Vectors Question (Vector Proof)

[Andiio]

Having trouble with question (d).

Ended up with:

YX.YE = YF.YX
As cos(theta) = a.b/|a||b|, thus a.b. = |a||b|cos(theta)

Thus, |YX||YE|cos(XYE) = |YF||YX|cos(FYX)
Since |YX|=|YE|=|YF|,
cos(XYE) = cos(FYX)

Thus angle(XYE) = angle(FYX) = 90º     <------ obtained this answer from the worked solutions, but am not sure exactly why. I understand it, but just am unsure.

Is it because XE.XF = 0 (from part a in the question)?

Thanks!

[kamil9876]

What is Y?

[Andiio]

What is Y?

Oops sorry I didn't specify that, I let Y = the midpoint of AB; I let YX || BC || AD, and since BC = BE, YX = YE (similar triangles)

I kinda looked at the worked solutions for a bit of help though, how would you approach the question kamil?

Thanks!

[kamil9876]

How did you get ? I don't think that is correct.

cos(XYE) = cos(FYX)


Thus angle(XYE) = angle(FYX) = 90º     <------ obtained this answer from the worked solutions, but am not sure exactly why. I understand it, but just am unsure.

It is because and so if you add this equation to your one you get:

which implies that

But yeah your first line I am not convinced of.

========================================================================================

My approach:

let

Now we see that:




However using the notation from part b:





However notice that





Howver since EF and v are parralel and u is not parralel to neither of them, it must mean that:





Now that we've got that:







(since )

so:







And so we are done.

[Andiio]

Thanks kamil!

Just wondering, could anyone please help me with another (relatively simpler) question?

for (b), I ended up with OP.BC = 1/2(c.c - b.b), which would = 0 for OP _|_ BC, but geometrically I can see that |c| = |b|, but how do you prove it mathematically? :\

Not sure what to do for (c), is it just asking to prove that the point of intersection between OR and OQ is O?

Haven't attempted (d) yet! :]

It would be greatly appreciated if you anyone run me through them xD

Thanks heaps!

Mod edit: Please address your questions to the whole forum rather than a specific person, we have others more than capable enough to help and we don't want to exclude them from a discussion

[kamil9876]

for b it seems as though the author of the problem wants you to prove it without using |b|=|c|, seeing as that you are supposed to do that for the last part. However we can prove |a|=|b|=|c| easily actually using pythagoras' theorem(and if you want to strictly stick to the "vector proofs and no other geometry" mentality, well you can provide a little vector proof of it! go!):



Using an analogous argument you can show:



and so . And look I involuntarily did the last part for you too!

for c we want to show that if we draw a perpendicular bisector on the third line, (coming from P), then it will NOT miss O, and actually intersect it.