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a) Find a parametrization for the plane $$E_1 = \\{ (x_1, x_2, x_3) \in \mathbb{R}^3 | 3x_1 - 2x_2 + x_3 = -1 \\} $$

b) Give a describing linear equation for the parametric representation

$$E_2 = (1, 2, 3) + \mathbb{R}\cdot (4,5,6) + \mathbb{R}\cdot (7,8,9)$$


a) We could set $x_2 = s$ and $x_3 = t$

This gives $3x_1 -2s + t = -1$. Solving for $x_1 = \frac{-1 + 2s - t}{3}$. So this gives

$$x(s,t) = (\frac{-1 + 2s - t}{3}, s, t)$$


b) The equation of a plane can be expressed as

$$n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0$$

where $n = (n_1, n_2, n_3)$ is a vector normal (perpendicular) to the plane, and $(x_0, y_0, z_0)$ is a point on the plane.

We are given the point $(1,2,3)$ that lies on the plane, so

$$(x_0, y_0, z_0) = (1,2,3)$$

To find a vector perpendicular to the plane, we can take the cross product of the vectors $(4,5,6)$ and $(7,8,9)$.

https://www.emathhelp.net/en/calculators/linear-algebra/cross-product-calculator/?u=4%2C5%2C6&v=7%2C8%2C9

This gives $(-3,6,-3)$. We can simplify to $(-1, 2, -1)$

So we can write our equation as

$$-1 \cdot (x - 1) + 2 \cdot (y - 2) + -1 \cdot (z - 3) = 0$$

This simplifies to

$$-x + 1 + 2y - 4 - z + 3 = 0$$

$$-x + 2y - z = 0$$


Is this correct ? We have no solutions to this exercise. Thank you for your feedback !

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  • $\begingroup$ Feel free to comment if anything is unclear $\endgroup$
    – wengen
    Commented Sep 30, 2023 at 12:32
  • $\begingroup$ What makes you think you made a mistake ??? $\endgroup$
    – Digitallis
    Commented Oct 2, 2023 at 16:09
  • $\begingroup$ @Digitallis I don't have a solution, so I don't know. If possible, could you post an answer, even just stating that everything looks correct ? So the question doesn't appear in the unanswered section anymore. I will accept your answer and upvote it $\endgroup$
    – wengen
    Commented Oct 2, 2023 at 17:01
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    $\begingroup$ Il ne s'agit pas de ne pas utiliser le tag. Il s'agit plutôt d'éviter de poser des questions de type 'Ma preuve est-elle correcte ?" sans préciser ce qui te fait hésiter sur si elle est correcte ou non. $\endgroup$
    – Digitallis
    Commented Oct 2, 2023 at 17:09
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Digitallis
    Commented Oct 2, 2023 at 17:14

1 Answer 1

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By definition$$ E_1 = \{(x_1,x_2,x_3) : 3x_1 -2 x_2 + x_3 = -1 \}$$ This means that

$$(x_1,x_2,x_3) \in E_1 \iff 3x_1 -2 x_2 + x_3 = -1 \iff x_1 = -1/3 + 2/3x_2 - 1/3x_3 $$ That is,

$\begin{align*} (x_1,x_2,x_3) \in E_1 \iff (x_1,x_2,x_3) &= (-1/3 + 2/3x_2 - 1/3x_3,x_2,x_3) \\ &= (-1/3,0,0) + x_2(2/3,1,0) + x_3(-1/3,0,1) \end{align*}$

So the parametric equation is

$$ E_1 = (-1/3,0,0) + \mathbb R \cdot (2/3,1,0) + \mathbb R (-1/3,0,1)$$ or alternatively

$$ \begin{cases}x(s,t) &= -1/3 &+ (2/3) s &-(1/3) t \\ y(s,t) &= 0 &+ s &+ 0 \\ z(s,t) &= 0 &+ 0 &+ t \end{cases} $$ In your attempted solution for $a)$ you only described the $x$ variable in terms of parameters $s,t$. You should of course include parametric descriptions for the other coordinates as well.

For $b$) the reasoning looks ok. As long as you did not make any computation mistakes the equation should be the correct one.

One way to verify your work on your own is to take points in one plane and verify that they satisfy both the parametric and the cartesian equations.

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