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June 21st, 2014, 04:00 PM   #1
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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
(x-3)/3 = y/-3 = (z-2)/9
Line B, as far as I know is:

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:

-x+y+3z=0
3x-3y+9z=0
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.
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June 21st, 2014, 05:34 PM   #2
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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)

$\displaystyle \ell(0,2,7)+\mu(3,0,2)+(4,-1,5)=(3\mu+4,2\ell-1,7\ell+2\mu+5)$

$\displaystyle\left\{\begin{matrix}x=3\mu+4\\y=2 \ell-1\\z=7 \ell+2\mu+5\end{matrix}\right.$

so plugging in the two first equation to the third one you get

$\displaystyle z=7\cdot \frac{y+1}{2}+2\cdot\frac{x-4}{3}+5$, do the computations because I want to go to sleep.........
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June 22nd, 2014, 10:11 AM   #3
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Thank you ZardoZ!
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June 22nd, 2014, 12:16 PM   #4
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Quote:
When I make the dot product for finding a vector perpendicular to both lines, I get this:

-x+y+3z=0
3x-3y+9z=0
Didn't you mean cross product?
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June 23rd, 2014, 05:09 AM   #5
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Probably "cross product" is what he meant!
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