A straight line has vector equation and a plane has equation , . Show that the angle between and is…

The equation of the line is $\vec{r} = 5\hat{i} + \lambda(5\hat{i} + \sin\theta \hat{j} + \cos\theta \hat{k})$. The direction vector of the line is $\vec{d} = 5\hat{i} + \sin\theta \hat{j} + \cos\theta \hat{k}$. The equation of the plane is $x = p \Rightarrow x + 0y + 0z = p$, so the normal vector to the plane is $\vec{n} = \hat{i}$. The angle $\phi$ between a line and a plane is given by: $\sin\phi = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}| |\vec{n}|}$. Here, $\vec{d} \cdot \vec{n} = 5(1) + (\sin\theta)(0) + (\cos\theta)(0) = 5$. $|\vec{d}| = \sqrt{5^2 + \sin^2\theta + \cos^2\theta} = \sqrt{25 + 1} = \sqrt{26}$. $|\vec{n}| = 1$. Thus, $\sin\phi = \frac{5}{\sqrt{26}} \Rightarrow \phi = \sin^{-1}\left(\frac{5}{\sqrt{26}}\right)$. Since this value contains neither $\theta$ nor $p$, the angle between $L_\theta$ and $\pi_P$ is independent of both $\theta$ and $p$.