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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: monkeywantsabanana on April 05, 2011, 08:20:17 pm

Title: Vector proofs
Post by: monkeywantsabanana on April 05, 2011, 08:20:17 pm: Using a vector method to prove if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.

Thanks in advance.
Title: Re: Vector proofs
Post by: m@tty on April 05, 2011, 08:29:37 pm: To make a parallelogram, you have two vectors, $\bold{a} \text{ and } \bold{b}$ .

For it to also be a rhombus, the sides must have the same magnitude ( $||\bold{a}||=||\bold{b}||$ )

The diagonals are $\bold{a} + \bold{b}$ and $\bold{a}-\bold{b}$

If they are perpendicular, their dot product is zero:

$(\bold{a} + \bold{b}) \cdot (\bold{a} - \bold{b}) = 0$

$\bold{a}^2 - \bold{b}^2 = 0$

$\implies ||\bold{a}|| = ||\bold{b}||$

And it is a rhombus, as defined above.

Therefore, if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.
Title: Re: Vector proofs
Post by: monkeywantsabanana on April 05, 2011, 08:33:14 pm: Thank you - again.
Title: Re: Vector proofs
Post by: monkeywantsabanana on April 05, 2011, 09:14:11 pm: How do i also prove that the midpoint of the hypotenuse of a right-angled triangle is equidistant from all vertices?
Title: Re: Vector proofs
Post by: moekamo on April 05, 2011, 09:26:23 pm: have the right angled corner as O, the others as A and B, the midpoint of hypotonuse is M.
vector OA is a, and vector BO is b.

Then BA=b+a
BM=0.5BA=0.5(b+a)
AM=-BM=-0.5(b+a)
OM=-OB+BM=0.5(a-b)

then you can show that these all have the same magnitude, done

note also that a is perpendicular to b, i.e a.b=0
Title: Re: Vector proofs
Post by: jasoN- on April 05, 2011, 09:29:38 pm: B
/ |
/ |
X /\ |
/ \ |
/__ \|
O A

Let OA = ai
Let AB = bj
$\therefore OB = ai + bj$
$OX = XB = \frac{1}{2}OB = \frac{1}{2}(ai+bj)$
$AX = AO + OX = -ai + \frac{1}{2}(ai+bj) = \frac{1}{2}(bj-ai)$
To prove that the midpoint of the hypotenuse is equidistant from all vertices |OX| = |XB| = |AX|
$|OX| = |XB| = \frac{1}{2}\sqrt{a^2+b^2} = \frac{\sqrt{a^2+b^2}}{{2}}$
$|AX| = \sqrt{(\frac{a}{2})^2+(\frac{-b}{2})^2} = \frac{\sqrt{a^2+b^2}}{2}$
$\therefore \mbox{Midpoint of the hypotenuse is equidistant from all vertices as}|OX| = |XB| = |AX|$

edit: correct me if im wrong lol
Title: Re: Vector proofs
Post by: moekamo on April 05, 2011, 09:41:48 pm: Quote from: jasoN- on April 05, 2011, 09:29:38 pm
B
/ |
/ |
X /\ |
/ \ |
/__ \|
O A

Let OA = a
Let AB = b
$\therefore OB = a + b$
$OX = XB = \frac{1}{2}OB = \frac{1}{2}(a+b)$
$AX = AO + OX = -a + \frac{1}{2}(a+b) = \frac{1}{2}(b-a)$
To prove that the midpoint of the hypotenuse is equidistant from all vertices |OX| = |XB| = |AX|
$|OX| = |XB| = \frac{1}{2}\sqrt{1+1} = \frac{\sqrt{2}}{{2}}$
$|AX| = \sqrt{(\frac{1}{2})^2+(\frac{-1}{2})^2} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$
$\therefore \mbox{Midpoint of the hypotenuse is equidistant from all vertices as}|OX| = |XB| = |AX|$

huh? you've somehow proven that the distance from the right angled corner to the midpoint of the hypotenuse is always sqrt(2)/2 !?!?!?

you cant just let a and be be 1(although you can make them in the i and j directions), for this to be general, they have to be any side lengths of any right angled triangle, you have to solve the magnitudes in terms of a and b.
Title: Re: Vector proofs
Post by: jasoN- on April 05, 2011, 09:42:55 pm: oh sorry, you're right. im thinking complex numbers (like i and j) lol.
ill edit
Title: Re: Vector proofs
Post by: monkeywantsabanana on April 05, 2011, 10:05:21 pm: Thanks guys !
Title: Re: Vector proofs
Post by: enpassant on April 06, 2011, 10:44:59 pm: Here is a simpler proof.
right-angle triangle OAB, M is the midpoint of AB the hypotenuse, vector OA=a, vector OB=b
vector OM=1/2 (a+b)
(OM)^2=1/2 (a+b) dot 1/2 (a+b)
=1/4 (a+b) dot (a+b)
=1/4 (a^2+b^2) .......... because a and b are perpendicular, a dot b =0
=1/4 (AB)^2
.: OM=1/2 AB, end of proof