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Find an equation of a plane which contains the points: $(0,0,3),(3,2,1)0$, and $(6,2,0)$

I know I need a vector in order to use the equation $d=ax_0$+b$y_0$+c$z_0$

Now, could I just select any two points as my terminal and initial, say, (0,0,3) and (3,2,1) subtract the two and create a "n" vector, and then just go from there? Or is there something I must do will all three points?

Overclock
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3 Answers3

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If the points are $P,Q,R$ you need to form vectors $\vec{v_1}=\overrightarrow{PQ}$ and $\vec{v_2}=\overrightarrow{PR}$ and then compute the cross product and thus the normal vector $$ \vec n=\vec{v_1}\times\vec{v_2} $$ The the equation you seek will emerge from expanding the dot product $$ \vec n\cdot \left( \begin{pmatrix} x\\y\\z \end{pmatrix}-\overrightarrow{OP}\right)=0 $$ where $\overrightarrow{OP}$ is the vector sharing the same coordinates as the point $P$, essentially pointing from the origin of the coordinate system to the point $P$.

String
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  • Would you mind telling me why I need to have two separate vectors then use the cross product? Why not the dot product? It makes sense that I need to use two because it covers all three points. – Overclock Feb 02 '15 at 21:19
  • The idea is that both vectors lie in the plane so they are parallel to the plane. The cross product is then perpendicular to the plane - hence we have a normal vector. Finally the dot product equation produces all vector starting from $P$ that are perpendicular to the normal vector. – String Feb 02 '15 at 21:38
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In vectorial form, the plane is:

$$(0,0,3)+t(3,2,-2)+s(6,2,-3)\;,\;\;t,s,\in\Bbb R$$

In rectangular form, you need to solve a system of linear equations:

$$\begin{align}&&&&&3C&+&D=0&I\\{}\\ &3A&+&2B&+&C&+&D=0&II\\{}\\ &6A&+&2B&&&+&D=0&III\end{align}$$

Do the coefficients matrix and reduce:

$$\begin{pmatrix}3&2&1&1\\ 6&2&0&1\\ 0&0&3&1\end{pmatrix}\rightarrow \begin{pmatrix}3&2&1&1\\ 0&\!\!-2&\!\!-2&\!\!-1\\ 0&0&3&1\end{pmatrix}$$

From here, we can choose

$$C=1\,,\,\,\,D=-3\;,\;\;2B=-2C-D=1\implies B=\frac12\;,\;\;3A=-2B-C-D=1\implies A=\frac13$$

and the plane is

$$\frac13x+\frac12y+z-3=0\iff 2x+3y+6z-18=0$$

Timbuc
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Alternatively, you can parameterize the plane by taking $r(u,v)=p_1+u(p_2-p_1)+v(p_3-p_1)$

and replace $p_1,p_2,p_3$ with your points. Notice $r(u,v)$ is in the form $(x(u,v),y(u,v),z(u,v))$. To see why this works, graph the points and draw vectors between $p_2$ and $p_1$ and $p_3$ and $p_1$. Where should the sum of these vectors be according the the vector addition rule ? Also what happens when $u=0$ or $v=0$ ? If you want to convert between this form and the explicit form, how would you do that ?

Vivek Kaushik
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