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When I have a plane, I can calculate two equations:

Parametric, of the form $$P + s\cdot \vec{q} + t \cdot \vec{r}$$ For some point $P$ in the plane, two direction arrows $\vec{q},\vec{r}$ that are not parallel, and some $s,r\in \mathbb{R}$.

Normal, of the form

$$ax+by+cz = d$$ With only one point and one direction arrow perpendicular to the plane.


I am wondering, is it possible, given one form, to transform it into the other?

Here I have two planes:

$$A: x - y + 2z = 10\\ B: (1,1,1)+t(1,-1,1)+s(0,1,0)$$

$A$ is normal and $B$ is parametric. How can I transform $A$ to parametric and $B$ to normal?

I happen to know that a normal form of $B$ is

$$-x+z=0$$

From a previous exercise, but I'm not sure what was done to achieve this form.

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  • $\begingroup$ Let $(x,y,z)=(1,1,1)+t(1,-1,1)+s(0,1,0)$. Then, solve these simultaneous equations. Then, you will get $-x+z=0$ and whatever $y$ suffices the equations. $\endgroup$ May 25 '14 at 7:26
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Let $$ A=\{(x,y,z)\in\Bbb R^3:x-y+2z=10\} $$ Note that the vector $(1,-1,2)$ is normal to $A$ and that $(10,0,0)\in A$. Since $$ \DeclareMathOperator{Null}{Null}\Null\begin{bmatrix}1&-1&2\end{bmatrix}= \DeclareMathOperator{Span}{Span}\Span\{ \begin{bmatrix}-2&0&1 \end{bmatrix},\begin{bmatrix}1&1&0\end{bmatrix} \} $$ we may write $$ A=\{(10,0,0)+r\cdot(-2,0,1)+s\cdot(1,1,0)\in\Bbb R^3:r,s\in\Bbb R\} $$

Now, let $$ B=\{(1,1,1)+r(1,-1,1)+s(0,1,0)\in\Bbb R^3:r,s\in\Bbb R\} $$ Note that $$ (1,-1,1)\times(0,1,0)=(-1,0,1) $$ Can you use this equation to finish putting $B$ into normal form?

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  • $\begingroup$ I'm kind of unfamiliar with that Null[] = Span{} thing you did there, what is it? $\endgroup$ May 25 '14 at 7:56

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