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I am given 3 vectors. $u=[1, 2, 3], v=[-3, 1, -2], w=[2,-3,-1]$. And the question is how do I know that $u, v, w$ lie in a plane? I can't understand what they mean. Because for me each of them fills some line. And only their pairwise linear combinations can fill a plane, and all triple linear combinations fill 3d space. So how should I answer the question? Thanks in advance.

UPD1: This is from book "Introduction to Linear Algebra - 4th edition", Gilbert Strang. Page 6, problem 5. enter image description here

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  • $\begingroup$ They're treating the vectors as points. In any interpretation a vector would never fill a line, it would at most fill a line segment. But most often they are treated as points in mathematics. $\endgroup$ – Matt Samuel Mar 25 '17 at 12:46
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    $\begingroup$ For any three vectors they can either be co-planar or not. Consider all vectors where on component is equal. $\endgroup$ – marshal craft Mar 25 '17 at 12:46
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    $\begingroup$ @MattSamuel, why? This linear combination: $cu$, where $c$ is some real number, it fill a line. Or you just mean vector itself, not linear combination? $\endgroup$ – Turkhan Badalov Mar 25 '17 at 12:48
  • $\begingroup$ Just the vector. $\endgroup$ – Matt Samuel Mar 25 '17 at 12:48
  • $\begingroup$ It seems you are confused about Cartesian/rectilinear vectors where the vector is implicitly implied to start at origin an stop at the coordinate. You are considering this as a coordinate, but vectors have a basis which is a unit direction. So the triple coordinates are coefficents of the scalar product of the basis vector. $\endgroup$ – marshal craft Mar 25 '17 at 12:58
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The question is whether or not these vectors are linearly independent. A nice practical criterion is a determinant of a matrix cinsisting of these vectors. They are linearly independent iff this determinant is non-zero. This is not the case here, so $u,v,w$ are complanar (i.e. linearly dependent) vectors. Try to express $u$ as a linear combination of $v,w$.

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  • $\begingroup$ Why do you say they are dependent? I can't find any way to obtain one vector from another. $\endgroup$ – Turkhan Badalov Mar 25 '17 at 12:56
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    $\begingroup$ @TurkhanBadalov: really? what did you find for $u+v+w$? $\endgroup$ – Taladris Mar 25 '17 at 14:26
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Three vectors $u,v,w$ always lie in the plane parallel to the subspace spanned by $v-u$ and $w-u$ that passes through $u$. It is the set of all combinations $$u+a(v-u)+b(w-u)$$ or equivalently $$au+bv+cw$$ where $a+b+c=1$.

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One way is to show the cross product of one vector with the other two yields two vectors which are parallel, ie. There unit vectors are equal or negative of the other.

You could even define a plane in $\Bbb R^3$ to be a set of vectors,

$ P = \{\mathbf x \in \Bbb R^3 \ | \ \mathbf a , \mathbf b \in P \ \Rightarrow \ \frac{\mathbf x \times \mathbf a}{|| \mathbf x \times \mathbf a ||} = \frac{\mathbf x \times \mathbf b}{|| \mathbf x \times \mathbf b ||} \lor \frac{\mathbf x \times \mathbf a}{|| \mathbf x \times \mathbf a ||} = -1\frac{\mathbf x \times \mathbf b}{|| \mathbf x \times \mathbf b ||} \}$.

Then three vectors $u,v,w$ are coplanar iff $u,v,w \in P$. That is they are contained in some plane $P$ defined above.

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