Question

Find the angle between vector \[\hat{i}+\hat{j}\] and vector \[\hat{i}-\hat{j}\] .

Answer 1

Hint: Now we know that the dot product of two vectors $ \vec{a}=\left( {{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k} \right) $ and $ \vec{b}=\left( {{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k} \right) $ is given by $ \vec{a}.\vec{b}=\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}} \right) $ . Also we know that $ \vec{a}.\vec{b}=|\vec{a}||\vec{b}|\cos \theta $ where $ \theta $ is the angle between two vectors $ \vec{a} $ and $ \vec{b} $ and $ |\vec{a}|=\sqrt{{{a}_{1}}^{2}+{{a}_{2}}^{2}+{{a}_{3}}^{2}} $ for any vector $ \vec{a}=\left( {{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k} \right) $ . Hence with equation (1) and equation (2) we can find the value of $ \theta $ .

Complete step-by-step answer:
Now the given vectors are \[\hat{i}+\hat{j}\] and vectors \[\hat{i}-\hat{j}\] .
Now let \[\vec{a}=\hat{i}+\hat{j}\]
We know that for any vector $ \vec{a}=\left( {{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k} \right) $ modulus of vector $ \vec{a} $ is $ |\vec{a}|=\sqrt{{{a}_{1}}^{2}+{{a}_{2}}^{2}+{{a}_{3}}^{2}} $
Hence using this we get
 $ |\vec{a}|=\sqrt{{{1}^{2}}+{{1}^{2}}}=\sqrt{2}.........................(1) $
Now consider \[\hat{i}-\hat{j}\]
Let $ \vec{b}=\hat{i}-\hat{j} $
Now again we know for any vector $ \vec{a}=\left( {{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k} \right) $ modulus of vector $ \vec{a} $ is $ |\vec{a}|=\sqrt{{{a}_{1}}^{2}+{{a}_{2}}^{2}+{{a}_{3}}^{2}} $
Hence we get
 $ |\vec{b}|=\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}}=\sqrt{2}....................(2) $
Now we know that the dot product of two vectors $ \vec{a}=\left( {{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k} \right) $ and $ \vec{b}=\left( {{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k} \right) $ is given by $ \vec{a}.\vec{b}=\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}} \right) $ .
Hence the dot product of \[\vec{a}=\hat{i}+\hat{j}\] and $ \vec{b}=\hat{i}-\hat{j} $ is given by
 $ \vec{a}.\vec{b}=1(1)+1(-1)=0 $
Now for dot product we also have that $ \vec{a}.\vec{b}=|\vec{a}||\vec{b}|\cos \theta $ where $ \theta $ is the angle between two vectors $ \vec{a} $ and $ \vec{b} $ .
Hence for we get of \[\vec{a}=\hat{i}+\hat{j}\] and $ \vec{b}=\hat{i}-\hat{j} $
 $ 0=|\vec{a}||\vec{b}|\cos \theta $ where $ \theta $ is the angle between two vectors $ \vec{a} $ and $ \vec{b} $ .
Now from equation (1) and equation (2) we substitute the value of $ |\vec{a}| $ and $ |\vec{b}| $ hence we get.
 $ 0=\sqrt{2}\sqrt{2}\cos \theta $ where $ \theta $ is the angle between two vectors $ \vec{a} $ and $ \vec{b} $ .
Hence we get
 $ 0=2\cos \theta $
Dividing the equation by 2 we get.
 $ \cos \theta =0 $
We know that if $ \cos \theta =0 $ then the value of $ \theta $ is $ \dfrac{\pi }{2} $
Hence we get the angle between vector \[\hat{i}+\hat{j}\] and vector \[\hat{i}-\hat{j}\] is $ \dfrac{\pi }{2} $ .

Note: Note that we have if the vectors are perpendicular then their dot product is 0 and vice versa. Hence if we get a dot product as 0 we can directly say that the vectors are perpendicular.
Also we can find the angle between two vectors with the help of $ \vec{a}\times \vec{b}=|\vec{a}||\vec{b}|\sin \theta $ where $ \theta $ is angle between two vectors and for vectors $ \vec{a}=\left( {{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k} \right) $ and $ \vec{b}=\left( {{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k} \right) $ cross product is given by \[\vec{a}\times \vec{b}=\left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
\end{matrix} \right|\]