# Homework Help: Parallel vectors

1. Apr 1, 2014

### negation

Mod note: Corrected a typo in the problem statement.
1. The problem statement, all variables and given/known data

[STRIKE]If[/STRIKE]Is the vector AP parallel to u?

u= (1,4,-3)

A(-2,-1,-2)

3. The attempt at a solution

I do know that if AP is parallel to u then there must be some scalar multiple such that Ax = u.
I think it's a matter of setting up the question that I'm muddled.

Anyway, my attempt.

Let A = a1,a2,a3

(a1,a2,a3)(-2,-1,-2) = (1,4,3)

-2a1= 1
-a2=4
-2a3 = 3

in augmented form:

-2,0,0|-1
0,2,0|-4
0,0,2|3

REF...

a1= -1/2
a2= -2
a3 = 3/2

Last edited by a moderator: Apr 1, 2014
2. Apr 1, 2014

### HallsofIvy

What do you mean by "AP"? Normally, a notation like that means "A" and "P" are points and "AP" is the vector from A to P. But there is no "P" given here.

You are being very confusing here. AP is parallel to u if and only if there exist sum scalar multiple x such that xAP= u. Is "A" here a point or a vector?

Didn't you just tell us that A was (-2, -1, -2)?

??what kind of multiplication is this? You should be trying to find a single real number, x, such that x(-2, -1, -2)= (-2x, -1x, -2x)= (1, 4, 3) (That is, assuming your "AP" is just the vector A).

I still have no clue what you are doing. Assuming that your "AP" is just the vector (-2, -1, -2) the you should be trying to find a single number x such that x(-2, -1, -2)= (-2x, -x, -2x)= (1, 4, 3). Yes, that gives -2x= 1 so that x= -1/2. But then -x= 1/2, not 4 so we can see immediately that this cannot be done.

In fact, the question "Is (-2, -1, -2) parallel to (1, 4, 2) can be answered quickly by noting that -2 divided by 1 is -2 but that -1 divided by 4 is -1/4. Those are not the same so one is not the multiple of the other.

3. Apr 1, 2014

### Staff: Mentor

Due to your unclear problem statement, it took me a while to figure out what the problem is asking. I think it should read something like this:
Find the coordinates of a point P so that the vector $\vec{AP}$ is parallel to u, where u and A are shown above.
You are given A. Why are you trying to solve for its coordinates? What you need to find are the coordinates of a point P.
No.
No.

The main concept to understand here is how two vectors can be equal. The dot product doesn't enter into things in this problem

4. Apr 1, 2014

### negation

I know it's unclear. I was trying to infer from what the problem was implying.
Now that you said, it make sense that A = (-1,-2,-2). When I typed A(-1,-2,-2), it was NOT a typo but a direct copy from the source.

5. Apr 1, 2014

### negation

After Mark44's clarification, I understand what the question is asking.
I believe the question meant to suggest A = (-2,-1,-2) instead of A(-2,-1,-2).

6. Apr 1, 2014

### Staff: Mentor

That was not my concern. Either way, A is a point. $\vec{AP}$ is a vector. You need to find the coordinates of P (not A) so that $\vec{AP}$ is parallel to your given vector u. Do you understand how to determine whether two vectors are parallel?

7. Apr 1, 2014

### negation

I think I do. Both vectors are parallel if every coordinates in u is a scalar multiple of A.

A = (-2,-1,-2) u= (1,4,-3)

if AP is parallel to u then AP cross u = zero vector
This requires the operation of determinant.

But as Hallsofivy suggested, and, what I initially thought so, was through the use of scalar multiples.
If AP is parallel to u then AP is a scalar multiple of u such that
γAP = u or AP 1/γ (u)

8. Apr 1, 2014

### Staff: Mentor

No. The only possible interpretation of A is that it is a point. Otherwise $\vec{AP}$ wouldn't be meaningful.
To distinguish between points and vectors, I usually write vectors the angle brackets; e.g., u = <1, 4, -3>.
There is no need whatsoever to do it this way.
Yes, although you are missing an = in the second equation.

9. Apr 1, 2014

### negation

Woops.

So....γ(-2,-1,-2) = <1,4,-3>
-2γ = 1
-γ = 4
-2γ = -3

as can be seen, for all γ, γ differ in values. Therefore, the vector AP is not parallel to the point <u>.

10. Apr 1, 2014

### Staff: Mentor

NO!!
You are still treating A as if it were a vector, which it most emphatically is not! Since the question is asking about $\vec{AP}$, you need to come up with an expression for $\vec{AP}$. Since you're not given the coordinates for P (a point), how can you represent this point? Once you have P, then you can write an expression for the vector $\vec{AP}$.

You have misunderstood the question in this thread from the very start.

11. Apr 1, 2014

### jaytech

Just do a dot product and solve for the angle. It has to be zero degrees to be parallel.

$\vec{A}$$\cdot$$\vec{u}$ = |A|*|u|cosθ

12. Apr 1, 2014

### Staff: Mentor

No. A is not a vector. The question clearly asks about $\vec{AP}$, the displacement vector between the points A and P.

13. Apr 1, 2014

### jaytech

Ah, I misread it then, my apologies. Can use that to check the solution though :).

14. Apr 1, 2014

### LCKurtz

@negation: Do you know how to write the equation of the line through A with direction vector u?

15. Apr 1, 2014

### negation

I must have forgotten. Could I have some assistance?
edit: is the displacement AP expressed as AX =P?

Last edited: Apr 1, 2014
16. Apr 1, 2014

### Staff: Mentor

What does that even mean? A is a point.

If P1(x1, y1, z1) and P2(x2, y2, z2) are points in R3, then the displacement vector between these points is $\vec{P_1P_2}$ = $<x_2 - x_1, y_2 - y_1, z_2 - z_1>$. I hope that looks familiar.

17. Apr 1, 2014

### Staff: Mentor

There's a much simpler way to check that two vectors are parallel. The dot product is not needed.

18. Apr 1, 2014

### LCKurtz

Don't you have a text book that gives the equation of a line through a point P with direction vector D?

19. Apr 2, 2014

### negation

I don't. My course book covers only convuluted theorem and proof which makes it rather painful to understand for someone relatively new to vector and subspace.
I would really appreciate a leg-up

20. Apr 2, 2014

### negation

In order the find the displacement, it is p2 - p1?

21. Apr 2, 2014

### Staff: Mentor

Did you read what I wrote? What I have above is how to find the displacement vector between two points.

22. Apr 2, 2014

### negation

Sounds like I need to find the magnitude. I know the point A. But I do not know the point P.

23. Apr 2, 2014

### Staff: Mentor

You don't need the magnitude. Let P(x, y, z) be the unknown point.

This is really a very simple problem with very few calculations, but it does require some minimal knowledge about vectors.

24. Apr 2, 2014

### negation

Hi Mark,

If p(x,y,z) is the arbitrary point at which the vector head is then, the displacement is A-P =
(-2-x,-1-y,-2-z).

And let the displacement (-2-x,-1-y,-2-z) be AP.

Then, if u is parallel to AP, then it must be the case that γu = AP.

On a related note, I noticed that your previous post contained more information but which I was unable to capture while in the uni's library.

25. Apr 2, 2014

### LCKurtz

Have you taken calculus, through 3D lines, planes, vectors, etc? Is the material at this link familiar to you?

http://www.math.oregonstate.edu/hom...estStudyGuides/vcalc/lineplane/lineplane.html