vectors (complex) (1 Viewer)

jnney · Dec 16, 2011

PQR is an isosceles triangle right-angled at P. If the complex number -3+2i represents Q and 4+5i represents R, find the complex numbers representing the possible solutions of P.

boredj · Dec 16, 2011

i came across this question (in terry lee i think) and i think u use the right angle bisector theorem

SpiralFlex · Dec 16, 2011

First of all draw a pretty diagram.

Before we start, we should note this: $\overrightarrow{PQ}=q-p$ (Where $q$ and $p$ represent complex numbers.)

So we know there are two possibilities. We just "simply sketch", $P$ can be on the top side or $P$ can be on the bottom side.

Case I: $\overrightarrow{PQ}$ is rotated $+90$ (ie. Anticlockwise) to get $\overrightarrow{PR}$ .

Case II: $\overrightarrow{PQ}$ is rotated $-90$ (ie. Clockwise) to get $\overrightarrow{PR}$ .

Case I: $(q-p)*i=r-p$

$(-3+2i-p)*i=4+5i-p$

$-3i-2-pi=4+5i-p$

$p-pi=6+8i$

$p(1-i)=6+8i$

$p=\frac{6+8i}{1-i}*\frac{1+i}{1+i}$

$=\frac{-2+14i}{2}$

$\therefore p=-1+7i$

Case II: $(q-p)*-i=r-p$

$(-3+2i-p)*-i=4+5i-p$

$3i+2+pi=4+5i-p$

$p+pi=2+2i$

$p(1+i)=2(1+i)$

$\therefore p =2$

Please note that when you first sketch it, it is virtually impossible to tell it is at 2! So just simply say it's on one of the sides.

boredj · Dec 16, 2011

i didnt read the question and thought p was variable ^^ spiralflex is right.

but doing that way is hard to get rid of the i on p to determine the point

bleakarcher · Dec 16, 2011

I did something stupid. Highly recommend Spiralflex's method anyway.

hayabusaboston · Dec 16, 2011

Guys just don't bother lol spiral is the polyglot of maths haha. Fluent in latex

SpiralFlex · Dec 16, 2011

boredj said:
i didnt read the question and thought p was variable ^^ spiralflex is right.

but doing that way is hard to get rid of the i on p to determine the point

This is the correct way to think if you are "starting" to do these questions.

boredj · Dec 16, 2011

spiral howd u get rid of the imaginary part attached to p

SpiralFlex · Dec 16, 2011

You simplify is by realising the denominator, you should get the perfect answer of 2!

boredj · Dec 16, 2011

(-3 + 2i - p) x I = -3I - 2 - Pi

-3I - 2 - pi = 4 + 5i - p

howd u get rid of the i

SpiralFlex · Dec 16, 2011

$-3i-2-pi=4+5i-p$

$p-pi=6+8i$

$p(1-i)=6+8i$

$p=\frac{6+8i}{1-i}*\frac{1+i}{1+i}$

$=\frac{-2+14i}{2}$

$=-1+7i$

SpiralFlex · Dec 16, 2011

bleakarcher said:
I did something stupid. Highly recommend Spiralflex's method anyway.

There's no such thing as stupidity in maths. Great inventions come from "stupidity".

bleakarcher · Dec 16, 2011

lol, my method was not what stupid, i meant i made a stupid error in my working.

Fus Ro Dah · Dec 17, 2011

bleakarcher said:
lol, my method was not what stupid, i meant i made a stupid error in my working.

wut.

vectors (complex) (1 Viewer)

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