Angle made by the position vector of the point ( 5 , - 4 , - 3 ) with the positive direction of X - axis is

The position vector of the point $\left(5, -4, -3\right)$ is $5\hat{i}-4\hat{j}-3\hat{k}$

We know that $\hat{i}$ is the unit vector along $x-$axis

So, the angle between the vector $5\hat{i}-4\hat{j}-3\hat{k}$ and $x-$axis will be same as the angle between the vector $5\hat{i}-4\hat{j}-3\hat{k}$ and $\hat{i}$.

Let the angle be $\theta$

$\Rightarrow \left(5\hat{i}-4\hat{j}-3\hat{k}\right)·\hat{i}=\left|5\hat{i}-4\hat{j}-3\hat{k}\right|\left|\hat{i}\right|\cos \theta$

$\Rightarrow 5=5\sqrt{2}\cos \theta$

$\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}$

$\Rightarrow \theta =\frac{\pi }{4}$