 0\le \theta \le \pi \Rightarrow ~\cos \frac{\theta }{2}\ge 0 \right]\]

Again,

\[{{\left| \hat{a}-\hat{b} \right|}^{2}}={{\left| {\hat{a}} \right|}^{2}}+{{\left| {\hat{b}} \right|}^{2}}-2\hat{a}.\hat{b}\]

\[\Rightarrow {{\left| \hat{a}-\hat{b} \right|}^{2}}=1+1-2\left| {\hat{a}} \right|\left| {\hat{b}} \right|\cos \theta ~~~~\left[ \because \left| {\hat{a}} \right|=\left| {\hat{b}} \right|=1 \right]\]

\[\Rightarrow {{\left| \hat{a}-\hat{b} \right|}^{2}}=2-2\cos \theta =2\left( 1-\cos \theta  \right)\]

\[\Rightarrow {{\left| \hat{a}-\hat{b} \right|}^{2}}=2\left( 2{{\sin }^{2}}\frac{\theta }{2} \right)=4{{\sin }^{2}}\frac{\theta }{2}\]

\[\Rightarrow \sin \frac{\theta }{2}=\frac{1}{2}\left| \hat{a}-\hat{b} \right|~~~~\left[ \because 0\le \theta \le \pi \Rightarrow ~\sin \frac{\theta }{2}\ge 0 \right]\]

Now,

\[\tan \frac{\theta }{2}=\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}=\frac{\frac{1}{2}\left| \hat{a}-\hat{b} \right|}{\frac{1}{2}\left| \hat{a}+\hat{b} \right|}=\frac{\left| \hat{a}-\hat{b} \right|}{\left| \hat{a}+\hat{b} \right|}\]

Prove that

\[{{\left( \frac{\overrightarrow{a}}{{{a}^{2}}}-\frac{\overrightarrow{b}}{{{b}^{2}}} \right)}^{2}}={{\left( \frac{\overrightarrow{a}-\overrightarrow{b}}{ab} \right)}^{2}}~~if~~\left| \overrightarrow{a} \right|=a~~and~~\left| \overrightarrow{b} \right|=b.\]

Solution:

\[{{\left( \frac{\overrightarrow{a}}{{{a}^{2}}}-\frac{\overrightarrow{b}}{{{b}^{2}}} \right)}^{2}}={{\left| \frac{\overrightarrow{a}}{{{a}^{2}}} \right|}^{2}}+{{\left| \frac{\overrightarrow{b}}{{{b}^{2}}} \right|}^{2}}-2\left| \frac{\overrightarrow{a}}{{{a}^{2}}} \right|\left| \frac{\overrightarrow{b}}{{{b}^{2}}} \right|\cos \theta \]

\[\left[ \because ~~\theta ~~is~~the~~angle~~between~~\overrightarrow{a}~~and~~\overrightarrow{b}\Rightarrow \theta ~~is~~also~~angle~~between~~\frac{\overrightarrow{a}}{{{a}^{2}}},\frac{\overrightarrow{b}}{{{b}^{2}}} \right]\]

\[\Rightarrow {{\left( \frac{\overrightarrow{a}}{{{a}^{2}}}-\frac{\overrightarrow{b}}{{{b}^{2}}} \right)}^{2}}=\frac{{{\left| \overrightarrow{a} \right|}^{2}}}{{{a}^{4}}}+\frac{{{\left| \overrightarrow{b} \right|}^{2}}}{{{b}^{4}}}-\frac{2\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos \theta }{{{a}^{2}}{{b}^{2}}}\]

\[\Rightarrow {{\left( \frac{\overrightarrow{a}}{{{a}^{2}}}-\frac{\overrightarrow{b}}{{{b}^{2}}} \right)}^{2}}=\frac{{{a}^{2}}}{{{a}^{4}}}+\frac{{{b}^{2}}}{{{b}^{4}}}-\frac{2ab\cos \theta }{{{a}^{2}}{{b}^{2}}}\]

\[\Rightarrow {{\left( \frac{\overrightarrow{a}}{{{a}^{2}}}-\frac{\overrightarrow{b}}{{{b}^{2}}} \right)}^{2}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}-\frac{2\cos \theta }{ab}\]

\[\Rightarrow {{\left( \frac{\overrightarrow{a}}{{{a}^{2}}}-\frac{\overrightarrow{b}}{{{b}^{2}}} \right)}^{2}}=\frac{{{b}^{2}}+{{a}^{2}}-2ab\cos \theta }{{{a}^{2}}{{b}^{2}}}\]

\[\Rightarrow {{\left( \frac{\overrightarrow{a}}{{{a}^{2}}}-\frac{\overrightarrow{b}}{{{b}^{2}}} \right)}^{2}}=\frac{{{\left| \overrightarrow{b} \right|}^{2}}+{{\left| \overrightarrow{a} \right|}^{2}}-2\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos \theta }{{{a}^{2}}{{b}^{2}}}\]

\[\therefore {{\left( \frac{\overrightarrow{a}}{{{a}^{2}}}-\frac{\overrightarrow{b}}{{{b}^{2}}} \right)}^{2}}=\frac{{{\left( \overrightarrow{a}-\overrightarrow{b} \right)}^{2}}}{{{a}^{2}}{{b}^{2}}}={{\left( \frac{\overrightarrow{a}-\overrightarrow{b}}{ab} \right)}^{2}}\]

Prove by vector method, for any triangle ABC

\[\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\]

Solution:

Here,

\[\left| \overrightarrow{BC} \right|=a,\left| \overrightarrow{CA} \right|=b,\left| \overrightarrow{AB} \right|=c\]

Now, from the triangle ABC we have,

\[\overrightarrow{CA}+\overrightarrow{AB}=-\overrightarrow{BC}\]

\[\Rightarrow {{\left( \overrightarrow{CA}+\overrightarrow{AB} \right)}^{2}}={{\overrightarrow{BC}}^{2}}\]

\[\Rightarrow {{\left( \overrightarrow{CA} \right)}^{2}}+{{\left( \overrightarrow{AB} \right)}^{2}}+2\overrightarrow{CA}.\overrightarrow{AB}={{\overrightarrow{BC}}^{2}}\]

\[\Rightarrow {{\left| \overrightarrow{CA} \right|}^{2}}+{{\left| \overrightarrow{AB} \right|}^{2}}+2\left| \overrightarrow{CA} \right|\left| \overrightarrow{AB} \right|\cos \left( \pi -A \right)={{\left| \overrightarrow{BC} \right|}^{2}}\]

\[\Rightarrow {{b}^{2}}+{{c}^{2}}-2bc\cos A={{a}^{2}}\]

\[\Rightarrow \cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\]

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