# Find vector equation of the plane spanned by $\langle -1, 0, 4\rangle, \langle 2, 3, 10\rangle$ that contains the point $P_0 = (2, 3, -5)$

$$\text{Remark:}$$ I notate vectors as $$\langle a_1, a_2,\ldots\rangle$$

I was given the following problem.

Find a vector equation for the plane $$S$$ spanned by $$\langle -1, 0, 4\rangle, \langle 2, 3, 10\rangle$$ and containing the point $$P_0 = (2, 3, -5)$$.

I am new to multivariate calculus and I self study. The lack of teachers forces me to post my solutions here in seek of correction/validation and suggestions. I will show what I did and explain a particular aspect of the problem that troubled me.

$$I$$. Let $$\vec{v} = \langle -1, 0, 4\rangle, \vec{w} = \langle 2, 3, 10\rangle$$. Then the normal vector of the plane spanned by $$\vec{v}, \vec{w}$$ is

\begin{align} \vec{n} &= \vec{v} \times \vec{w} \\ &= \langle -12, 18, -3\rangle \end{align}

$$II$$. Let $$\vec{r_0} = \langle 2, 3, -5\rangle$$ be the position vector of $$P_0$$. Then $$\vec{r_0} - \vec{w} = \langle 0, 0, -15\rangle$$ is the vector describing the line $$\overline{P_0P_1}$$, with $$P_1 = (2, 3, 10) \in S$$.

$$III$$. A vector equation for $$S$$ is then given by

$$\vec{n}\vec{r_0} = \vec{n}\vec{w}$$

Besides from wanting to know if my solution is correct at all, I wonder why I was given two vectors and a point to find a vector equation for $$S$$. It seems to me such equation could be found with the two vectors alone, by $$i$$. finding the normal vector of $$S$$ and $$ii$$. using the vector $$\vec{v} - \vec{w}$$ to describe a vector on the plane (instead of using $$\vec{r_0} - \vec{w}$$ as I did). Indeed, at least in my mental representation, if two vectors span a plane, a third vector defined as their difference is on the plane. Therefore I see no need for the presence of $$P_0$$ in this problem and I presume an equivalent solution is given by

$$\vec{n}(\vec{v} - \vec{w}) = 0$$ or equivalently $$\vec{n}\vec{v} = \vec{n}\vec{w}$$

Am I confusing something here? Thanks in advance.

• $\|\vec{v}\times\vec{w}\|\neq1 : \vec{v}\times\vec{w}$ is not normal, which means with norm $=1$. $\vec{v}\times\vec{w}$ is said to be orthogonal to the plane spanned by $\vec{v}$ and $\vec{w}$. Right ? Oct 27, 2022 at 11:17
• @StéphaneJaouen: in the context of American multivariable calculus courses, "normal" = "orthogonal", not "of unit length". So one talks about "finding a normal vector to the plane" and so on.
– JBL
Oct 27, 2022 at 11:57
• The vector plane spanned by $v=(-1,0,4)$ and $w=(2,3,10)$ is by definition $\mathbb R v+\mathbb R w$; it goes through (0,0,0). The plane spanned by $v=(-1,0,4)$ and $w=(2,3,10)$ and containing $P_0$ is by definition $S=P_0+\mathbb R v+\mathbb R w$ and you can prove easily that that plane does not go through $(0,0,0)$. We then speak of an affine plane for such planes. Right ? Oct 27, 2022 at 11:58
• @StéphaneJaouen: You're welcome; it's a terrible usage of "normal", I don't know where it comes from, but it seems to be standard (at least in the particular context I mentioned). About your second comment, if "we speak of" means "JBL, an American professor of mathematics, communicating with doctoral students or above" then yes I would say "affine plane"; if "we" means "JBL, a professor of mathematics, communicating with students in his standard American multivariable calculus course", then no, we don't teach them the word "affine" (regrettably).
– JBL
Oct 27, 2022 at 12:00
• In what comes to the meaning of "affine plane", I must say @JLB is correct and I have not been taught such concept at any point. However, as I understand your comments, it is the case that the vectors $\vec{w}, \vec{v}$ span a plane $S'$ that touches the origin, while the requirement that $S$ passes through $P_0$ makes $S \neq S'$. In my mental representation, $S$ and $S'$ are "parallel" (if such thing can be said of planes), or in other words $S$ is $S'$ shifted in a certain direction. Is that correct? Oct 27, 2022 at 12:23

Let me share with you an intuitive way of how I would approach this question.

Let us recall - To span a plane, we need a position vector, and two directional vectors. However, we are given two vectors spanning the plane and consisting of a point. So, what you did right is finding the normal $$\hat n$$ by taking the cross product of the two vectors. From $$\hat n$$, it can already tell you the cartesian equation of the plane.

$$1.$$ From $$[-12 \space 18 \space -3]^T$$, the cartesian equation is given by: $$-12x+18y-3z=c$$ where $$c$$ is a constant, also called the level set of the function $$f$$.

$$2.$$ We then substitute the point $$P_0$$ in the equation to find our constant $$c$$. $$c = -12(2) + 18(3) -3(-5) = 45$$ $$\implies -12x + 18y -3z = 45$$

$$3.$$ We can scale by a factor of $$3$$. $$\iff -4x+6y-z=15$$

$$4.$$ Now, we can convert the cartesian equation into a vector equation. For simplicity, we isolate $$z$$ by making it the subject. \begin{align} x& = x \\ y & =y \\ z & = -15 - 4x + 6y \\ \end{align} By letting $$x,y$$ to be $$s,t$$ respectively, this is the vector equation of our plane: $$\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ -15 \\ \end{bmatrix} + s \begin{bmatrix} 1 \\ 0 \\ -4 \\ \end{bmatrix} + t \begin{bmatrix} 0 \\ 1 \\ 6 \\ \end{bmatrix}: \forall s,t \in \mathbb{R}$$ Hope it helps.

• A vector equation, not "the" vector equation, isn't it ? Oct 27, 2022 at 13:41

I. $$\vec{n}$$ is indeed $$(-12,18,-3)$$;

II. $$(0,0,15)$$ is $$\color{red}{one}$$ vector directing the line $$\bar{P_0P_1}$$;

III. It seems to me that the best way to fix OP's error here is to give the expected answer (hope I'm not wrong): A vector equation for the plane S spanned by $$\vec{v}$$ and $$\vec{w}$$ and containing the point $$P_0$$ is $$S=P_0+\mathbb R\vec{v}+\mathbb R\vec{w}$$.

This deserves some explanation given the questions asked in the post :

• even if we have a very concrete interpretation of our calculations(which OP is building), we nevertheless work in the real vector space $$\mathbb R^3$$, of which we recall that the elements are the triplets of real numbers $$(x,y,z)$$, with $$x\in \mathbb R, y\in \mathbb R$$ and $$z\in \mathbb R$$. That we prefer to speak of points for some of these elements or of vectors for other elements depending on what we have in mind as a physical interpretation, the fact remains that these are just elements of $$\mathbb R^3$$ from the point of view that interests us in the posed exercise (not from the point of view of the OP who constructs his mental representations), that is to say from the point of view of the algebraist or the calculator.
• Note that $$\vec{P_0P_1}.\vec{n}=(0,0,15).(-12,18,-3)\neq0$$.
• The provided vector equation gives a parametrization of the plane $$S=\{(2,3,-5)+t(-1,0,4)+u(2,3,10):t,u \in \mathbb R\}=\{(2-t+2u,3+3u,-5+4t+10u)\}$$.
• One last tip: to build the right representations, start by working on similar exercises in $$\mathbb R ^2=\mathbb C$$ : it's much easier to draw.