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How do you determine the plane which contains the line \begin{align} x & = -1 + 3t \\ y & = 5 + 2t \\ z & = 2 + t \end{align} and passes through the point $P = (2,4,-1)$?

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Please tag as homework! –  John Engbers May 29 '12 at 20:34
    
Provided, of course, that it is homework. –  Gerry Myerson May 30 '12 at 0:13
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2 Answers

the line is \begin{equation} (-1,5,2) + t(3,2,1) \end{equation} Then the plane contains the segment $\overline{(-1,5,2)(2,4,-1)}$ $\quad((-3,1,3) = (-1,5,2) -((2,4,-1))$ and the plane $\Pi$ has normal $(3,2,1) \times (-3,1,3) = (5,-12,9)$. Hence the plane is $5x -12y + 9z + d = 0$. As $(2,4,-1) \in \Pi$, we have $d = 47$. Hence, the plane is $5x -12y +9z +47=0$.

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Does it make a difference if i get 3,-1,-3? –  Nick May 29 '12 at 21:57
    
Not at all. All you need is any nonzero scalar multiple of the normal. –  ncmathsadist May 29 '12 at 23:50
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Here's another way (it's always good to know more than one way to solve a problem).

Your plane goes through $(-1,5,2)$ and $(2,7,3)$ (obtained by taking $t=0$ and $t=1$, respectively) and also $(2,4,-1)$. If your plane is $$ax+by+cz+d=0$$ then each of these three points gives you an equation relating the four unknowns $a,b,c,d$. So you get three homogeneous linear equations in 4 unknowns. Presumably, you know how to solve such a system. You'll get a one-parameter family of solutions (because $akx+bky+ckz+dk=0$ is the same plane, for any non-zero value of $k$); just pick any member of this family.

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