Vectors - Proving Parallelograms

[ninbam1k]

Hi guys, so I was just wondering about all the ways of being able to prove that a quadrilateral is a parallelogram using vectors.

Help please?

[MBBS]

I guess two directed line segments would be parallel vertically. The other two would be parallel horizontally?

Work out the gradient of each of the directed line segments?

This seems far too easy an answer for it to be right.

[luffy]

I only did Spesh Units 1/2 this year. But in order to prove two a shape is a parallelogram, we were told to do the following:
1. Find the magnitude of all 4 vectors. (Obviously, if they all have the same magnitude, then it is in fact, a rhombus.)
2. For each corresponding pair with the same magnitude, prove the two vectors are parallel. I'm assuming you know how to do that.

[pi]

I only did Spesh Units 1/2 this year. But in order to prove two a shape is a parallelogram, we were told to do the following:
1. Find the magnitude of all 4 vectors. (Obviously, if they all have the same magnitude, then it is in fact, a rhombus.)
2. For each corresponding pair with the same magnitude, prove the two vectors are parallel. I'm assuming you know how to do that.

+1,

And with point 2, the proof would be that they two vectors would be colinnear (ie. two vectors, one of which is a non-zero scalar multiple of the other) and the other two would also be collinear.

If you want to get a bit more technical though: http://mathworld.wolfram.com/ParallelogramLaw.html

[taiga]

One set of the vectors are of same magnitude AND parallel.

If you can prove that, logic dictates that the other set of vectors must also be of same magnitude and parallel

If you think about it, if you were to draw 2 parallel lines of same magnitude, to make it into a 4 sided shape, if you were to fill in the other 2 lines, it's impossible to get anything else but a parallelogram.

If the magnitudes are all the same, it's a rhombus. If the diagonals intersect perpendicular ontop of that, it's a square.