Author Topic: Vector Proofs (Read 2105 times) Tweet Share

dcc · « **on:** January 09, 2008, 07:21:35 pm »

Hey I'm doing some vector proofs, I just wanted to check that my proofs are logical

Proof 1: Prove that the midopint of the hypotenuse of a right-angled triangle is equidistant from all vertices.

$\mbox{Let $\triangle$OAB be a right-angled triangle, and let M be the midpoint of the line segment AB.}$

$\mbox{Let $\vec{OA}$ = a, $\vec{OB}$ = b}$

$\mbox{Where }a = x\tilde{\imath}\mbox{ and } b = y\tilde{\jmath}$

$\mbox{The vector $\vec{OM}$ can be written as such:}$

$\vec{OM} = \vec{OA} + \dfrac{1}{2}\vec{AB}$

$\mbox{Therefore }\vec{OM} = a + \dfrac{1}{2}(-a + b)$

$\mbox{Simplifying we get:}$

$\vec{OM} = \dfrac{1}{2}a + \dfrac{1}{2}b$

$\mbox{Since we are trying to prove that }|\vec{OM}|=|\vec{AM}|=|\vec{BM}|\mbox{, we should find $\vec{AM}$ and $\vec{BM}$ also.}$

$\vec{AM} = \vec{AO} + \vec{OM}$

$\vec{AM} = \dfrac{1}{2}b - \dfrac{1}{2}a$

$\mbox{Using the same methods, finding $\vec{BM}$ is easy:}$

$\vec{BM} = \dfrac{1}{2}a - \dfrac{1}{2}b$

$\mbox{Now finding the magnitudes of all these vectors:}$

$|\vec{OM}| = \sqrt{(\dfrac{x}{2})^2 + (\dfrac{y}{2})^2}$

$|\vec{OM}| = \sqrt{\dfrac{x^2}{4} + \dfrac{y^2}{4}}$

$|\vec{AM}| = \sqrt{(\dfrac{-x}{2})^2 + (\dfrac{y}{2})^2}$

$|\vec{AM}| = \sqrt{\dfrac{x^2}{4} + \dfrac{y^2}{4}}$

$|\vec{BM}| = \sqrt{(\dfrac{x}{2})^2 + (\dfrac{-y}{2})^2}$

$|\vec{BM}| = \sqrt{\dfrac{x^2}{4} + \dfrac{y^2}{2}}$

$\mbox{Therefore as we can see, }|\mbox{$\vec{OM}$}|\mbox{ = }|\mbox{$\vec{AM}$}|\mbox{ = }|\mbox{$\vec{BM}$}|\mbox{, therefore the midpoint of the hypotenuse of a right angled triangle is equidistant from all the vertices of the triangle.}$

Is that a sufficient proof? Have i made an error in my calculations or? Well I have some more

Proof 2: Prove that if a triangle has 3 sides where c^2 = a^2 + b^2 then the triangle is right angled.

$\mbox{Use vector methods to prove that, for $\triangle$ABC, if }c^2 = a^2 + b^2 \mbox{ then $\angle$ACB is a right angle.}$

$\mbox{Let the hypotenuse be side c, with opposite angle C.}$

$\mbox{Let a = $\vec{AB}$, b = $\vec{AC}$, c = $\vec{BC}$}$

$c^2 = a^2 + b^2 \mbox{can be expressed as: }c^2 - a^2 - b^2 = 0$

$\vec{BC} = \vec{BA} + \vec{AC}$

$\mbox{Now taking the vector dot product of $\vec{BC}$ with $\vec{BC}$}$

$\vec{BC}.\vec{BC} = (\vec{BA}+\vec{AC}).(\vec{BA}+\vec{AC})$

$\vec{BC}.\vec{BC} = \vec{BA}.\vec{BA} + 2(\vec{BA}.\vec{AC}) + \vec{AC}.\vec{AC}$

$c^2 = a^2 + 2(\vec{BA}.\vec{AC}) + b^2$

$c^2 - a^2 - b^2 = 2(\vec{BA}.\vec{AC}),\mbox{ but since }c^2 - a^2 - b^2 = 0,\mbox{ then } 2(\vec{BA}.\vec{AC}) = 0$

$\mbox{Since $\vec{BA}$ and $\vec{AC}$ are both non-zero vectors, the angle between them must be 90$^{\circ}$ (as }cos90^{\circ}\mbox{ = 0), and since $\angle$ACB is the angle between $\vec{BA}$ and $\vec{AC}$, }$

$\mbox{Then by the definition of the dot product (}a.b = |a||b|cos\theta\mbox{, where $\theta$ is the angle between a and b) it is shown that $\triangle$ABC must be a right angled triangle!}$

Sorry if i mixed up like, the names of the sides, but it was hard because the book gives crap names to things.

One last one (sorry if im being lengthy in the explanations but if i screw something up I want to be able to tell where i stuff up)

Proof 3: ABCD is a rectangle, use the property of linear independence to prove that the diagonals of a rectangle bisect each other.

$\mbox{ABCD is a rectangle. Use the property of linear independence to prove that the diagonals of a rectangle bisect each other.}$

$\mbox{Let }\vec{AB} = a\mbox{, }\vec{AD} = b\mbox{ (Try to imagine a rectangle, top-left being A, top-right being B, bottom-right being C and bottom-left being D)}$

$\vec{AC} = \vec{AD} + \vec{DC}$

$\vec{AC} = b + a$

$\vec{DB} = \vec{DA} + \vec{AB}$

$\vec{DB} = -b + a = a - b$

$\mbox{Let O be the point of intersection of the diagonals AC and DB}$

$\vec{AO} = x(\vec{AC})\mbox{ and }\vec{DO} = y(\vec{DB})$

$\vec{DA} = \vec{DO} + \vec{OA}$

$\vec{DA} = y(\vec{DB}) - x(\vec{AC})$

$-b = y(a - b) - x(b + a)$

$0 = ya - yb - xb - xa + b$

$0 = (y-x)a - (y + x - 1)b$

$\mbox{Since a and b are linearly independent (because they aren't parallel)}$

$y - x = 0\mbox{ and }y + x - 1 = 0$

$y = x\mbox{ and }y + y - 1 = 0$

$2y - 1 = 0\mbox{ :. }2y=1\mbox{ therefore }y=\dfrac{1}{2}\mbox{ and since }y=x\mbox{ therefore }x=\dfrac{1}{2}$

$\mbox{Therefore O is the midpoint of the diagonals AC and DB, so the diagonals do bisect each other.}$

Is that a sufficient proof?

Thanks in advance, robbo.

Ahmad · « **Reply #1 on:** January 09, 2008, 08:46:15 pm »

Looked over it briefly, looks good.

Extension, use vector methods:
The midpoints of the sides AB, BC, CD, DA of the parallelogram ABCD are M, N, P, Q respectively. Each midpoint is joined to the two vertices not on its side. Show that the area outside the resulting 8-pointed star is 2/5 the area of the parallelogram.

dcc · « **Reply #2 on:** January 09, 2008, 11:46:28 pm »

This one is annoying! Not sure if the logic makes sense here or whatever:

$\mbox{Assume the parallelogram ABCD is a square (a special case of the parallelogram) with side length x}$

$\mbox{Let }|\mbox{$\vec{DA}$}|\mbox{ = x}$

$\mbox{We only need to find the area of one of the triangles and then multiply by 8 (as everything outside the star has the same area).}$

$\mbox{Consider point Q (which is the midpoint of DA), which therefore has a side length = $\dfrac{x}{2}$}$

$\mbox{Now, consider the point of intersection between AP and BQ (which I will call X)}$

$\mbox{Let }p = |AX|$

$\mbox{Let y = }|\mbox{GX}|$

$\mbox{We have a set of similar triangles here (like a million of the buggers), but i think we are most interesting the the triangles $\triangle$ABQ and $\triangle$AQX (which are both have right angles and thus the same ratio for side lengths}$

$\mbox{It can be seen that the length of GX is y, thus the length of GQ is $\dfrac{y}{2}$, as can be seen by the similarity of the triangles.}$

$\mbox{Using Pythagoras, we can find the length of QX, which is $\sqrt{y^2 + \dfrac{y^2}{4}}$=$\sqrt{\dfrac{5y^2}{4}}$}$

$\mbox{Again using similar triangles, it can be seen that $\dfrac{p}{\sqrt{\dfrac{5y^2}{4}}}$=$\dfrac{x}{\dfrac{x}{2}}$=2}$
$\mbox{Therefore: }p = 2\sqrt{\dfrac{5y^2}{4}}$

$\mbox{Then, using Pythagoras (since we have p in terms of y):}$

$(\dfrac{x}{2})^2 = p^2 + (\sqrt{\dfrac{5y^2}{4}})^2$

$\mbox{Substituting for p: } \dfrac{x^2}{4} = (2\sqrt{\dfrac{5y^2}{4}})^2 + \dfrac{5y^2}{4}$

$\dfrac{x^2}{4} = \dfrac{20y^2}{4} + \dfrac{5y^2}{4}$

$\dfrac{x^2}{4} = \dfrac{25y^2}{4}$

$x^2 = 25y^2$

$x = 5y$

$y = \dfrac{x}{5}$

$\mbox{Area of a single triangle = }(\dfrac{1}{2})(\dfrac{x}{2})(\dfrac{x}{5})$

$Area_{single}=\dfrac{x^2}{20}$

$\mbox{Finding the total area of all 8 triangles (count em!), we multiply by 8:}$

$Area_{all triangles}=\dfrac{8x^2}{20}=\dfrac{2}{5}x^2$

$\mbox{But remember the area of a square:}$

$Area_{square}=x^2$

$Area_{all triangles}=\dfrac{2}{5}(Area_{square)}$

$\mbox{The area outside the 8 pointed star is given by the above formula, which shows the area is $\dfrac{2}{5} the area of the square. }$

Now, am i right in assuming, that if you were to stretch out the sides of the square to make a more traditional case of a parallelogram, then the areas would all still be within the same ratio? Because I first tried attacking this problem using a traditional parallelogram but I got some dirty dirty equations which I didnt want to go near, so a square makes more sense lol.

Oh and I didn't know how to do this with more of a focus on vectors, so I did it this way, what would be the way to solve it using vector equations?

Ahmad · « **Reply #3 on:** January 10, 2008, 12:18:37 am »

You're looking at the right triangles, but you can use vectors to prove it straight from the paralellogram! (In a very systematic, no intelligence required sort of way).

Hint: Call the sides of the parallelogram vectors a and b. In your diagram let the intersection be K. Show in general terms that QK/QB is a certain ratio, then head from there.

dcc · « **Reply #4 on:** January 10, 2008, 02:30:20 am »

Ok second try, but with more vectors this time

Imagine parallelogram ABCD (A - bottom left, B - bottom right, C - top right, D - top left).

Let P be the midpoint of line segment CB.

Let Q be the midpoint of line segment AB

Let the point of intersection between AP and QD be designated K.

Let the bit making the height of the triangle be L (its to the left of Q)

$\mbox{Let a = $\vec{AB}$, b = $\vec{AD}$}$

$\mbox{Let }\vec{QK} = x(\vec{QD}), \vec{AK} = y(\vec{AP})$

$\vec{QK} = x(\vec{QD}) = x(b - \dfrac{1}{2}a)$

$\vec{AK} = y(\vec{AP}) = y(a + \dfrac{1}{2}b)$

$\vec{AQ} = \dfrac{1}{2}a = \vec{AK} + \vec{KQ}$

$\dfrac{1}{2}a = ya + \dfrac{1}{2}by - xb + \dfrac{1}{2}xa$

$\mbox{Making LHS = 0 we get:}$

$0 = (y + \dfrac{1}{2}x - \dfrac{1}{2})a + (\dfrac{1}{2}y - x)b$

$\mbox{But since a and b are not parallel then in general, xa + yb can only equal 0 when x = y = 0, so:}$

$y + \dfrac{1}{2}x - \dfrac{1}{2} = 0\mbox{ and }\dfrac{1}{2}y - x = 0$

$x = \dfrac{1}{2}y$

$y = 2x\mbox{, substituting into the first equation: }$

$2x + \dfrac{1}{2}x - \dfrac{1}{2} = 0$

$\dfrac{5}{2}x = \dfrac{1}{2}$

$5x = 1\mbox{, }x = \dfrac{1}{5}\mbox{ :. y=\dfrac{2}{5}}$

$\mbox{Now Let M be the point vertically underneath D (which would represent the height of the parallelogram)}$

$\mbox{Since length }|\mbox{QK}|\mbox{ = $\dfrac{1}{5}}|\mbox{QD}|\mbox{ then }\dfrac{|QD|}{|QK|}=5$

$\mbox{Therefore, }\dfrac{|MD|}{|LK|} = 5$

$|LK| = \dfrac{1}{5}|MD|$

$\mbox{Now, the area of the single triangle is given by $\dfrac{1}{2}$bh, therefore: }$

$A_{single}=(\dfrac{1}{2})(\dfrac{|a|}{2})(|LK|)$

$A_{single}=(\dfrac{|a|}{4})(|LK|)=(\dfrac{|a|}{4})(\dfrac{1}{5}|MD|)=\dfrac{1}{20}|a||MD|$

$\mbox{Then, finding it for all 8 triangles:}$

$A_{all\triangle}=\dfrac{8}{20}|a||MD|=\dfrac{2}{5}|a||MD|$

$\mbox{Now consider the area of the parallelogram, which is found by A = bh}$

$A_{parallelogram}=bh=|a||MD|$

$\mbox{Therefore the area of all 8 triangles can be expressed as:}$

$A_{all\triangle}=\dfrac{2}{5}A_{parallelogram}$

Im just wondering, is there any way to simplifying |MD|, because it feels sloppy just leaving it there, even if it does work.

Ahmad · « **Reply #5 on:** January 10, 2008, 10:04:17 am »

You don't need to find MD, once you find QK/QB you know that's the ratio of the areas [AKQ] and [AQB], and [AQB] is trivial to find. To show this, for example, you could use similarity (same base). Well done though, you should be fine with vector proofs.

Mao · « **Reply #6 on:** January 10, 2008, 03:27:26 pm »

I AM SO LOST, after reading the first few lines in proof no.1

maybe i should open my textbook and start reading...

fredrick · « **Reply #7 on:** January 10, 2008, 04:22:14 pm »

Quote from: Obsolete Chaos on January 10, 2008, 03:27:26 pm

I AM SO LOST, after reading the first few lines in proof no.1

maybe i should open my textbook and start reading...

That makes two of us

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Author Topic: Vector Proofs (Read 2105 times) Tweet Share

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