Plane described by three vectors?

Question

Suppose the vectors $a = [1, 0, 1]^T$ , $b = [2, 7, -2]^T$ and $c = [3, 1, 5]^T$ are lying on a plane in $\mathbb{R}^3$.

Prove that $d = [4, 8, 2]^T$ also lies on that plane. (Hint: A plane can be expressed as a particular solution plus the span of two linearly independent vectors.)

What does it mean those three vectors lie on a plane? Do they represent points? Otherwise they are clearly not on the same plane.

Answer 1

We can regard vectors as corresponding to their terminal points (assuming initial point at the origin).

Following the hint provided, the plane $P$ containing the points corresponding to the vectors $a,b,c$ is the set of points corresponding to the set of vectors $$\{a+su+tv\mid s,t\in\mathbb{R}\}$$ where $u=b-a$ and $v=c-a$.

Thus, to show that the point corresponding to the vector $d$ is on the plane $P$, it suffices to find $s,t\in\mathbb{R}$ satisfying the vector equation $$ \langle{4,8,2}\rangle=\langle{1,0,1}\rangle+s\langle{1,7,-3}\rangle+t\langle{2,1,4}\rangle $$ or equivalently, to find $s,t\in\mathbb{R}$ satisfying the system \begin{align*} 4&=1+s+2t\\[4pt] 8&=0+7s+t\\[4pt] 2&=1-3s+4t\\[4pt] \end{align*} of $3$ equations in $2$ unknowns.

Solve the subsystem consisting of two of the three equations for $s,t$, and then verify that the values found for $s,t$ also satisfy the remaining equation.

Answer 2

Find the normal of the vector which is proportional to

$$ \vec{n} = \vec{a}\times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} $$ _{where $\times$ is the vector cross product.}

$$ \vec{n} = \pmatrix{31 \\ -10 \\ -13} $$

Now project the vector $\vec{d}$ along the normal to find that its projection equals the projection of any of the defining points

$$ \vec{d} \cdot \vec{n} = \vec{a} \cdot \vec{n} = \vec{b} \cdot \vec{n} = \vec{c} \cdot \vec{n} $$ _{where $\cdot$ is the vector inner product.}

$$ \begin{aligned} \vec{d}\cdot\vec{n} & = 18 & \vec{a}\cdot\vec{n} & = 18 & \vec{b}\cdot\vec{n} & = 18 & \vec{c}\cdot\vec{n} & = 18 & & \checkmark \end{aligned} $$

Answer 3

To answer your specific question, it’s common when working in $\mathbb R^n$ to identify points with their displacement vectors from the origin. This unfortunately does create an ambiguity when talking about vectors lying on a plane in $\mathbb R^3$: is this talking about a two-dimensional subspace of the vector space $\mathbb R^3$ that contains the vectors, or is it an affine subset of $\mathbb R^3$ that contains the “endpoints” of the vectors? You’re just going to have to determine this from context. In the particular case that you’re asking about, it would appear that the latter is intended since the vectors $a$, $b$ and $c$ span all of $\mathbb R^3$.

As to how one might solve this problem, here’s a related but slightly different approach to the others that have been suggested. The common plane, if it exists, has an implicit Cartesian equation of the form $Ax+By+Cz+D=0$. If you substitute the coordinates of the four known points into this equation, you get a system of homogeneous linear equations in the unknown coefficients. For the four points to lie on the same plane, this system must have an infinite number of solutions (do you see why?). This is equivalent to the condition $$\det\begin{bmatrix}a^T&1\\b^T&1\\c^T&1\\d^T&1\end{bmatrix}=0.$$ Note, though, that this approach allows the possibility that $a$, $b$ and $c$ are colinear, in which case every other point lies on some common plane with them, but I’m taking for granted that they’re not based on the problem statement.

Should you want to know the unknown coefficients, any nonzero null vector of this matrix provides them, or you can use the same idea to produce an implicit Cartesian equation directly: for any point $d=(x,y,z)$, the above determinant vanishes iff $d$ is coplanar to $a$, $b$ and $c$, so that expression is an equation of the plane defined by the latter three points.