If $\vec{A}=2\hat{i}-2\hat{j}+2\hat{k}$ and $\vec{B}=\hat{i}+\hat{j}$ then \[\]
(a)Find the angle between $\vec{A}$ and $\vec{B}$ (b) Find the resultant vector of $\vec{A}$ and $\vec{B}$on $x-$axis.\[\]

Hint: We use the formula for smaller angle between two vectors written in 3 components $\left( \vec{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k},\vec{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k} \right)$ that is ${{\cos }^{-1}}\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}} \right)$. We use the resultant of $\vec{a}$ and $\vec{b}$ which is given by vector$\vec{R}=\vec{a}+\vec{b}=\vec{a}=\left( {{a}_{1}}+{{b}_{1}} \right)\hat{i}+\left( {{a}_{2}}+{{b}_{2}} \right)\hat{j}+\left( {{a}_{3}}+{{b}_{3}} \right)\hat{k}$ and then find the projection of $\vec{R}$ along $x-$axis which same as along $\hat{i}$ with formula $\vec{R}\cdot \dfrac{{\hat{i}}}{\left| {\hat{i}} \right|}$.\[\]

Complete step-by-step answer:
We know that $\hat{i}$,$\hat{j}$ and $\hat{k}$ are unit vectors(vectors with magnitude 1) along $x,y$ and $z$ axes respectively. So the magnitude of these vectors $ \left| {\hat{i}} \right|=\left| {\hat{j}} \right|=\left| {\hat{k}} \right|=1$. The vectors just like their axes are perpendicular to each other which means any angle among $\hat{i}$,$\hat{j}$ and $\hat{k}$is ${{90}^{\circ }}.$ So $\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1\cdot 1\cdot \cos {{0}^{\circ }}=1$ and $\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{i}=\hat{j}\cdot \hat{k}=\hat{k}\cdot \hat{j}=\hat{i}\cdot \hat{k}=\hat{k}\cdot \hat{i}=1\cdot 1\cdot \cos {{90}^{\circ }}=0$ \[\]
We also know that the dot product of two vectors $\vec{a}$ and $\vec{b}$ is denoted as $\vec{a}\cdot \vec{b}$ and is given by $\vec{a}\cdot \vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta $ where $\theta $ is the angle between the vectors $\vec{a}$ and $\vec{b}$.So if two vectors written in 3 components $\vec{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k},\vec{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k},$ their dot product is given by $\vec{a}\cdot \vec{b}={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}$ and the angle between them is given by ${{\cos }^{-1}}\left( \vec{a}\cdot \vec{b} \right)={{\cos }^{-1}}\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}} \right)$. The resultant of $\vec{a}$ and $\vec{b}$ is given by vector$\vec{R}=\vec{a}+\vec{b}=\vec{a}=\left( {{a}_{1}}+{{b}_{1}} \right)\hat{i}+\left( {{a}_{2}}+{{b}_{2}} \right)\hat{j}+\left( {{a}_{3}}+{{b}_{3}} \right)\hat{k}.$ \[\]

We also know that projection of $\vec{a}$ along $\vec{b}$ is the component of $\vec{a}$ along $\vec{b}$ which is given by $\vec{a}\cdot \dfrac{{\vec{b}}}{\left| {\vec{b}} \right|}$ where $\left| {\vec{b}} \right|$ is the magnitude $\vec{b}$. \[\]
(a)\[\]
The given two vectors are $\vec{A}=2\hat{i}-2\hat{j}+2\hat{k}$ and $\vec{B}=\hat{i}+\hat{j}$ so we find the angle between them with inverse cosine of sum of product of respective components. The angle between them is
$\theta ={{\cos }^{-1}}\left( \vec{A}\cdot \vec{B} \right)={{\cos }^{-1}}\left( 2\left( 1 \right)+\left( -2 \right)1+\left( -1 \right)0 \right)={{\cos }^{-1}}\left( 0 \right)={{90}^{\circ }}$
We add the respective components to find the resultant of the vectors $\vec{a}$ and $\vec{b}$ . So we have,
\[ \vec{R}=\vec{A}+\vec{B}=\left( 2+1 \right)\hat{i}+\left( -2+1 \right)\hat{j}+\left( 0+1 \right)k=3\hat{i}-\hat{j}+\hat{k}\]
(b)\[\]
The projection of the resultant vector along the $x-$axis is the same as projection along the unit orthogonal vector $\hat{i}$ which is a vector in the direction of the $x-$axis . So the projection of $\vec{R}$ along $x-$axis (denoted as P) is
\[ P=\vec{R}\cdot \dfrac{{\hat{i}}}{\left| {\hat{i}} \right|}=\vec{R}\cdot \dfrac{{\hat{i}}}{1}=\left( 3\hat{i}-\hat{j}+\hat{k} \right)\cdot \hat{i}=3\hat{i}\cdot \hat{i}=3 \]

Note: We need to be careful of the confusion between dot product and the resultant . The dot product involves multiplication of respective components while resultant involves addition. We can also find the angle from cross product as ${{\sin }^{-1}}\left( \vec{a}\times \vec{b} \right)$