Having a bit of trouble with some trigonometry in a vector question.
Plane $\Pi$ has equation $5x-3y-z=1$. Point P $(2,1,6)$ lies in $\Pi$. Point Q has coordinates $(7,-1,2)$. Find the exact value of the sine of the angle $\theta$ between $(PQ)$ and $\Pi$.
This is my solution so far:
$\overrightarrow{PQ}$ = $\begin{pmatrix} 5 \\-2 \\-4\\ \end{pmatrix}$. This is the direction vector of (PQ), call this $\overrightarrow{d}$.
The normal vector of $\Pi$ is $\overrightarrow{n}$ = $\begin{pmatrix} 5\\-3\\-1\\ \end{pmatrix}$
We can find the angle $\phi$ between the normal vector and the direction vector of (PQ).
$cos\phi$ = $\frac {\overrightarrow{d} \cdot \overrightarrow{n}} {\lVert \overrightarrow{d}\rVert \lVert \overrightarrow{n}\rVert} = \frac {\sqrt{7}} {3}$
$\rightarrow \phi = arccos(\frac {\sqrt7} {3})$
$\theta = \pi - \phi \rightarrow sin\theta = sin(\pi - arccos(\frac {\sqrt7} {3}))$
How do I get an exact value from this? The answer is $(\frac {\sqrt7} {3})$ so I'm guessing there's some trigonometric identity/rule involved that I either don't know or have overlooked. Please help!
Hints:
1) $\sin(\pi -\theta)=\sin \theta\,$: this is a basic property of the sine function, no addition formula required for that.
2) $\cos(\arccos x)=x$ by definition of $\arccos x$, and $\arccos x\in[o,\pi]$. On this interval, the sine is non-negative, hence …