Calculus 3 (vector calculus): Let u=<5,-2,3> and v=<-2,1,4>. Find the value of c which will force the vector w=<2c,3,c-1> to lie in the plane of u and v.
The three vectors should lie in the same plane if they are dependent.
Yeah I got that part. So what I did was I did the cross product of u and v and that should get me the norm of the plane i got <-11,-26,1>. Then I crossed u and w and got <-2c-7, c+5, 15+4c>. Because if all of the vectors are in the same plane then the norm of u and w should be the same as u and v. So I did -11=-2c-7, -26=c+5, 1=15+4c. and then plug the values of c into w. Then Im trying to get the triple scalar because when I do that and it equals 0 I know all of the vectors are co-planar. But its not working out so :(
If they are dependent, then you should be able write w=au+bv
\[<2c,3,c-1>=a<5,-2,3>+b<-2,1,4>\] => 2c=5a-2b 3=-2a+b c-1=3a+4b
http://www.vitutor.com/geometry/vectors/linearly_dependent.html If you don't like that one, then there is another example here. Which I think you wanted to do cross product anyways.
Do you know how to find the determinant of a matrix?
I actually prefer that example over what I did.
but either way will work
Please let me know if you need anymore help. I have to clean the litter boxes real quick so I will answer when I get back if you do respond when I do get back.
Ok I'm leaving but bump if you have questions so other people can see.
OMG thank you so much for replying. And sorry for my absence I needed a break the frustation was too much.
Im just not understanding it. I know the mechanics of it but I just dont really see it if you know what I mean. And professor is no help >.<
Take the cross product of the two vectors that are in the plane. This will give you a vector normal to the plane! Take this new vector and take the dot product of this vector with the vector you want to fall in the plane. The dot product of these two vectors has to be zero, so now you can solve for your variables!
Wow I totally didn't read lol. Oh well, good luck.
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