Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer
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)

share|cite|improve this answer

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

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

share|cite|improve this answer
Very good step by step explanation, @miracle173 thank you. – meiryo Jun 10 '11 at 10:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.