Condition to three vectors being collinear

[LCSphysicist]

Homework Statement

Find a condition, just using operation of vector space, such that the vectors u,v,w belong to the subspace E be colinear

Relevant Equations

\n

Now i am rather confused, the answer apparently is that $(w-u) = \lambda(u-v)$

But, i could find a way that disprove the answer, that is:
Be u v and w vectors belong to R2, a subspace of R3:

What do you think? This is rather strange.

 

[Mark44]

Homework Statement:: Find a condition, just using operation of vector space, such that the vectors u,v,w belong to the subspace E be colinear
Relevant Equations:: \n

Now i am rather confused, the answer apparently is that $(w-u) = \lambda(u-v)$

What this is saying is that $\vec w - \vec u$ is a scalar multiple of $\vec u - \vec v$. In other words, the two vector differences point in the same or opposite directions.

But, i could find a way that disprove the answer, that is:
Be u v and w vectors belong to R2, a subspace of R3:

View attachment 272763

What do you think? This is rather strange.

Your example is not a counterexample: it does not disproved the book's solution.
From your drawing $\vec w - \vec u$ can be drawn from the common point (the point where all three vectors start), and pointing straight up. $\vec u - \vec v$ can be drawn also from the common point and pointing straight up.
To draw $\vec w - \vec u$, that's really the same as $\vec w + (-1)\vec u$, so go out to the end of $vec w$ and then go backwards the length of ##\vec u. That should take you to a point directly above the common point.

BTW, in English we don't say "Be u v and w vectors" -- we say "Let u, v, and w be vectors that belong to ..."

 

[LCSphysicist]

What this is saying is that $\vec w - \vec u$ is a scalar multiple of $\vec u - \vec v$. In other words, the two vector differences point in the same or opposite directions.

Your example is not a counterexample: it does not disproved the book's solution.
From your drawing $\vec w - \vec u$ can be drawn from the common point (the point where all three vectors start), and pointing straight up. $\vec u - \vec v$ can be drawn also from the common point and pointing straight up.
To draw $\vec w - \vec u$, that's really the same as $\vec w + (-1)\vec u$, so go out to the end of $vec w$ and then go backwards the length of ##\vec u. That should take you to a point directly above the common point.

BTW, in English we don't say "Be u v and w vectors" -- we say "Let u, v, and w be vectors that belong to ..."

Yes, reading again with your information makes me realize that the question is not saying that $\vec w - \vec u$ is a scalar multiple of $\vec u - \vec v$ IMPLIES the collinear condition, but it is saying that if the vectors are collinear, so this is true.
While i was using "iff", the book was using the right "if"

"BTW, in English we don't say "Be u v and w vectors" -- we say "Let u, v, and w be vectors that belong to ..."
Thank you for this clarification.

 

[haruspex]

the question is not saying that w→−u→ is a scalar multiple of u→−v→ IMPLIES the collinear condition

It is saying that.
"a condition such that [if satisfied by the vectors then] the vectors u,v,w [are] colinear"
You may be confused by the way collinearity is being used here. They mean that the points represented by the vectors are collinear, as in your diagram.