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

1. CliffSedge

Isn't it dot product which finds the area?

2. CliffSedge

You can use the origin (0,0) as the common start point for those two vectors; that would make it easiest to draw.

3. henpen

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))

4. henpen

Cross product most likely, but forget about the result being a vector.

5. CliffSedge

Oh, that's right, cross product because it is a*b*sin(Θ).

6. CliffSedge

b*sin(Θ) is the height of the parallelogram. Duh, should have remembered the basic trig on that one.

7. Copythat

Is there any way to solve for the area without trig?

8. CliffSedge

Not really...

9. CliffSedge

Well, doing the cross-product has the trig built-in, so you wouldn't actually be using a trig function explicitly.

10. henpen

I suppose you could try to use pythagoras

11. CliffSedge

^ what do you mean? By stacking up right triangles and using subtraction?

12. henpen

|dw:1350764732234:dw| find B perpendicular with A by finding the 'gradient' of A... you can do it, but it's really fiddly

13. henpen

No- there must be an elegant way of finding B perp A

14. henpen

as a function of Ax, Bx, Ay and By

15. CliffSedge

I'd rather just do the cross-product.

16. CliffSedge

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

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