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Find the parametric equations of the plane $$3x+4y-2z = 4$$

I found two points which are $(0, 0, -2)$ and $(0, 1, 0)$ then I subtracted from each other

$(0, 1, 0) - (0, 0, -2)$ would be $(0, 1, 2)$

I used the vector equation $(x, y, z) = (0, 0, -2) + (0, 1, 2)$ and then got my parametric equation by just reading it off the vector.

However, in the answer key it shows that they found three points did the same thing as I did above with the points, which uses two scalar multiples instead of one

$$(x,y,z) = x0 + t1v1 + t2v2$$

where $v1$ and $v2$ are vectors

What I'm wondering is my answer still correct? Do I need a two multiples like the answer key?

Yash Jain
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Pwer331
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    https://math.stackexchange.com/questions/152467/parametric-form-of-a-plane An excellent description of a very similar problem – JuliusL33t Feb 02 '18 at 22:26

4 Answers4

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HINT

  1. Find 3 point not aligned in the plane $\vec {OP}, \vec {OQ}, \vec {OR}$.
  2. Compute the (linearly independent) vectors $\vec {PQ}$ and $\vec {PR}$.
  3. Write down the plane equation $\vec {OP}+t \vec {PQ}+s \vec {PR}$
user
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  • Oh every plane that is parallel requires two parameters, and every line parallel to a point requires one parameter? – Pwer331 Feb 02 '18 at 22:16
  • @JakeGinobi Each point in a plane can be expressed by a linear combination of two linearly independent vectors in the plane, if the plane is not through the origin we need also to fix a point of reference. – user Feb 02 '18 at 22:20
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Without going into the details of the arithmetic for your particular plane,the plane you are looking for is a $2$ dimensional object so the parametric equations for that plane MUST have TWO free parameters.

Furthermore, the "answer" $(x,y,z)=x0+t1v1+t2v2$, must be WRONG as it is the set of all linear combinations of the vectors v1 and v2 and therefore is a subspace of $R^3$ and therefore a plane passing through the origin, while the origin $(0, 0, 0)$ is definitely NOT on the set of all points $(x,y,z)$ satisfying $3x + 4y -2z = 4$.

Finally, since you never state what parametric equations you got from your method it is not possible to tell if they are correct or not, but if you have only one free parameter (I think that is what you mean by "needing two scalar multiples as in the answer key"), then it cannot be correct by the first paragraph above.

Yash Jain
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Arunas Rudvalis
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Since you haven’t shown your final result, I’m going to guess that you either ended up with a single-parameter equation of a line, or a two-parameter equation of the wrong plane. A plane is a two-dimensional object, so you need two linearly-independent direction vectors to describe it, but you’ve only found one so far.

You don’t really have to find more than one point on the plane to solve this, though. You can pretty much read a point on the plane and two independent direction vectors for it directly from the plane’s implicit Cartesian equation. Proceed as you might in solving any other system of linear equations: form the augmented coefficient matrix and bring it into row-reduced echelon form. In this case, there’s only one equation, so this matrix will only have one row: $$\left[\begin{array}{ccc|c}3&4&-2&4\end{array}\right].$$ This matrix is already in echelon form, so one can read a solution from it with no further manipulation, but I’ll go through the rest of the process for clarity. Divide the first row by $3$ to set the pivot to $1$, resulting in $$\left[\begin{array}{ccc|c} 1 & \frac43 & -\frac23 & \frac43\end{array}\right].$$ From this you should be able to find that a general solution to the equation is $$\begin{bmatrix}\frac43\\0\\0\end{bmatrix}+s\begin{bmatrix}\frac43\\-1\\0\end{bmatrix}+t\begin{bmatrix}-\frac23\\0\\-1\end{bmatrix},$$ which is also a parametric equation of this plane. (Add a couple of rows of zeros to the rref if you’re having trouble seeing how to get these vectors from it.)

amd
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Here is a quick solution based on some simple algebraic facts (very practical for exams):

Find point $P$ in the plane: $$(0,1,0)$$

Find two linearly independent vectors orthogonal to the normal vector of the plane (the coefficients in the equation): $$(3,4,-2) \rightarrow \vec{v1} = (-4,3,0), \vec{v2} = (2,0,3)$$

Done: $(0,1,0) + t_1(-4,3,0) + t_2(2,0,3)$

trancelocation
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