Vector equation of plane $3x − 5y + 2z = 1$ in $\mathbb{R}^{3}$

Question

It seems like it shouldn't be that difficult, but it's completely stumped me. Nothing on this exact format of question in lecture notes or in the textbook. If anybody can explain how one gets the vector equation from the formula of a 3D plane, it would be much appreciated, thanks.

I understand that that is the question. I'm wondering what your book / lecturer / notes means when it says "vector equation". — Arthur, Sep 24 '18 at 20:33
"Give a vector equation of the plane 3x - 5y + 2z = 1 in R^3", I don't understand it myself what it's specifically asking. — Harmonia, Sep 24 '18 at 20:38
This is a wonderful question, very similar to yours: https://math.stackexchange.com/questions/1194582/how-to-find-the-vector-equation-of-a-plane-given-the-scalar-equation. It uses dot products to find 2 perpendicular vectors to the normal vector of the plane. Those two vectors lie in the plane, and thus span the entirety of the plane. I hope you find it useful! — D.R., Sep 24 '18 at 22:43
Was interesting, and took a while, but think I figured it out, thanks — Harmonia, Sep 25 '18 at 02:42

score 0 · Answer 1 · answered Sep 24 '18 at 20:03

0

HINT

find three not collinear points $P,Q,R$ on the plane
the parametric equation is: $$P+t(Q-P)+s(R-P)$$

answered Sep 24 '18 at 20:03

user

162,563

I have not seen this covered before, so I do not know how points are found, the meaning of not collinear, or what a parametric equation even is... – Harmonia Sep 24 '18 at 20:25
@Harmonia Have you encountered the concept of parametric equation for a line? – user Sep 24 '18 at 20:29
Guess I was kinda an idiot when I said that. I know what parametric equations are, but seem to have forgotten that... – Harmonia Sep 24 '18 at 20:36
@Harmonia Well here we are asked to find the parametric equation for the plane. Then recall that a plane through the origin is spanned by any two linearly independent vectors which lie on the plane. Since the plane here is not through the origin we also need to add a point onto the plane, – user Sep 24 '18 at 20:39
I am unsure though how one would find points on a plane from the equation of the line given, the meaning of collinear, or the relevance of that parametric equation – Harmonia Sep 24 '18 at 20:40
And isn't it asking for a vector equation, not the parametric equation for the plane? – Harmonia Sep 24 '18 at 20:41
@Harmonia For example the xy plane is described by $z=0$ is cartesian equation and for example as $t(1,0)+s(0,1)$ in parametric form. – user Sep 24 '18 at 20:42
@Harmonia I guess that vector equation is used as a synonymous of parametric equation. – user Sep 24 '18 at 20:44
z = 0 in xy plane makes sense, but the t(1, 0) + s(0, 1) is completely new to me – Harmonia Sep 24 '18 at 20:45
@Harmonia The latter is exactly the parametric equation for the $xy$ axis where $s$ and $t$ are real numbers. Indeed $(1,0,0)$ and $(0,1,0)$ (sorry in the previous I forgot the z coordinate) are linearly independent and they span the xy plane. – user Sep 24 '18 at 20:49
Alright, but I still don't understand how they are used, what their purpose here is. I'm sorry, this is all just new to me. – Harmonia Sep 24 '18 at 20:52
@Harmonia I can suggest to revise the subject on your material. As an alternative take also a look here VIDEO. – user Sep 24 '18 at 20:55
Yeah, after reviewing the various material, was able to figure it out, thanks – Harmonia Sep 25 '18 at 02:43
@Harmonia You are welcome! If you want show your work here I can take a look to it! Bye – user Sep 25 '18 at 02:55

Vector equation of plane $3x − 5y + 2z = 1$ in $\mathbb{R}^{3}$

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