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I can convert something in the 2nd dimension fine, but I'm having difficulty with something like $4x - 3y + 6z = 12$. Any help?

EDIT: Solve using only algebra, no matrices yet.

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3 Answers 3

up vote 5 down vote accepted

Since there are three variables and one equation, you just denote the secondary variables as parameters, i.e. $y=s,z=t$ and then $x=\frac{12+3s-6t}{4}$. Then one parametric form is $(\frac{12+3s-6t}{4},s,t)$.

In the general case of a set of linear equations, it helps thinking of the equations that need parametrization as a system with more variables than equations. The key is to find how many secondary variables are there, and take them as parameters.

[edit:] You should be more precise in your formulation... The parametric vector form is very easy to obtain from the parametric vorm. Separate in three vectors separating $s,t$ and the constant term like this $(\frac{12+3s-6t}{4},s,t)=(3,0,0)+s(\frac{3}{4},1,0)+t(\frac{-6}{4},0,1)$.

No parametric form is unique. If you replace $s,t$ by any linear combination of other two parameters, you get another parametric form, with different coefficients.

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Sorry, I meant parametric vector form. Does this still apply? The answer says x = (3 0 0) + s(-3 0 2) + t (3 4 0) with proper formatting. How do I get here? –  meiryo Jun 6 '11 at 13:34

Desribe one of the variables as composition of other two: x = (12 + 3y - 6z) / 4, and that gives you the parametric form of ( (12 + 3y - 6z) / 4, y, z)

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I cannot add a comment, therefore I have to post an answer. refering to the answer of Beni Bogosel and the question of meiro: selecting $s=4v$ and $t=2u$ we get from the solution of Beni Bogosel $$ x = (3, 0, 0) + u(-3, 0, 2) + v(3, 4, 0) $$

there is a slightly different technic to get the parameter form

$$\vec{x} = \vec{p} + s*\vec{a} + t*\vec{b}$$

$\vec{p}$ is a point of the given plane that means that the coordinates of $\vec{p}$ must satisfy the equation. for example if set the y and z coordinate of $\vec{p}$ to $0$ you get

$$ 4x-3*0+6*0=12$$

and therefor $x=3$ and $\vec{p} = \left( \begin{array}{c} 3\\ 0\\ 0 \end{array} \right) $ . $ \vec{a} $ and $ \vec{b} $ are vectors parallel to the given plane. The coefficients

$$ \vec{n}=\left( \begin{array}{c} 4\\ -3\\ 6 \end{array} \right) $$ of the equation represent a normal vector of the plane. Therefore $\vec{a}$ and $\vec{b}$ must be normal to $\vec{n}$. We can find a vector normal to a threedimensional vector $\vec{n}$ by taking $\vec{n} $ setting one coordinate to $0$, interchanging the remaining coordinates and changing the sign of one of the interchanged coordinates (the inner product of this new vector with the original vector is now $0$, this means that they are normal). For example: $$ \left( \begin{array}{c} 4\\ -3\\ 6 \end{array} \right) $$ set one coordinate to $0$, e.g. the $y$ coordinate

$$ \left( \begin{array}{c} 4\\ 0\\ 6 \end{array} \right) $$ interchange the remaining coordinates

$$ \left( \begin{array}{c} 6\\ 0\\ 4\end{array} \right) $$

and change the the sign of one of the interchanged coordinates, e.g. the sign of $6$. We get

$$ \left( \begin{array}{c} -6\\ 0\\ 4\end{array} \right) $$

which can be written as

$$ 2\left( \begin{array}{c} -3\\ 0\\ 2\end{array} \right) $$

We dont worry about length and orientation of the vector normal to $\vec{n}$ therefor we can use

$$ \vec{a}=\left( \begin{array}{c} -3\\ 0\\ 2\end{array} \right) $$

Taking $\vec{n}$ and setting the $z$-coordinate to $0$ and doing the analogous procedure we get the vector

$$ \vec{b}=\left( \begin{array}{c} 3\\ 4\\ 0\end{array} \right) $$

using the above equation for the vector $\vec{x}$ we get

$$ \vec{x}= \left( \begin{array}{c} 3\\ 0\\ 0\end{array} \right) + s\left( \begin{array}{c} -3\\ 0\\ 2\end{array} \right) + t\left( \begin{array}{c} 3\\ 4\\ 0\end{array} \right) $$

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Very good step by step explanation, @miracle173 thank you. –  meiryo Jun 10 '11 at 10:11

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