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I am supposed to show that the inscribed angle in a half circle is $90^\circ$ using dot product. The problem is written below.

If I have a circle with midpoint $M$ and radius $r$, and three points $R$, $P$ , $Q$ on the circle so that $\mid PR\mid = 2r $. How can I show that $PQR$ is a right angle?

Thanks for any help!

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    What have you tried? We request askers to show that they have given thought to the problem. – 2017-02-23

2 Answers 2

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If you need to use the dot product...

We can translate the circle such that M lies on the origin. and rotate it such that $|PR|$ lies on the $x-$axis

$P = (-r,0), R = (r,0), Q = (r\cos t, r\sin t)$

$\vec {QP} = r(-1 - \cos t, -\sin t)\\ \vec {QR} = r(1 - \cos t, -\sin t)\\ \vec {QP} \cdot \vec {QR} = r^2 ((\cos^2 t-1) +\sin^2 t) = 0$

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The dot product formula is:

$a\cdot b = \left | a \right |\left | b \right |cos\theta$

So if $\theta = \pi/2$, then RHS = 0

We need to show that:

$$\vec{PQ}\cdot \vec{RQ} = 0$$

Let $\vec{MP} = \vec{p}, \vec{MR} = \vec{r}, \vec{MQ} = \vec{q}$

Then we have:

$\vec{PQ}\cdot \vec{RQ} = (\vec{q} - \vec{p} )\cdot (\vec{q} - \vec{r})$

Note that $\vec{r} = -\vec{p}$ and $\left | \vec{p} \right | = \left | \vec{q} \right | = \left | \vec{r} \right |$

Therefore we can simplify the RHS as shown:

$\vec{PQ}\cdot \vec{RQ} = \left | \vec{q} \right |^2 - \left | \vec{p} \right |^2 = 0$ as required