Basics of Vectors -- Parallel versus Co-Planar

[WombatWithANuke]

Homework Statement

Ex. 2. Express each of the following by an equation:
(a) Two vectors A and B are parallel.
(b) Three vectors A, B and C are co-planar.

Homework Equations

C = A + B

3. The Attempt at a Solution
I understand (a) the answer being B = (alpha)*A because that is a scalar transformation, but for part (b) the answer in the end of the book is C = (alpha)*A + (beta)*B where (alpha) and (beta) are both scalar quantities. Wouldn't it still be correct to say that C = A + B without the scalar transformation. I just don't understand why the book chose to use this answer. Thank you for any help.

 

[robphy]

Let $\vec A$ be the unit vector $\hat x$.
Let $\vec B$ be the unit vector $\hat y$.
Consider any vector $\vec C$ in the xy-plane. Is $\vec C=\vec A+\vec B$?

 

[scottdave]

While C = A + B does describe 3 vectors which will be coplanar, there are many other A B C combinations, where they do not form a triangle.
For example, let C = A + B then what can we say about the vector D = A + C? This vector also lies in the same plane, but can you get B from adding these (and not use scalar multipliers)?
We know that C = A + B, substitute that in, and get D = A + (A + B) = 2A + B. This is why you need the scalar multipliers. Take any two non-parallel vectors in a plane, and you can add scalar multiples of them to produce a new 'triangle' which lies in the plane and the 3rd leg lies in the same plane. I hope this helps.
Take the special case of vectors i and j, which happen to be length 1 and orthogonal. You can scale each of those and add to get any vector in the i-j plane. But your two vectors, used to build other vectors, do not have to be orthogonal for it to work (but they cannot be parallel).

 

[haruspex]

C = (alpha)*A + (beta)*B

Amusingly, that is not the answer either. If A and B are parallel then the three are necessarily coplanar, but you would only be able to write C as a linear combination of A and B if all three are parallel.

A correct answer is that there exist three scalars, a, b and c, not all zero, such that $a\vec A+b\vec B+ c\vec C=0$.

Technically, there is the same problem with a). A better answer is, likewise, that there exist two scalars, not both zero...