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June 21st, 2014, 04:00 PM  #1 
Member Joined: Nov 2013 Posts: 34 Thanks: 1  I can't find a vector perpendicular to two vectors
I have two lines: LINE A x=t y=2+t z=7+3t LINE B Line B, as far as I know is:(x3)/3 = y/3 = (z2)/9 x=3t+3 y=3t z=9t+2 I need to find the equation of the plane which contains LINE A and LINE B. First of all, I should find a vector perpendicular to both lines, which will let me make a equation of the plane I'm looking for. So, Direction vector for LINE A is <1,1,3> Direction vector for LINE B is <3,3,9> When I make the dot product for finding a vector perpendicular to both lines, I get this: What I normally do is to solve the system and then find the equation of the plane with a point. But this system, can't be solved. What do you recommend me to do? Thank you very much. 
June 21st, 2014, 05:34 PM  #2 
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus 
Line A for t=0 you get (0,2,7) Line B for t=0 you get (3,0,2) for t=1/3 you get (4,1,5) $\displaystyle \ell(0,2,7)+\mu(3,0,2)+(4,1,5)=(3\mu+4,2\ell1,7\ell+2\mu+5)$ $\displaystyle\left\{\begin{matrix}x=3\mu+4\\y=2 \ell1\\z=7 \ell+2\mu+5\end{matrix}\right.$ so plugging in the two first equation to the third one you get $\displaystyle z=7\cdot \frac{y+1}{2}+2\cdot\frac{x4}{3}+5$, do the computations because I want to go to sleep......... 
June 22nd, 2014, 10:11 AM  #3 
Member Joined: Nov 2013 Posts: 34 Thanks: 1 
Thank you ZardoZ!

June 22nd, 2014, 12:16 PM  #4  
Global Moderator Joined: May 2007 Posts: 6,710 Thanks: 675  Quote:
 
June 23rd, 2014, 05:09 AM  #5 
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus 
Probably "cross product" is what he meant! 

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find, perpendicular, vector, vectors 
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