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

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Isn't it dot product which finds the area?
You can use the origin (0,0) as the common start point for those two vectors; that would make it easiest to draw.
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))

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Other answers:

Cross product most likely, but forget about the result being a vector.
Oh, that's right, cross product because it is a*b*sin(Θ).
b*sin(Θ) is the height of the parallelogram. Duh, should have remembered the basic trig on that one.
Is there any way to solve for the area without trig?
Not really...
Well, doing the cross-product has the trig built-in, so you wouldn't actually be using a trig function explicitly.
I suppose you could try to use pythagoras
^ what do you mean? By stacking up right triangles and using subtraction?
|dw:1350764732234:dw| find B perpendicular with A by finding the 'gradient' of A... you can do it, but it's really fiddly
No- there must be an elegant way of finding B perp A
as a function of Ax, Bx, Ay and By
I'd rather just do the cross-product.
\[\left[\begin{matrix}i & j & k \\ 4 & 6 & 0 \\ 3 & 5 & 0\end{matrix}\right] = 4\times5 - 6\times3=2\]

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