Thread: Three vectors lie on the same line

Initial point on the origin

v

1 = (−1, 2, 3), v2 = (2,−4,−6) and v3 = (−3, 6, 0)

How do we do this?

I know they are linearly dependent as 2v1 + v2 + 0v3 = 0

Originally Posted by adam_leeds

Initial point on the origin

v

1 = (−1, 2, 3), v2 = (2,−4,−6) and v3 = (−3, 6, 0)

How do we do this?

I know they are linearly dependent as 2v1 + v2 + 0v3 = 0

Two vectors are collinear (parallel) if they are multiples of each other.

Originally Posted by Plato

Two vectors are collinear (parallel) if they are multiples of each other.

So they don't lie on the same line as they are linearly dependent

No, saying that three vectors are linearly dependent means they lie in the same plane so it is still possible that they lie on the same line.

If they were independent, then they could not be on the same line but knowing that they are dependent doesn't tell you whether they lie on the same line or not. Do as Plato suggested- are they all multiples of one another?

Originally Posted by HallsofIvy

No, saying that three vectors are linearly dependent means they lie in the same plane so it is still possible that they lie on the same line.

If they were [b]independent[b], then they could not be on the same line but knowing that they are dependent doesn't tell you whether they lie on the same line or not. Do as Plato suggested- are they all multiples of one another?

being multiples of each other is linear dependency though?

Yes- but it doean't work the other way: if three vectors are linear dependent, they don't have to be multiples of one another.

Originally Posted by HallsofIvy

Yes- but it doean't work the other way: if three vectors are linear dependent, they don't have to be multiples of one another.

So they lie on the same line as 2v1 + v2 + 0v3 = 0

Sorry it doesnt, as only 2 of them are multiples of each other.

Originally Posted by adam_leeds

Initial point on the origin
v

1 = (−1, 2, 3), v2 = (2,−4,−6) and v3 = (−3, 6, 0)

How do we do this?

Let’s answer this one and for all
If the three were collinear then the vectors would be parallel (multiples of each other).

I hate to add to a "once and for all" but the initial post said that these were vectors with "Initial point on the origin". So it is sufficient to determine whether, say, (2, -4, -6) and (-3, 6, 0) are multiples of (-1, 2, 3).

Of course (2, -4, -6)= -2(-1, 2, 3) so that is a multiple. But (-3, 6, 0) is NOT a multiple of (-1, 2, 3). -3= 3(-1) and 6= 3(2) but 0 is NOT 3 (3). These vectors are do NOT lie on a single line.

One more time, adam leeds, showing that three vectors are "dependent", that is showing that 2v1 + v2 + 0v3 = 0, only shows they lie in the same plane, not that they along the same line.

It is also, of course, true, as Plato says, that and [tex]v_3- v_2= (-5, 10, 6) are not multiples of one another. -5= (5/3)(-3) and 10= (5/3)(6) but 6is NOT equal to (5/3)(9)