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$\begingroup$

But I am wondering why isit $PQ \perp QR$ and not $QP \perp QR$ as shown below?

UPDATE

How do I get the equation: $(i-1)b=ic-a=i(1-2i)-(-1+4i)=3-3i$?

Where does $(i-1)$ come from? I dont really get what they did after that either :(

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    OK, I updated my post.2011-11-22

2 Answers 2

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The author is thinking of $p-q$ as the vector from $Q$ to $P$ and $r-q$ as the vector from $Q$ to $R$.

As Jyrki notes, the author is saying: by looking at the picture, that $p-q=i(r-q)$ (because multiplication of $r-q$ by $i$ rotates it ccw by $90^\circ$).

He starts by saying the vectors $PQ$ and $QR$ have the same length and are perpendicular. You are right in your comment. He should have written $|QP|=|QR|$ (and $QP\perp QR$ to avoid confusion, in my opinion).


For your update, I'm not sure where he got that... You could do the following however: From

$a-b =i(c-b)$, substitute $a=-1+4i$ and $c=1-2i$: $ (-1+4i)-b =i( (1-2i)-b) $ This gives: $ -1+4i-b= i+2-bi \iff -b+bi = i+2+1-4i\iff b(i-1)=3-3i. $

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    It looks like from the author's perspective, "PQ" and "QR" aren't vectors, they're line segments. That's why the || notation was omitted.2011-11-22
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From $a-b=i(c-b)$ you get $a-b=ic-ib,\quad ib-b=ic-a,\quad (i-1)b=ic-a$ Is there another part that you don't see?