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I'll mark the vectors in bold.

$p_1 = \bf{i} - 2 \bf{j} + \bf{k}$

$p_2 = 2 \bf{i} + \bf{j} - \bf{k}$

$p_3 = \bf{i} + \bf{j} + \bf{k}$

Could someone please explain to me the way of finding the equation of a plane in $\bf{r}\cdot n = \bf{d}$ form for a plane containing $p_1, p_2, p_3$?

Thanks

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1 Answer 1

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Presumably, what you are looking for is an equation in the form $$ax + by + cz = d$$ with $a,b,c,d\in\mathbb{R}$, which can be expressed in the form $$(x,y,z)\cdot\mathbf{n} = d$$ with $\mathbf{n}=(a,b,c)$ the normal vector of the plane.

To find the normal vector of the plane, construct two vectors that are in the plane, are not collinear, and take their cross product.

Since you have three points that are not collinear that are in the plane, note that $\mathbf{p1}-\mathbf{p2}$, $\mathbf{p1}-\mathbf{p3}$, and $\mathbf{p2}-\mathbf{p3}$ are all vectors that (after suitable translation) lie in the plane. Once you have the normal vector, finding $d$ is a matter of figuring out what value is required by plugging in at least one point in the plane.


Example. Suppose you want to find the plane that contains $\mathbf{p}_1=(1,1,1) = \mathbf{i}+\mathbf{j}+\mathbf{k}$, $\mathbf{p}_2=(-3,-2,0) = -3\mathbf{i}-2\mathbf{j}$, and $\mathbf{p}_3 = (0,3,1) = 3\mathbf{j}+\mathbf{k}$.

The plane contains vectors parallel to $$\mathbf{p}_1-\mathbf{p}_2 = (1,1,1) - (-3,-2,0) = (4,3,1)$$ and to $$\mathbf{p}_1 - \mathbf{p}_3= (1,1,1)-(0,3,1) = (1,-2,0).$$ So then the vector $(4,3,1)\times(1,-2,0)$ is perpendicular to plane: $$(4,3,1)\times(1,-2,0) = \left|\begin{array}{rrr} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 4 & 3 & 1\\ 1 & -2 & 0 \end{array}\right| = \Bigl( 0+2, -(0-1), -8-3\Bigr) = (2,1,-11).$$ Therefore, the plane that contains the three points will have equation of the form $$2x + y - 11z = d$$ for some $d$. Plugging in $(1,1,1)$, which we know is in the plane, we get $$d = 2(1) + 1 - 11(1) = -8$$ so the plane in question is the plane with equation $$2x + y - 11z = -8.$$ Indeed, you can verify that $(1,1,1)$, $(-3,-2,0)$, and $(0,3,1)$ all satisfy the equation. To write in "$\mathbf{r}\cdot \mathbf{n}=\mathbf{d}$ form", we simply write $$(x,y,z)\cdot(2,1,-11) = -8.$$

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@Arturo In order to find, say, p1 - p2, do I cross-product p1 and p2 or? –  Sorin Cioban May 14 '11 at 20:51
1  
@Sorin: To find $\mathbf{p1}-\mathbf{p2}$, you take $\mathbf{p1}$, and you subtract $\mathbf{p2}$. What did you think the - stood for? –  Arturo Magidin May 14 '11 at 20:53
    
@Sorin Cioban: p1 - p2 can be found by component-by-component subtraction, so is -3i-3j+2k. –  Henry May 14 '11 at 20:54
    
Great! Thanks! I got n = 6i + 3k. Thus, d = 6x + 3z. –  Sorin Cioban May 14 '11 at 21:09
    
@Sorin: No, that cannot be correct. $d$ must be a constant, not an expression involving $x$s, $y$s, or $z$s. It has to be a number. –  Arturo Magidin May 14 '11 at 21:10

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