# So, the title is: Parallel Vectors: Is AP Parallel to u?

• negation
In summary, the problem is asking to find the coordinates of a point P such that the vector AP is parallel to the given vector u. This can be done by finding a scalar multiple x, such that xAP = u, and using the coordinates of A and u to find the coordinates of P. The dot product is not necessary for this problem.
negation
Mod note: Corrected a typo in the problem statement.

## Homework Statement

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

u= (1,4,-3)

A(-2,-1,-2)

## 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:
negation said:

## Homework Statement

If the vector AP parallel to u?

u= (1,4,-3)

A(-2,-1,-2)
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.

## 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.
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?

I think it's a matter of setting up the question that I'm muddled.

Anyway, my attempt.

Let A = a1,a2,a3
Didn't you just tell us that A was (-2, -1, -2)?

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

-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
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.

1 person
negation said:

## Homework Statement

If the vector AP parallel to u?

u= (1,4,-3)

A(-2,-1,-2)
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.
negation said:

## 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
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.
negation said:
(a1,a2,a3)(-2,-1,-2) = (1,4,3)
No.
negation said:
-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

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

1 person
Mark44 said:
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

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.

HallsofIvy said:
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.

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).

negation said:
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).
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?

Mark44 said:
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?

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)

negation said:
I think I do. Both vectors are parallel if every coordinates in u is a scalar multiple of A.
No. The only possible interpretation of A is that it is a point. Otherwise ##\vec{AP}## wouldn't be meaningful.
negation said:
A = (-2,-1,-2) u= (1,4,-3)
To distinguish between points and vectors, I usually write vectors the angle brackets; e.g., u = <1, 4, -3>.
negation said:
if AP is parallel to u then AP cross u = zero vector
This requires the operation of determinant.
There is no need whatsoever to do it this way.
negation said:
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)
Yes, although you are missing an = in the second equation.

Mark44 said:
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.

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>.

negation said:
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>.
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.

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

jaytech said:
Just do a dot product and solve for the angle. It has to be zero degrees to be parallel.
No. A is not a vector. The question clearly asks about ##\vec{AP}##, the displacement vector between the points A and P.
jaytech said:
$\vec{A}$$\cdot$$\vec{u}$ = |A|*|u|cosθ

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

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

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

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

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

jaytech said:
Ah, I misread it then, my apologies. Can use that to check the solution though :).
There's a much simpler way to check that two vectors are parallel. The dot product is not needed.

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

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

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

LCKurtz said:
Don't you have a textbook that gives the equation of a line through a point P with direction vector D?

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

Mark44 said:
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.

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

Mark44 said:
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.
negation said:
In order the find the displacement, it is p2 - p1?
Did you read what I wrote? What I have above is how to find the displacement vector between two points.

Mark44 said:
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.

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

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

Mark44 said:
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.

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.

LCKurtz said:
Don't you have a textbook that gives the equation of a line through a point P with direction vector D?

negation said:
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

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

LCKurtz said:
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

I did but the whole parametric topic was only briefly covered. In fact, it was covered under precalculus, not calculus.

My current unit revolves around vector, subspace, multivariable, etc. I do not think it is a difficult topic because looking at books like Thomas calculus, the concepts can be easily understood and readily applied. But the way my course book is authored, it focuses heavily on theorem and proof.
It is painful, very very painful for a new entree.

negation said:
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).
Actually, this would be ##\vec{PA}##, the displacement vector from P to A.
negation said:
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.
OK, put in the coordinates for u and ##\vec{AP}##.

Also, I wouldn't use y as the scalar multiple. I would use k or c, but not x or y.
negation said:
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.
What information was that? Are you referring to my post or someone else's?

Well, you have a point A and a direction vector u and the link I gave you shows you how to write the equation of the line through A with direction vector u. Any point on that line except A itself solves your original problem.

Mark44 said:
Actually, this would be ##\vec{PA}##, the displacement vector from P to A.
OK, put in the coordinates for u and ##\vec{AP}##.

Also, I wouldn't use y as the scalar multiple. I would use k or c, but not x or y.

What information was that? Are you referring to my post or someone else's?

#16.

I wasn't able to view the itex in my school's library.

LCKurtz said:
Well, you have a point A and a direction vector u and the link I gave you shows you how to write the equation of the line through A with direction vector u. Any point on that line except A itself solves your original problem.

I will work through it.
Edit: coursebook has a similar feel to 'elementary differential equations with linear algebra' written by Albert L. Rabenstein.

The vector PA: <p1+2, p2+1,p3+2>

<p1+2, p2+1,p3+2> = k<1,4,-3>
<p1+2, p2+1,p3+2> = <γ,4γ,3γ>

p1+2 = k
p2+1 = 4k
p3+2 = -3k

Suppose <p1,p2,p3> = <1,2,3>

The k values are inconsistent.

negation said:
The vector PA: <p1+2, p2+1,p3+2>
This is actually ##\vec{AP}##.

##\vec{PA}## would be <-2 - p1, -1 - p2, -2 - p3>. These two vectors point in the opposite directions, so ##\vec{AP} = - \vec{PA}##.
negation said:
<p1+2, p2+1,p3+2> = k<1,4,-3>
<p1+2, p2+1,p3+2> = <γ,4γ,3γ>
From the previous equation, the right side above should be <k, 4k, -3k>.
negation said:
p1+2 = k
p2+1 = 4k
p3+2 = -3k

Suppose <p1,p2,p3> = <1,2,3>
##\vec{AP}## and ##\vec{u}## are parallel, so ##\vec{AP} = k\vec{u}## for some scalar k, with k ≠ 0.

Instead of picking some random values for the coordinates of ##\vec{AP}##, pick a value for k.
negation said:
The k values are inconsistent.

Last edited:
negation said:
The vector PA: <p1+2, p2+1,p3+2>

That isn't PA, it is AP, which is what you want.

<p1+2, p2+1,p3+2> = k<1,4,-3>
<p1+2, p2+1,p3+2> = <γ,4γ,3γ>

p1+2 = k
p2+1 = 4k
p3+2 = -3k

Suppose <p1,p2,p3> = <1,2,3>

The k values are inconsistent.

Big surprise they are inconsistent. Where did <1,2,3> come from? Did you just pick it out of the air?? It's not surprising that guessing values for the p's doesn't work.

To solve the problem, you need to find any ##p_1,p_2,p_3## values and ##k## that do work.

Took too long typing this, Mark beat me to it.

Mark44 said:
This is actually ##\vec{AP}##.

##\vec{PA}## would be <-2 - p1, -1 - p2, -2 - p3>. These two vectors point in the opposite directions, so ##\vec{AP} = - \vec{PA}##.
From the previous equation, the right side above should be <k, 4k, -3k>.
##\vec{AP}## and ##\vec{u}## are parallel, so ##\vec{AP} = k\vec{u}## for some scalar k, with k ≠ 0.

Instead of picking some random values for the coordinates of ##\vec{AP}##, pick a value for k.

let k = 1

p1+2 = 1 ∴p1=-1
p2+1=4 ∴p2=3
p3+2=-3 ∴p3=-5

I'm very sure it doesn't end here.

Last edited by a moderator:
Way back in the OP, the question was, "Is the vector ##\vec{AP}## parallel to u?" Apparently the coordinates of P were not given (or at least, they weren't shown in this thread). I believe the answer would be showing the conditions on the coordinates of P, something like this:
p1 = ...
p2 = ...
p3 = ...

These three equations would involve the parameter k and the coordinates of u.

It has taken a long while and many posts in this thread to get to this point, so a summary of where you're going might be useful.

1. Two vectors are parallel if either one is a scalar multiple of the other. In light of this problem, the equation is ##\vec{AP} = k\vec{u}##, where A is the point (-2, -1, -2) and u is the vector <1, 4, -3>. The point P wasn't given, so we used variables to represent it: P(p1, p2, p3).
2. Write the position vector ##\vec{AP}##, which is <p1 - (-2), p2 - (-1), p3 - (-2)>, and set this equal to ku, which is <k, 4k, -3k>.
3. Solve the equation for the unknowns p1, p2, and p3.

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