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I was solving the following problem:

Find the vector equation of the line representing the intersection of the planes

$2x + y - z = 10~$ and $~3x + 4y +2z = 30$

and although I eventually realized how to do it by setting up an augmented matrix, before I did, I tried two different methods and got two different answers and was just wondering what the equations I got using the incorrect method represent?

Method 1:

I made both equations $= 0$ by subtracting the constant and set them equal to each other. I simplified to get:

$x + 3y + 3z = 20$

and then using two parameters I converted it into a vector equation:

$ \left(\begin{array}{cc} x\\ y\\ z\\ \end{array}\right)= $ $ \left(\begin{array}{cc} 20\\ 0\\ 0\\ \end{array}\right) $ $+~\lambda \left(\begin{array}{cc} -3\\ 1\\ 0\\ \end{array}\right) $ $+~\mu \left(\begin{array}{cc} -3\\ 0\\ 1\\ \end{array}\right) $

So, I got an equation of a plane instead of a line. What is this plane? does it relate to anything or is it just a random plane because I did it incorrectly?

Method 2:

I solved for $y$ in each equation and then set them equal and converted to vector form to get:

$ \left(\begin{array}{cc} x\\ z\\ \end{array}\right) = $ $ \left(\begin{array}{cc} 2\\ 0\\ \end{array}\right) + \lambda~$ $ \left(\begin{array}{cc} 6/5\\ 1\\ \end{array}\right) $

Same questions as above? Although it isn't the provided answer (which I get by solving an augmented matrix of coefficients), does it have any relevance to anything or is it just a garbage answer?

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up vote 2 down vote accepted

Method 1. The reason you got a plane instead of a line is that when you took the two equations and set them equal to each other, you added a whole bunch of solutions.

Why? Because while it is certainly true that any values of $x$, $y$, and $z$ that make both $2x+y-z-10 = 0$ and $3x+4y+2z-30=0$ correct will also make $2x+y-z-10=3x+3y+2z-30$ correct, there are many points that will make the latter true without making the original two equations true. For example, if if $2x+y-z-10=3$ and $3x+4y+2z-30=3$, then the point is not in both original planes, but the point will satisfy the new equation $2x+y-z-10=3x+3y+2z-30$.

For a specific example, the point $(2,3,3)$ does not satisfy $2x+y-z-10=0$, nor does it satisfy $3x+4y+2z-30=0$, but it does satisfy $2x+y-z-10 = 3x+4y+2z-30$.

The problem is that while it is true that two things that are both equal to zero must be equal to each other, it is not true that two things that are equal to each other must both be equal to $0$.

Method 2: Same problem; you are including all your original answers, but also any other answers that happen to yield the same value for $y$, even if that value does not satisfy the original equations. While among your answers are the correct answers, it is almost impossible to pick them out from among the extraneous solutions.

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Regarding method 1: the plane you calculated is a "bisector" of the two give planes. I put the word "bisector" in quotes because the bisector plane doesn't necessarily make equal angles with the two given planes. But it does pass through their line of intersection, so in some sense it is "in between" them. Given two plane equations $P(x,y,z)=0$ and $Q(x,y,z)=0$, it's clear that any plane of the form $hP + kQ=0$ passes through their line of intersection (because at any point where $P=0$ and $Q=0$, then certainly $hP + kQ$ will be zero, also). This family of planes is called a "pencil" of planes, sometimes. You calculated the family member that has $h = 1$ and $k= -1$. As the previous answer explained, this plane contains the line you're looking for, but it contains lots of other points, too.

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