A straight line L_θ has vector equation r⃗ = 5î + λ(5î + sinθ ĵ + cosθ k̂) and a plane π_P has equation x = p, p ∈ mathbbR. Show that the angle between…

The line has direction vector d⃗ = 5î + sinθ ĵ + cosθ k̂, so |d⃗| = √(25 + sin²θ + cos²θ) = √(26). The plane x = p, i.e. r⃗·î = p, has normal n⃗ = î. If φ is the angle between the line and the plane, sinφ = (|d⃗·n⃗|)/(|d⃗||n⃗|) = 5/(√(26)), so φ = sin⁻¹(5/(√(26))). This contains neither θ nor p, so the angle is independent of both.

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A straight line L_θ has vector equation r⃗ = 5î + λ(5î + sinθ ĵ + cosθ k̂) and a plane π_P has…

A straight line $L_\theta$ has vector equation $\vec{r} = 5\hat{i} + \lambda(5\hat{i} + \sin\theta \hat{j} + \cos\theta \hat{k})$ and a plane $\pi_P$ has equation $x = p$, $p \in \mathbb{R}$. Show that the angle between $L_\theta$ and $\pi_P$ is independent of both $\theta$ and $p$.

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The line has direction vector $\vec{d} = 5\hat{i} + \sin\theta\,\hat{j} + \cos\theta\,\hat{k}$, so $|\vec{d}| = \sqrt{25 + \sin^2\theta + \cos^2\theta} = \sqrt{26}$. The plane $x = p$, i.e. $\vec{r}\cdot\hat{i} = p$, has normal $\vec{n} = \hat{i}$. If $\phi$ is the angle between the line and the plane, $\sin\phi = \frac{|\vec{d}\cdot\vec{n}|}{|\vec{d}||\vec{n}|} = \frac{5}{\sqrt{26}}$, so $\phi = \sin^{-1}\left(\frac{5}{\sqrt{26}}\right)$. This contains neither $\theta$ nor $p$, so the angle is independent of both.