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

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CliffSedgeBest ResponseYou've already chosen the best response.1
Isn't it dot product which finds the area?
 one year ago

CliffSedgeBest 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.
 one year ago

henpenBest 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))
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Cross product most likely, but forget about the result being a vector.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
Oh, that's right, cross product because it is a*b*sin(Θ).
 one year ago

CliffSedgeBest 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.
 one year ago

CopythatBest ResponseYou've already chosen the best response.0
Is there any way to solve for the area without trig?
 one year ago

CliffSedgeBest 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.
 one year ago

henpenBest ResponseYou've already chosen the best response.1
I suppose you could try to use pythagoras
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
^ what do you mean? By stacking up right triangles and using subtraction?
 one year ago

henpenBest 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
 one year ago

henpenBest ResponseYou've already chosen the best response.1
No there must be an elegant way of finding B perp A
 one year ago

henpenBest ResponseYou've already chosen the best response.1
as a function of Ax, Bx, Ay and By
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
I'd rather just do the crossproduct.
 one year ago

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