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Mathematics 19 Online
OpenStudy (anonymous):

Draw a parallelogram whose adjacent sides are determined by the vectors [4,6] and [3,5]. Find the area of the parallelogram.

OpenStudy (anonymous):

Isn't it dot product which finds the area?

OpenStudy (anonymous):

You can use the origin (0,0) as the common start point for those two vectors; that would make it easiest to draw.

OpenStudy (anonymous):

No, I don't think it's dot product, as dot product is a measure of how much the vectors are in each other's direction, so it would have high dottiness if they were like this|dw:1350760137445:dw|but very low area (dottiness=cos(angle between vectors))

OpenStudy (anonymous):

Cross product most likely, but forget about the result being a vector.

OpenStudy (anonymous):

Oh, that's right, cross product because it is a*b*sin(Θ).

OpenStudy (anonymous):

b*sin(Θ) is the height of the parallelogram. Duh, should have remembered the basic trig on that one.

OpenStudy (anonymous):

Is there any way to solve for the area without trig?

OpenStudy (anonymous):

Not really...

OpenStudy (anonymous):

Well, doing the cross-product has the trig built-in, so you wouldn't actually be using a trig function explicitly.

OpenStudy (anonymous):

I suppose you could try to use pythagoras

OpenStudy (anonymous):

^ what do you mean? By stacking up right triangles and using subtraction?

OpenStudy (anonymous):

|dw:1350764732234:dw| find B perpendicular with A by finding the 'gradient' of A... you can do it, but it's really fiddly

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