Suppose $t, u, v, w \in \mathbb{R}^3$. If $(t \times u) \times (v \times w) = 0$, are $t,u,v,w$ on the same plane?

Question

Suppose $t, u, v, w \in \mathbb{R}^3$. If $(t \times u) \times (v \times w) = 0$, are $t,u,v,w$ on the same plane?

My Answer No. Let $p_1$ be the plane described by $x + y + z = 3$ and $p_2$ the plane described by $x + y + z = 4$. Then the planes do not intersect but they have the same norm, $n = (1,1,1)$. Letting $t,u \in p_1$ and $v,w \in p_2$, it follows that they are not on the same plane but

$$(t \times u) \times (v \times w) = n \times n = 0$$

Answer 1

If $t$ and $u$ are linearly dependent, then it is trivial that $$(t\times u)\times (v\times w)=0\,.\tag{*}$$ If $v$ and $w$ are linearly dependent, then (*) is obviously true. From now on, suppose that $t$ and $u$ are linearly independent, and $v$ and $w$ are also linearly independent.

If the span of $t$ and $u$ coincides with the span of $v$ and $w$, then let $p$ denote the span of all vectors $t$, $u$, $v$, and $w$. Thus, both $t\times u$ and and $v\times w$ are vectors normal to $p$. Hence, $t\times u$ and $v\times w$ are linearly dependent, whence (*) is true.

Suppose now that the span of $t$ and $u$ does not coincide with the span of $v$ and $w$. Therefore, the span of $t$, $u$, $v$, and $w$ is the whole $\mathbb{R}^3$, which is a $3$-dimensional vector space. Thus, one of the vectors is linearly dependent of the other three. Without loss of generality, assume that $$t=au+bv+cw$$ for some $a,b,c\in\mathbb{R}$. This means $u$, $v$, and $w$ are linearly independent. Note that $$t\times u=-b(u\times v)-c(u\times w)\,.$$ That is, $$(t\times u)\times (v\times w)=-b(u\times v)\times (v\times w)-c(u\times w)\times (v\times w)\,.$$ Now, $$(u\times v)\times (v\times w)=\big((u\times v)\cdot w\big)\,v-\big((u\times v)\cdot v\big)\,w=\rho\,v\,,$$ where $\rho:=(u\times v)\cdot w$. Similarly, $$(u\times w)\times(v\times w)=\big((u\times w)\cdot w\big)\,v-\big((u\times w)\cdot v\big)\,w=\rho\,w\,.$$ Consequently, $$(t\times u)\times (v\times w)=-b\,\rho\,v-a\,\rho\,w\,.$$ Since $u$, $v$, and $w$ are linearly independent, $\rho\neq 0$. Thus, (*) is true if and only if $a=b=0$. Therefore, $t=au$. However, this contradicts the assumption that $t$ and $u$ are linearly independent.

In conclusion, we have shown that (*) is true if and only at least one of the following conditions is true:

• $t$ and $u$ are linearly dependent,

• $v$ and $w$ are linearly dependent, and

• the span of $t$ and $u$ is the same as the span of $v$ and $w$.

Answer 2

So, your example relies on some notion that vectors have three things:

1. a magnitude (also called a norm or length),
2. a direction, and
3. a location.

Unfortunately, vectors conventionally only have the first two. That is, the vector from $[0, 0, 0]$ to $[0, 0, 1]$ is held to be the exact same vector as the vector from $[1, 1, 1]$ to $[1, 1, 2]$ even though they are in different places.

That’s not to say that the location of the vector is never important: a good example is in calculating a torque where you need to know two vectors, first a force vector for the force involved, and second a displacement vector from some prespecified point of rotation about which we are calculating to the place where the force is applied. So when it is important, we generally include that as a second vector.

In this case it’s not 100% clear what exactly is intended, but I think it is perhaps likely the question I would restate as this:

Points $T,U,V,W$ in $\mathbb R^3$ have displacement vectors from the origin $\mathbf t,\mathbf u,\mathbf v, \mathbf w$ respectively, such that $(\mathbf t \times \mathbf u)\times (\mathbf v \times \mathbf w) = \mathbf 0.$ Are the points $T,U,V,W$ co-planar? (And, if so, does this plane include the origin $O$?)

In this case I think the question is perhaps a sort of trick question—something that requires more insight than at first glance? So if the vectors are nondegenerate then you have a plane defined by $OTU$ with normal $\mathbf n_{TU}$ and a plane defined by $OVW$ with normal $\mathbf n_{VW}$ and the zero cross product between these two normals should suffice to make those two vectors parallel which should make the two planes equal: thinking geometrically, this makes a lot of sense.

But for instance if you were to consider a tetrahedron, $$ \begin{align} \mathbf t &= \begin{bmatrix}1\\ 0\\ 0\end{bmatrix}& \mathbf u &= \begin{bmatrix}2\\ 0\\ 0\end{bmatrix}\\ \mathbf v &= \begin{bmatrix}3/2\\\sqrt{\frac12}\\+\frac12\end{bmatrix}& \mathbf w &= \begin{bmatrix}3/2\\\sqrt{\frac12}\\-\frac12\end{bmatrix}, \end{align} $$ then obviously $TUVW$ are not coplanar as we are talking about a tetrahedron, but also obviously the cross product requested is trivially zero as $\mathbf t \times \mathbf u = \mathbf 0.$

I am only telling you this because it’s also possible that this is not a trick question and your teacher came up with the plausible intuitive argument and you have the opportunity to give a devastating rebuttal, heh.