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Calculus 3 problem: lines and planes in space

  1. Sep 15, 2014 #1
    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. I did the cross product of u and v, then i crossed u and w, then I equal the product of u and v with what I got for w. But for some reason when I try doing the triple scalar of u,v, and w; it does not give me zero which would prove that w is in the plane of u and v.
     
  2. jcsd
  3. Sep 16, 2014 #2

    Fredrik

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    It will be much easier for us to help you if you post the calculation that got the wrong result. Here's a dot product symbol and a cross product symbol that you can copy and paste: · ×
     
  4. Sep 16, 2014 #3
    I think you may be doing the problem incorrectly. The scalar triple product formula is a • (b x c). So I do not believe you need to cross u and v and u and w and equate them. I'm not 100% certain though (I'm currently taking calc 3 myself) so perhaps someone can confirm or deny my suspicion.
     
  5. Sep 16, 2014 #4

    LCKurtz

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    You could just write the equation$$
    \vec w = a\vec u + b\vec v$$out and set the components equal. 3 equations in 3 unknowns.

    Alternatively, and probably easier, just dot ##\vec w## with ##\vec u \times \vec v##, which is normal to the plane, and and set it equal to zero. Then you can just solve for ##c##.
     
    Last edited: Sep 16, 2014
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