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Give the parametric equations of the line of intersection of the planes $$4x + 2y + 2z = -1$$ and $$3x + 6y + 3z = 7$$

Also, give the equation of the plane that passes through the point $(2,-1,4)$ and is perpendicular to the line found above.

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    It is usually recommended that you give your thoughts on the matter, so that your fellow users might be directed properly when answering your inquiry. It is also considered ungentle to use the imperative.2012-05-29
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    Alright so I tried it. I found the cross product of the 2 planes to find a vector, which is (6,-6,18). Im kind of stuck though...2012-05-30
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    Shouldn't you get solution as in $ax+by=c$?2012-05-30
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    How do I find a point on the plane? If i set x = 0, and isolate for y or z in 1 eq and plug it into the other eqn, does it work?2012-05-30
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    Not really. Note that the two equations you have impart a condition on what $x$ and $y$ should be. Multiply the first one through $3$ and subtract the other equation multiplied by $2$. The $z$ will cancel out.2012-05-30
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    But then I am still left with two variables, x and y.. if i solve for one, and plug it into the other eqn, it will have either x and z or y and z ...?2012-05-30
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    Let one of the variables be $t$, and then solve for the others in terms of $t$. Then you should be left with the equation of a line in the form $(p_x,p_y,p_z) + t (d_x,d_y,d_z)$.2012-05-30

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