Find the angle between the lines whose direction ratios are proportional to 4, -3, 5 and 3, 4, 5.

Hint: Determine the line vectors using the direction ratios, we know for the vector $\overrightarrow{r}=a\overset{\hat{\ }}{\mathop{i}}\,+b\overset{\wedge }{\mathop{j}}\,+c\overset{\wedge }{\mathop{k}}\,$, the direction ratios are (a, b, c) and the proportionality given in the question, then find their scalar product.

Complete step-by-step answer:
Let us assume the line vector with direction ratios 4, -3, 5 to be $\overrightarrow{a}$.
Then, $\overrightarrow{a}=4\overset{\hat{\ }}{\mathop{i}}\,-3\overset{\wedge }{\mathop{j}}\,+5\overset{\wedge }{\mathop{k}}\,$.
Let us assume the line vector with direction ratios 3, 4, 5 to be $\overrightarrow{b}$.
Therefore, the x component is 3.
Therefore, the y component is 4.
Therefore, the z component is 5.
Then, $\overrightarrow{b}=3\overset{\hat{\ }}{\mathop{i}}\,+4\overset{\wedge }{\mathop{j}}\,+5\overset{\wedge }{\mathop{k}}\,$.
We know that the scalar product of $\overrightarrow{x}\ and\ \overrightarrow{y}$is
$\overrightarrow{x}.\overrightarrow{y}=\left| \overrightarrow{x} \right|\left| \overrightarrow{y} \right|\cos \theta $,
Where $\overrightarrow{x}\ and\ \overrightarrow{y}$ are the magnitudes of $\overrightarrow{x}\ and\ \overrightarrow{y}$ respectively and $\theta $ is the angle between $\overrightarrow{x}\ and\ \overrightarrow{y}$.
Let us assume the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ i.e. the two given line to be $\phi $, then
$\left( \overrightarrow{a}.\overrightarrow{b} \right)=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos \phi $
This can also be written as,
$\cos \phi =\dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|}..........(i)$
Now we will find the values separately,
$\begin{align}
  & \left| \overrightarrow{a} \right|=\sqrt{{{4}^{2}}+{{\left( -3 \right)}^{2}}+{{5}^{2}}}=\sqrt{50}=5\sqrt{2} \\
 & \left| \overrightarrow{b} \right|=\sqrt{{{3}^{2}}+{{4}^{2}}+{{5}^{2}}}=\sqrt{50}=5\sqrt{2} \\
\end{align}$
Substituting these values in equation (i), we get
$\begin{align}
  & \cos \phi =\dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|} \\
 & \Rightarrow \cos \phi =\dfrac{\left( 4\overset{\hat{\ }}{\mathop{i}}\,-3\overset{\wedge }{\mathop{j}}\,+5\overset{\wedge }{\mathop{k}}\, \right)\left( 3\overset{\hat{\ }}{\mathop{i}}\,+4\overset{\wedge }{\mathop{j}}\,+5\overset{\wedge }{\mathop{k}}\, \right)}{5\sqrt{2}\times 5\sqrt{2}} \\
 & \Rightarrow \cos \phi =\dfrac{4\times 3-3\times 4+5\times 5}{50} \\
 & \Rightarrow \cos \phi =\dfrac{25}{50} \\
 & \Rightarrow \cos \phi =\dfrac{1}{2} \\
\end{align}$
Therefore,
$\phi ={{\cos }^{-1}}\dfrac{1}{2}$
$\phi =\dfrac{\pi }{3}$
Therefore, the angle between the two lines is $\dfrac{\pi }{3}$.

Note:Another approach to find the angle between two lines is by first finding the unit vectors of the given lines by using the formula $\overset{\wedge }{\mathop{a}}\,=\dfrac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}$. And the angle between two unit vector is given by $\cos \phi =\overset{\wedge }{\mathop{a}}\,.\overset{\wedge }{\mathop{b}}\,$