Question

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

Answer 1

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 $\vec{a}\times \vec{b}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\\ \end{matrix} \right|$