Find a vector parallel to two planes

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To find a vector parallel to the two planes defined by the equations 3x+y+z=1 and 3x+z=0, the normal vectors are identified as <3,1,1> and <3,0,1>. The cross product of these normal vectors yields a vector that is perpendicular to them, thus parallel to the planes. The calculated vector <1,0,-3> is valid, and other multiples like <2,0,-6> are also acceptable solutions. Multiple correct answers exist for this problem, confirming that various scalar multiples can represent the same direction.
UrbanXrisis
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there are two planes:

3x+y+z=1
3x+z=0

find a vector U with positive first coordinate that is parallel to both planes.

the way I went about solving this:
the normal vectors are: <3,1,1> and <3,0,1>, the cross product of the vectors will give a vector perpendicular to the normal, which means it would be parallel to the two planes right?
 
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Yes, that's right.
 
when I did this, I get <1,0,-3> and this is not correct. so i do not know where I went wrong
 
(1, 0, -3) is correct. Maybe you are comparing it to a vector that is a multiple of (1, 0, -3)--there is more than one answer. (2, 0, -6) is also correct.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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