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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.
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))
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?
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.
\[\left[\begin{matrix}i & j & k \\ 4 & 6 & 0 \\ 3 & 5 & 0\end{matrix}\right] = 4\times5 - 6\times3=2\]