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i believe this has something to do with crossproducts.
for the part below use:
u= 8i +2j and v = 2i+j-k
find the area of the parallelogram formed by u and v
ok so that was the problem
i started it by trying to find the P, Q, and R.
i got P(8,2,0) Q(2,1,0) and R(0,-1,0)
are the points correct for P Q and R
i m kinda stuck here.
will give the area ...
Code:| i j k |
| 8 2 0 | = 10
| 2 1 -1 |
Quote:
Originally Posted by skeeter
Code:| i j k |
| 8 2 0 | = 10
| 2 1 -1 |
not sure how 10 came out of that.. i see where you got the numbers and how u put them in that chart thing. but i dont see how 10 comes outta that hm.
10 is the determinant of the matrix.
Quote:
Originally Posted by skeeter
well ty but i still dont get it. i dont want to frustrate you.
The area of a parallelogram is the magnitude of the cross product. So do the cross product, then take the magnitude of that vector.
the magnitude of cross product of the two vectors gives the area of the parallelogram.Quote:
Originally Posted by Legendsn3verdie
when working in rectangular coordinate systems, the cross product of vectors a and b given by
http://emweb.unl.edu/math/mathweb/vectors/Image550.gif
can be evaluated using the rule
http://emweb.unl.edu/math/mathweb/vectors/Image551.gif
Quote:
Originally Posted by skeeter
ah ha! you the man an a life savor, let got land you a hot broad!