I asked this question a few days ago, where the question was this:

I have a task stating this:

Determine if the following vectors are coplanar. Assume that $v_1$, $v_2$ and $v_3$ are not coplanar.

$w_1=4\vec v_1+3\vec v_2$

$w_2=\vec v_2+4\vec v_3$

$w_3=-\vec v_1-3\vec v_3$

I don't quite understand how I'll do this when I do not know the values of any of the vectors. Also, what significance does the information "$v_1$, $v_2$ and $v_3$ are not coplanar" have in terms of the solution? I'm guessing knowing that helps decide whether they're coplanar or not, but I can't see how.

In that question we arrived at the conclusion that they are coplanar. However, since then, I've attempted to find a way to mathmatically prove that they are coplanar, but instead I seem to arrive at the conclusion that they are in fact NOT coplanar. What I've done looks like this:

If the sum of the three vectors equals the zero vector in any way other than by multiplying every vector by zero, then they are coplanar. So if I can prove that one of the vectors can be represented as a multiple of the other, then they are coplanar. So to try and prove this, I’ll try to add two vectors.

$$\vec w_1+\vec w_2=(4\vec v_1+3\vec v_2)+(\vec v_2+4\vec v_3)$$

From there, I realized I could do this instead:

$$\vec w_1-3\vec w_2=(4\vec v_1+3\vec v_2)-3(\vec v_2+4\vec v_3)$$ $$\vec w_1-3\vec w_2=4\vec v_1+3\vec v_2-3\vec v_2-12\vec v_3$$ $$\vec w_1-3\vec w_2=4\vec v_1-12\vec v_3$$

Now the two vectors are written in the same terms as the third equation. However, this poses the problem that I need to something to multiply $(4-12)$ with to make it $(-1-3)$. Which simply isn't possible, meaning there is no way I can represent $\vec w_1$ and $\vec w_2$ as a linear combination of $\vec w_3$. This again ultimately means the vectors are linearly independent, and thus also not coplanar.

So, did I do something wrong here? Am I trying to prove this the wrong way, or was the answer for the other question simply wrong? As far as I can tell now, they are not coplanar, but in the other question we concluded that they were.

  • $\begingroup$ I'm sorry, but your statement and example are wrong. Of course are $0, e_1, e_2$ coplanar since you can write $100 \cdot 0 + 0\cdot e_1 + 0\cdot e_2 = 0. Instead of $100$ you can use any non-zero constant. We also have the result that any three non-zero vectors are linear dependent if and only if they are coplanar. (See portal.tpu.ru/SHARED/k/KONVAL/Sites/English_sites/V/…) $\endgroup$
    – Marc
    Jan 14, 2014 at 21:20
  • $\begingroup$ @Marc: My apologies, I was out to lunch! I read collinear instead of coplanar. I will delete my comment. $\endgroup$
    – copper.hat
    Jan 15, 2014 at 5:24
  • $\begingroup$ @Marc: The question seems a little bogus to me. Any three vectors must be coplanar. Clearly $S=\operatorname{sp}\{v_1-v_3,v_2-v_3\}$ is at most a two dimensional space, hence a plane (or contained on one), and $S+\{v_3\}$ is just a translated plane. $\endgroup$
    – copper.hat
    Jan 15, 2014 at 5:54
  • $\begingroup$ No, three vectors must not be coplanar. Coplanarity of 3 vectors in 3D space is identical to linear dependency of 3 vectors in 3D space. If 3 vectors can be linearly independent in 3D space, then they can be non-coplanar too. $\endgroup$
    – Threethumb
    Jan 15, 2014 at 10:05

2 Answers 2


You did the right thing in general, but trying out specific cases is a bad way to go -- you might miss the one combination of the $w$s that actually comes out to zero.

Instead, what you want to do is say this:

Let's suppose that the $w$ are coplanar. Then there are numbers $a$, $b$, and $c$ such that

$$ aw_1 + b w_2 + c w_3 = 0 $$

Now I can substitute in the expressions for the $w$s to get that $$ a(4v_1 + 3v_2) + b (v_2 + 4v_3) + c (-v_1 - 3 v_3) = 0 $$ Expanding, you get $$ 4a v_1 + 3a v_2 + bv_2 + 4b v_3 - c v_1 -3c v_3 = 0 \\ (4a - c) v_1 + (3a + b) v_2 + (4b -3c) v_3 = 0 \\ $$

So IF you could find a (not all zeros) combination of the $w$s that's zero, you'd also have a combination of the $v$s that's zero. You might think that'd mean that the $v$s are coplanar, but it only means that if the coefficients aren't all zero.

What would make all three of those coefficients be zero? Let's do some algebra. We'll assume that $$ 4a - c = 0 \\ 3a + b = 0 \\ 4b - 3c = 0 $$ Multiply the first equation by 3 to get $$ 12a - 3c = 0 \\ 3a + b = 0 \\ 4b - 3c = 0 $$ Subtract the thrid equation from the first to get $$ 12a - 4b = 0 \\ 3a + b = 0 \\ 4b - 3c = 0 $$ Multiply the second equation by $4$ to get $$ 12a - 4b = 0 \\ 12a + 4b = 0 \\ 4b - 3c = 0 $$ Subtract the first from the second to get $$ 12a - 4b = 0 \\ 8b = 0 \\ 4b - 3c = 0 $$ So $b = 0$, which (from the first equation) says that $a = 0$, which from the third equation tells us $c = 0$.

In summary: if the three $w$s were coplanar, we'd have found $a,b,c$, not all zero, with $aw_1 + bw_2 + cw_3 = 0$. That would in turn make a combination of the $v$s be zero. That latter combination has all-zero coeffs only if the former does, so you'd have found a not-all-zero combination of the $v$s that's zero, which would imply the $v$s are coplanar. Since they're not, we've got a contradiction. The assumption that the $w$s are coplanar must be wrong. Hence they're non-coplanar.

  • $\begingroup$ Ah yes, I actually did think of this problem too. So I tried every combination of the $w$s to make sure I didn't miss anything. Your method is a lot more foolproof though, and leaves no doubt that they are not coplanar. Thanks! $\endgroup$
    – Threethumb
    Jan 14, 2014 at 21:04
  • $\begingroup$ Could you expand a bit on why you make every of the parenthesized expressions equal zero? I don't quite get how doing that proves that there are no non-zero scalars to multiply the terms with to make their sum equal zero. $\endgroup$
    – Threethumb
    Jan 14, 2014 at 21:29
  • $\begingroup$ The big structure of this argument is this: If the $w$s are coplanar, then the $v$s are coplanar. To show that, I assume the $w$s are coplanar, and I get that a certain combination of the $v$s is zero. That's of no use to me unless I know that it's a nontrivial combination (i.e., at least one of the three coeffs is nonzero). So I ask "Is it possible that all three are zero? If they were, what would that imply?" To answer that rhetorical question, I set all three to zero, and discover that it would mean that $a,b,c$ were all zero. So since $a,b,c$ are not all zero, nor are all three coeffs. $\endgroup$ Jan 14, 2014 at 21:33
  • $\begingroup$ Ah, I get the argument you're posing now. I'm still not 100% how you determine that the $a, b, c$ are not all zero though. $\endgroup$
    – Threethumb
    Jan 14, 2014 at 21:46
  • $\begingroup$ We ASSUME that $a,b,c$ not all zero (i.e., that the $w$s are coplanar). From this, we conclude (by the argument in my last comment) that the coeffs in the $v$ combination are not all zero. That makes the $v$s coplanar, a contradiction. So our assumption (that there exist $a,b,c$ not all zero, such that $aw_1 + bw_2 + cw_3 = 0$) was false, i.e., our assumption that the $w$s are coplanar was false. $\endgroup$ Jan 14, 2014 at 22:03

I think you are correct: if one is a linear combination of the other two, it can be any one. You picked $\vec{w}_3$ and showed that it is not.

  • $\begingroup$ That statement (that one vector being a combination of two others is unaffected by the vector you pick) is not generally true: if $v_1$ is a combination of $v_2$ and $v_3$, it's possible that $v_2$ is not a combination of $v_1$ and $v_3$. Example: $v_1 = [2, 0, 0]; v_2 = [0, 1, 0], v_3 = [1, 0, 0]$. You then have $$ v_1 = 0v_2 + 2v_3 $$ but $v_2$ cannot be expressed as a combination of $v_1$ and $v_3$. $\endgroup$ Jan 15, 2014 at 12:26

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