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Consider the straight line l in $\mathbb R^3$, defined by the system of two linear equations:$$x-y=1,y-z=2$$ Write an equation of the plane in $\mathbb R^3$ passing through the line l and the point $(1,2,3)$

My thinking is find two other points on the plane ($z=0$ and $x=0$)
Find the difference between the two sets of vectors
Finaly find the normal to the plane

Is this correct logic?

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There is a more straightforward way. If plane passing through the line $l$, then its equation is $$ \alpha(x-y-1) + \beta(y-z-2)=0. $$ Point $(1, 2, 3)$ lies on the plane; hence, $$ \alpha(1-2-1) + \beta(2-3-2) = 0\Longrightarrow 2\alpha + 3\beta = 0\Longrightarrow \alpha =3\gamma, \beta=-2\gamma $$ for a arbitrary $\gamma$. Substitute it and $$ 3(x-y-1) - 2(y-z-2)=0 $$ is your plane equation.

We can use your idea. We have two points on line (for example, $(3, 2, 0)$ and $(0, -1, -3)$) and point $(1, 2, 3)$; and it's simple to write plane equation now (but longer then the method above required).

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  • $\begingroup$ Yes, this make more sense; and reduced working. Thanks for the help. $\endgroup$ – Daniel Jul 14 '15 at 13:50
  • $\begingroup$ @Jack, i edited post $\endgroup$ – Michael Galuza Jul 14 '15 at 13:51
  • $\begingroup$ This looks like a great method, but the justification for your first equation is not clear. Could you explain that or point to a reference? $\endgroup$ – Rory Daulton Jul 14 '15 at 16:52
  • $\begingroup$ @RoryDaulton, with pleasure, but my English is not so rich for it:) I can point to it $\endgroup$ – Michael Galuza Jul 14 '15 at 17:06
  • $\begingroup$ Pointing to it would be fine: that's what I meant by "point to a reference." Thanks and +1! $\endgroup$ – Rory Daulton Jul 14 '15 at 17:14
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But your idea will work. Taking z= 0, y= 2. Then x- 2= 1 so x= 3. One point on the given line, so on the plane, is (3, 2, 0). Taking x= 0, y= -1 so -1- z= 2 and z= -3. Another point on the line, so on the plane, is (0, -1, -3).

So you want the plane defined by the three points (3, 2, 0), (0, -1, -3), and (1, 2, 3). The vector from (0, -1, -3) to (3, 2, 0), <3, 3, 3>, and the vector from (0, -1, -3) to (1, 2, 3), <1, 3, 6>, both lie in the plane so their cross-product is a normal vector to the plane.

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  • $\begingroup$ Yeah, and then once the normal vector is found you can substitue one of the original vectors back in to calculate d (ax+by+cz=d) if I remember correctly $\endgroup$ – Daniel Jul 14 '15 at 14:06
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Here is another method utilizing linear algebra and parametric equations:

The parametric form of a line is $\mathbf x=\mathbf p+t\mathbf u$. From the equations $x-y=1$ and $y-z=2$, we can form the following equation:

$$\begin{bmatrix}1 & -1 & 0 \\0 & 1 & -1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}=\begin{bmatrix}1 \\ 2\end{bmatrix}$$

After you solve the augmented matrix with row reduction you get the parametric equation:

$$\mathbf x=\begin{bmatrix}3+t \\ 2+t\\t\end{bmatrix}=\begin{bmatrix}3\\ 2\\ 0\end{bmatrix}+t\begin{bmatrix}1\\ 1\\ 1\end{bmatrix}$$

Here, $\mathbf p$ corresponds to a point $P(3,2,0)$. You can think this point as the new origin of the plane we want to produce. Now, we need another direction vector to define the surface. Since we want this surface to contain the point $Q(1,2,3)$, we may use the vector

$$PQ=\mathbf q-\mathbf p=\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}-\begin{bmatrix}3\\ 2\\ 0\end{bmatrix}=\begin{bmatrix}-2\\ 0\\ 3\end{bmatrix}$$

as the other direction vector $\mathbf v$ of the surface. Because $\mathbf v$ and $\mathbf u$ are linearly independent, they can be used to define a surface. Thus, parametric equation of the surface becomes:

$$\mathbf x=\mathbf p+t\mathbf u+s\mathbf v=\begin{bmatrix}3+t-2s\\ 2+t\\ t+3s\end{bmatrix}$$

Transforming the parametric equation into the general equation yields:

$$3x-5y+2z=-1$$

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