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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!

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    $\begingroup$ $\sin( \pi -x) = \sin(\pi)\cos(-x) + \cos(\pi)\sin(-x) = \sin(x)$. Your question then becomes, how do I get an exact value from $\sin( \arccos(\sqrt{7}/3))?$ $\endgroup$ – joeb Mar 1 '17 at 23:25
  • $\begingroup$ The first part makes perfect sense, thanks. Do the sin and arccos 'cancel' (in lack of a better term)? $\endgroup$ – user9750060 Mar 1 '17 at 23:37
  • $\begingroup$ They don't cancel but there are ways to calculate $\sin \arccos t$ explicitly $\endgroup$ – joeb Mar 1 '17 at 23:47
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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 …

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