A problem about parallel vectors

Problem
If u = (-3, 2, 1) and v = (5, 0, -3) and w = (6, 1, -4) is it possible to find a scalar t such that u + tv is parallel to w?

Solution
If 2 vectors P and Q are parallel then there is a number or scalar k so that P = kQ.
For example, (3, 6, 12) and (1, 2, 4) are parallel and (3, 6, 12) = 3 (1, 2, 4).
So this question is asking us whether we can find t so that

(-3, 2, 1) + t (5, 0, -3) = k (6, 1, -4). This equation can only be satisfied if each separate component matches up. So we have, in fact, 3 equations to look at. We would need
-3 + 5t = 6k
2 = k
1 - 3t = -4k From the second, k = 2. Then, from the third, we get t = 3.
Now, the crucial step is to find out whether these values for k and t fit the first equation. Substituting in, we find that they do, and therefore we know that u + 3v is parallel to w.

Geometrically, this means that the vectors u, v and w can all lie in the same plane in 3-D space. The vector w is called linearly dependent on u and v. If you pick any 3 vectors in 3-D space, they won't usually have this property. Generally, if you move their 3 tails to the origin then it will be possible to find some combination of them that will take you to any chosen point in 3-D space. The 3 vectors i, j and k are the most important example of this. However, if u, v and w have their tails moved to the origin, combinations of them will only take you to points in the 2-D plane in which they all lie. No combination of u, v and w will allow you to escape from this plane.

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