Are collinear vectors always scalar multiples of each other?

[Chris L]

Two quick questions:

1. Are two vectors considered collinear if the angle between them is 180°? (My guess would be yes, but it can't hurt to double check)

2. Let's say you're given a problem like this:

"given A = (1, 2, 3) and B = (x, 5, 6), find the value of x such that A and B are collinear"

The first method that comes to mind is to recognize that the cosine of the angle between the two has to be 1 (or -1 as well, depending on the answer to my first question), and from there using the definition of the dot product to conclude that A dot B = |A||B|. However, using this method, you have now turned what appears to be a reasonably straightforward problem into one that involves using the quadratic equation and generates an extraneous solution, requiring you to check both values to determine which one is actually valid.

Is there a simpler way to solve a problem like that, or is there no choice but to do all of that ugly algebra?

 

[Mark44]

Two quick questions:

1. Are two vectors considered collinear if the angle between them is 180°? (My guess would be yes, but it can't hurt to double check)

Yes.

2. Let's say you're given a problem like this:

"given A = (1, 2, 3) and B = (x, 5, 6), find the value of x such that A and B are collinear"

The first method that comes to mind is to recognize that the cosine of the angle between the two has to be 1 (or -1 as well, depending on the answer to my first question), and from there using the definition of the dot product to conclude that A dot B = |A||B|. However, using this method, you have now turned what appears to be a reasonably straightforward problem into one that involves using the quadratic equation which generates an extraneous solution, requiring you to check both to find the one that is actually valid.

Is there a simpler way to solve a problem like that, or is there no choice but to do all of that ugly algebra?

Vectors that are collinear are scalar multiples of each other. For this problem if u and v are collinear, then u = kv for some scalar k, and with u = <1, 2, 3> and v = <x, 5, 6>.

 

[Chris L]

So for that particular problem, there doesn't happen to be a solution since 2 = 5k and 3 = 6k, and obviously there isn't a value for k that satisfies both of those. Thanks for your response

 

[Hitchens]

Vectors that are collinear are scalar multiples of each other. For this problem if u and v are collinear, then u = kv for some scalar k, and with u = <1, 2, 3> and v = <x, 5, 6>.

Not that I disagree, but does co-linearity necessarily follow? IIRC two lines are co-linear iff they are parellel and share a point.

 

[Mark44]

The question was about vectors, which can be moved around so that they begin at an arbitrary point. If we're talking about direction vectors for two lines, it's possible for the lines to be parallel (with the vectors being scalar multiples of each other) so that the lines don't share a common point.