Find the angle between the lines \[\overrightarrow{r}=3\hat{i}+2\hat{j}-4\hat{k}+\lambda (\hat{i}+2\hat{j}+2\hat{k})\,and\,\,\overrightarrow{r}=(5\hat{j}-2\hat{k})+\mu (3\hat{i}+2\hat{j}+6\hat{k})\]
A- \[\theta ={{\sin }^{-1}}(\dfrac{19}{21})\]
B- \[\theta ={{\cos }^{-1}}(\dfrac{19}{21})\]
C- \[\theta ={{\cos }^{-1}}(\dfrac{20}{21})\]
D- None of these

Hint: Let vector equation of lines be $\overrightarrow{r}={{\overrightarrow{a}}_{1}}+\lambda {{\overrightarrow{b}}_{1}}\,and\,\,\overrightarrow{r}={{\overrightarrow{a}}_{2}}+\mu {{\overrightarrow{b}}_{2}}$, these two lines are parallel to vector ${{\overrightarrow{b}}_{1}}\,and\,{{\overrightarrow{\,b}}_{2}}$
Therefore, angle between these two lines is an angle between ${{\overrightarrow{b}}_{1}}\,and\,{{\overrightarrow{\,b}}_{2}}$
$\cos \theta =\dfrac{{{\overrightarrow{b}}_{1}}.{{\overrightarrow{b}}_{2}}}{|\overrightarrow{{{b}_{1}}}|.|\overrightarrow{{{b}_{2}}|}}$


Complete step by step solution:
We are given with four options in which three have some angle while the fourth option says none of the above. Let $\overset{'}{\mathop{\theta }}\,$be the angle between the two lines
\[\begin{align}
  & \overrightarrow{r}=3\hat{i}+2\hat{j}-4\hat{k}+\lambda (\hat{i}+2\hat{j}+2\hat{k}) \\
 & \overrightarrow{r}=(5\hat{j}-2\hat{k})+\mu (3\hat{i}+2\hat{j}+6\hat{k}) \\
\end{align}\]
The given lines are parallel to vector
\[{{\overrightarrow{b}}_{1}}=\hat{i}+2\hat{j}+2\hat{k}\,\,and\,\,{{\overrightarrow{b}}_{2}}=3\hat{i}+2\hat{j}+6\hat{k}\,\]respectively.
So, the angle $'\theta '\,$between them is given by
$\cos \theta =\dfrac{\overrightarrow{{{b}_{1}}}.\overrightarrow{{{b}_{2}}}}{|\overrightarrow{{{b}_{1}}}|.|\overrightarrow{{{b}_{2}}}|}$
$=\dfrac{(\hat{i}+2\hat{j}+2\hat{k}).(3\hat{i}+2\hat{j}+6\hat{k})}{|\hat{i}+2\hat{j}+2\hat{k}||3\hat{i}+2\hat{j}+6\hat{k})}$
Since we know
\[\begin{align}
  & \hat{i}.\hat{i}=1,\hat{i}.\hat{j}=0,\,\hat{i}.\hat{k}=0\,\,\, \\
 & \hat{j}.\hat{i}=0,\hat{j}.\hat{j}=1,\hat{j}.\hat{k}=0 \\
 & \hat{k}.\hat{i}=0,\hat{k}.\hat{j}=0,\hat{k}.\hat{k}=1 \\
\end{align}\]
and $|\overrightarrow{r}|=\sqrt{{{(coeff\,of\,\hat{i})}^{2}}+{{(coeff\,of\,\hat{j})}^{2}}+{{(coeff\,of\,\hat{k})}^{2}}}$
Use this formula in above and by substituting the values, we get,
\[\dfrac{=(1.3)(i.i)+1.2(i.j)+1.6(i.k)+(2.3)(j.i)+(2.2)(j.j)+(2.6)(j.k)+(2.3)(\hat{k}.\hat{i})+(2.2)(k.\hat{j})+(2.6)(\hat{k}.\hat{k})}{\sqrt{{{(1)}^{2}}+{{(2)}^{2}}+{{(2)}^{2}}}\sqrt{{{(3)}^{2}}+{{(2)}^{2}}+{{(6)}^{2}}}}\]
\[\begin{align}
  & =\dfrac{3+0+0+4+0+0+12+0+0}{\sqrt{1+4+4}\sqrt{9+4+36}} \\
 & =\dfrac{3+4+12}{\sqrt{9}.\sqrt{49}} \\
 & =\dfrac{19}{3\times 7} \\
 & =\dfrac{19}{21} \\
 & \cos \theta =\dfrac{19}{21} \\
 & \Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{19}{21} \right)\, \\
\end{align}\]
Hence the correct option is (B)


Additional Information- The angle between two lines is the angle between direction vectors of the lines. If two lines are perpendicular to each other than their direction vectors are also perpendicular. This means that the scalar product of the direction vectors is equal to zero. If two lines are parallel then their direction vectors are proportional. We can also write the result in the form of slopes that is the lines are parallel if their slopes are equal.


Note: To find an angle between two vectors, $\overrightarrow{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$, $\overrightarrow{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$then
\[\theta ={{\cos }^{-1}}\left\{ \dfrac{{{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}}{\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}\sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}} \right\}\]
Here, $\overrightarrow{a}$and$\overrightarrow{b}$ are two non-zero vectors inclined at an angle$\theta $. Scalar product of $\overrightarrow{a}$and$\overrightarrow{b}$ is
\[\overrightarrow{a}.\overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos \theta \]. A perpendicular vector to a line is also called a normal vector to the line. the angle formed by the intersection of two lines cannot be calculated if one of the lines is parallel to the y-axis