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Copythat
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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.
 2 years ago
 2 years ago
Copythat Group Title
Draw a parallelogram whose adjacent sides are determined by the vectors [4,6] and [3,5]. Find the area of the parallelogram.
 2 years ago
 2 years ago

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CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
Isn't it dot product which finds the area?
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
You can use the origin (0,0) as the common start point for those two vectors; that would make it easiest to draw.
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
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 thisdw:1350760137445:dwbut very low area (dottiness=cos(angle between vectors))
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Cross product most likely, but forget about the result being a vector.
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
Oh, that's right, cross product because it is a*b*sin(Θ).
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
b*sin(Θ) is the height of the parallelogram. Duh, should have remembered the basic trig on that one.
 2 years ago

Copythat Group TitleBest ResponseYou've already chosen the best response.0
Is there any way to solve for the area without trig?
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
Not really...
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
Well, doing the crossproduct has the trig builtin, so you wouldn't actually be using a trig function explicitly.
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
I suppose you could try to use pythagoras
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
^ what do you mean? By stacking up right triangles and using subtraction?
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
dw:1350764732234:dw find B perpendicular with A by finding the 'gradient' of A... you can do it, but it's really fiddly
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
No there must be an elegant way of finding B perp A
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
as a function of Ax, Bx, Ay and By
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
I'd rather just do the crossproduct.
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
\[\left[\begin{matrix}i & j & k \\ 4 & 6 & 0 \\ 3 & 5 & 0\end{matrix}\right] = 4\times5  6\times3=2\]
 2 years ago
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