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I'm preparing for an upcoming test and wanted to validate my solution to the following problem:

Find a vector equation of the plane determined by $3x+3y+z=1$

Here's what I did


$I$. Let $x=y=1$. Then from $3x+3y+z=1$ it follows that $\vec{r_0}=[1, 1, -5]$ is the position vector of a point in the plane.

$II$. From the cartesian equation it follows that $\vec{n}=[3, 3, 1]$ is a vector normal to the plane. Then for any vectors $\vec{v}, \vec{w}$ such that $\vec{v} \neq t \vec{w}$ (linearly independent), if $\vec{v}$ and $\vec{w}$ are normal to $\vec{n}$ they are direction vectors of two lines (or vectors) in the plane. Notice that

$$\begin{align} \vec{n} \cdot \vec{a}=0 &\iff 3a_1+3a_2+a_3=0\end{align} $$

If $a_1=a_2$, then $-6a_1=a_3$ is a solution. If $a_1=a_3$ then $-\frac{4}{3}a_1=a_2$ is a solution. We can then define

$$\vec{v}:=[1, 1, -6], \vec{w}:=[1, -\frac{4}{3}, 1]$$

knowin both $\vec{v}$ and $\vec{w}$ are normal to $\vec{n}$ (this is easy to verify algebraically) and therefore direction vectors for two lines (or vectors) in the plane.

$III$. This is enough to give

$$\begin{align} \vec{g}(t, s)&=\vec{r}_0 + t\vec{v}+s\vec{w} \\\vec{g}(t, s)&=[1, 1, -5] + t[1, 1, -6]+s[1, -\frac{4}{3}, 1]\end{align} $$

as a vector equation of the plane.


Is this solution correct? If not, why? If so, were there simpler approaches to the problem? Thanks in advance.

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1 Answer 1

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Too long for a comment, so….

Yes, your answer is correct.

An alternative way is to find three non-collinear points in the plane, $A, B$ and $C$, and then you can construct the vector equation of the plane as $$\underline{r}=\overrightarrow{OA}+\lambda\overrightarrow{AB}+\mu\overrightarrow{AC}$$

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  • $\begingroup$ Much appreciated! $\endgroup$
    – lafinur
    Commented Nov 18, 2022 at 17:52
  • $\begingroup$ You are welcome $\endgroup$ Commented Nov 18, 2022 at 18:15

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