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anonymous
 3 years ago
Draw a parallelogram whose adjacent sides are determined by the vectors [4,6] and [3,5]. Find the area of the parallelogram.
anonymous
 3 years ago
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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Isn't it dot product which finds the area?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You can use the origin (0,0) as the common start point for those two vectors; that would make it easiest to draw.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, 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 thisdw:1350760137445:dwbut very low area (dottiness=cos(angle between vectors))

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Cross product most likely, but forget about the result being a vector.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, that's right, cross product because it is a*b*sin(Θ).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0b*sin(Θ) is the height of the parallelogram. Duh, should have remembered the basic trig on that one.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is there any way to solve for the area without trig?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, doing the crossproduct has the trig builtin, so you wouldn't actually be using a trig function explicitly.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I suppose you could try to use pythagoras

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0^ what do you mean? By stacking up right triangles and using subtraction?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350764732234:dw find B perpendicular with A by finding the 'gradient' of A... you can do it, but it's really fiddly

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No there must be an elegant way of finding B perp A

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0as a function of Ax, Bx, Ay and By

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'd rather just do the crossproduct.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\left[\begin{matrix}i & j & k \\ 4 & 6 & 0 \\ 3 & 5 & 0\end{matrix}\right] = 4\times5  6\times3=2\]
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