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Determine components of a vector without knowing the angle

  • #1
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Homework Statement



(I roughly translate the problem statement from German)
Given the vectors a = (1,-2,3) and b = (1,1,1), divide the vector a in two components a1 (parallel to b) and a2 (perpendicular to b).

Homework Equations



In a previous question of the problem, I found that:
a.b = 2
a x b = (-5, 2, 3)
|a| = √(14) ≈ 3.74
|b| = √(3) ≈ 1.73

So far, our class has been mentioning vector addition, multiplication by a real number, components of a vector, scalar product, vector product, as well as diverse properties of the operations I just listed (commutativity, distributivity, associativity and homogeneity).

The Attempt at a Solution



My problem here is that the angle between the vectors is not given in the problem. I found several ways to calculate it with the use of arccos, but since that was not yet mentioned in my class, I am reluctant to use it.

What I know so far is that:
(I'll call ∂ the angle between a and b)
a1 = |a| cos ∂
a2 = |a| sin ∂
a1 x b = 0 (because a1 and b are parallel, their cross product is equal to 0)
a2.b = 0 (because a2 and b are perpendicular, their scalar product is equal to 0)
since a.b = |a|.|b|.cos ∂, I also find that cos ∂ = 2/√(14).√(3) ≈ 0.31
I also know that a = √(a1^2 + a2^2) since the angle between a1 and a2 is π/2 rad.

What am I missing to solve this problem? I just started a program of physic after many years without maths, so I need to refresh a bit :)


Thank you very much for your answers, I appreciate it.


Julien.
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
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First you can find the angle because [itex]\vec{a}\cdot\vec{b}= |\vec{a}||\vec{b}| cos(\theta)[/itex].

But you don't need to find the angle, directly, because component of [itex]\vec{a}[/itex] parallel to [itex]\vec{b}[/itex] has length [itex]|\vec{a}| cos(\theta)[/itex] and then replace that cosine with [itex]\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}[/itex] to get [itex]\frac{\vec{a}\cdot\vec{b}}{|\vec{b}}[/itex] as the length. Of course the vector component parallel to [itex]\vec{b}[/itex] is that times a unit vector in the [itex]\vec{b}[/itex] direction.
 
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  • #3
DEvens
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Remember the definition of dot product. Remember it both in terms of components of the vector, and in terms of the cosine of the angle between the two vectors.
 
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  • #4
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Remember the definition of dot product.
Slight correction: definitions.
DEvens lists both definitions: the component definition, and the coordinate-free definition; i.e. ##\vec{a} \cdot \vec{b} = |\vec{a} | |\vec{b} | \cos(\theta)##.
 
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  • #5
Ray Vickson
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Homework Statement



(I roughly translate the problem statement from German)
Given the vectors a = (1,-2,3) and b = (1,1,1), divide the vector a in two components a1 (parallel to b) and a2 (perpendicular to b).

Homework Equations



In a previous question of the problem, I found that:
a.b = 2
a x b = (-5, 2, 3)
|a| = √(14) ≈ 3.74
|b| = √(3) ≈ 1.73

So far, our class has been mentioning vector addition, multiplication by a real number, components of a vector, scalar product, vector product, as well as diverse properties of the operations I just listed (commutativity, distributivity, associativity and homogeneity).

The Attempt at a Solution



My problem here is that the angle between the vectors is not given in the problem. I found several ways to calculate it with the use of arccos, but since that was not yet mentioned in my class, I am reluctant to use it.

What I know so far is that:
(I'll call ∂ the angle between a and b)
a1 = |a| cos ∂
a2 = |a| sin ∂
a1 x b = 0 (because a1 and b are parallel, their cross product is equal to 0)
a2.b = 0 (because a2 and b are perpendicular, their scalar product is equal to 0)
since a.b = |a|.|b|.cos ∂, I also find that cos ∂ = 2/√(14).√(3) ≈ 0.31
I also know that a = √(a1^2 + a2^2) since the angle between a1 and a2 is π/2 rad.

What am I missing to solve this problem? I just started a program of physic after many years without maths, so I need to refresh a bit :)


Thank you very much for your answers, I appreciate it.


Julien.
Look up "orthogonal projection"; see, eg., https://en.wikipedia.org/wiki/Vector_projection or
https://en.wikibooks.org/wiki/Linear_Algebra/Orthogonal_Projection_Onto_a_Line
 
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  • #6
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To begin, thank you very much for all of your answers. That is really helpful!

I understand now the concept better, but I must still be making a mistake somewhere:

Since a.b = |a|.|b|.cos ∂ & a1 = |a|.cos ∂,
a.b = |b|.a1
So a1 = (a.b)/|b| = 2/√3 (if my first answers for the scalar product and for the magnitude were correct)

But there is still something wrong, because it is not a unit vector, and I cannot seem to find the values of the vector that respect the property of two perpendicular vectors a1.b = 0. Did I get the wrong result, or am I misunderstanding what I am supposed to find?


Sorry to ask again, I am sure it is pretty easy and I must quickly integrate it in my brain :)
 
  • #7
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Maybe I am misunderstanding what a unit vector is supposed to be... Here I have a definition that says "A unit vector is any vector with a magnitude of 1." But what I found was that a1 = 2/√3. And when I try to figure out what would be a1(x,y,z), I go like that:

Since a1 and b are parallel, there exists a real number k so that a1 = kb;
a1 = k(1 + 1 + 1) = 3k
k = a1/3 = (2/√3)/3 = (2√3)/9
thus defining the vector a1 = ((2√3)/9,(2√3)/9,(2√3)/9)

Ehem... Not only does this result look very strange, but it is also incompatible with the property of parallel vectors a1.b = 0:
a1.b = (2√3)/9 + (2√3)/9 + (2√3)/9 = (6√3)/9 = (2√3)/3 ≠ 0!

Is it in the calculus that I went wrong? Or is it my reasoning that is flawed? From another way of thinking that does not really rely on any theorem, I instinctively come to a result a1 = (1/3,1/3,1/3). Not only does that verify the property of parallel vectors a1.b = 0, but such vector would also have a magnitude of 1!


Thank you very much if you can clear up my confusion, I appreciate your help.


Julien.
 
Last edited:
  • #8
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Maybe I am misunderstanding what a unit vector is supposed to be... Here I have a definition that says "A unit vector is any vector with a magnitude of 1." But what I found was that a1 = 2/√3.
No. a1 should be a vector, not a scalar.
JulienB said:
And when I try to figure out what would be a1(x,y,z), I go like that:

Since a1 and b are parallel, there exists a real number k so that a1 = kb;
a1 = k(1 + 1 + 1) = 3k
No again. a1 should be a scalar multiple of b, so a1 = k<1, 1, 1> = <k, k, k>.
I think you might be confusing the ideas of a vector vs. the magnitude of a vector. IOW, confusion between v and |v|.
JulienB said:
k = a1/3 = (2/√3)/3 = (2√3)/9
thus defining the vector a1 = ((2√3)/9,(2√3)/9,(2√3)/9)
This vector has the right direction, but not the right length (magnitude).
JulienB said:
Ehem... Not only does this result look very strange, but it is also incompatible with the property of parallel vectors a1.b = 0:
Right, because a1 is supposed to be parallel to b, not perpendicular to it. If these two vectors were perpendicular, their dot product would be zero.
JulienB said:
a1.b = (2√3)/9 + (2√3)/9 + (2√3)/9 = (6√3)/9 = (2√3)/3 ≠ 0!

Is it in the calculus that I went wrong? Or is it my reasoning that is flawed? From another way of thinking that does not really rely on any theorem, I instinctively come to a result a1 = (1/3,1/3,1/3). Not only does that verify the property of parallel vectors a1.b = 0, but such vector would also have a magnitude of 1!


Thank you very much if you can clear up my confusion, I appreciate your help.


Julien.
 
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  • #9
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Thank you Mark for your answer. There must indeed be some confusion in my formulas. I still don´t get how to obtain a vector out of a scalar product in a fraction, in that case I am stuck at a1 = (a.b)/IbI.
 
  • #10
408
12
Is the magnitude I calculated |a1| = 2√3 wrong? Because whatever I try, I always come back to this very same result :(

|a1| = |a| cos ∂
a.b = |a|.|b|.cos ∂
cos ∂ = |a1|/|a|
Therefore, a.b = |a|.|b|.(|a1|/|a|) and |a1| = 2√3

That is a slight variation from the method suggested by Hallsofivy previously, and it gives me the same magnitude. I really dont get where is the mistake there..
 
  • #11
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Thank you Mark for your answer. There must indeed be some confusion in my formulas. I still don´t get how to obtain a vector out of a scalar product in a fraction, in that case I am stuck at a1 = (a.b)/IbI.
This formula doesn't make sense as you have written it. The left side is a vector, but the right side is a scalar.
The corrected formula is ##|\vec{a_1}| = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}##.
This gives you the length of the projection of a in the direction of b, i.e., the length of a1. To get the actual vector a1, multiply |a1| by a unit vector with the same direction. IOW, multiply by ##\frac{\vec{a_1}}{|\vec{a_1}|}##.
Is the magnitude I calculated |a1| = 2√3 wrong?
Yes, it's wrong. See above.
JulienB said:
Because whatever I try, I always come back to this very same result :(

|a1| = |a| cos ∂
a.b = |a|.|b|.cos ∂
cos ∂ = |a1|/|a|
Therefore, a.b = |a|.|b|.(|a1|/|a|) and |a1| = 2√3
It's hard to follow your thinking here.
First equation: Not useful, since you don't know |a1| and you don't know ##\theta##.
Second equation: Can be used to find ##\theta##, since you know a and b.
Third equation: Not useful, since you don't know ##\theta##.
Fourth equation: The equation is a true statement, but I don't know what you did to arrive at |a1| = 2√3. You can use what I said at the top of my post here to find |a1|, and then multiply by a suitable unit vector to get a1 itself.
JulienB said:
That is a slight variation from the method suggested by Hallsofivy previously, and it gives me the same magnitude. I really dont get where is the mistake there..
 
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  • #12
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I mistyped it actually, the answer I was getting is 2/√3, because the homework asked earlier to calculate the dot product a.b and the length of |b| (amongst other things).

a.b = a1b1 + a2b2 + a3b3 = 1 -2 + 3 = 2
|b| = √(b1^2+b2^2+b3^2) = √(1 + 1 + 1) = √3

So in the equation |a1| = a.b/|b|, I just replace a.b by 2 and |b| by √3.

I will take a deep look again at what you wrote and try to figure it out. I'm trying to understand why the length |a1| should be multiplied by a unit vector.


Thank you very much for your patience, I appreciate your efforts of explaining me the process. I'm working hard to soon not get stuck on such problems anymore. :)


Julien.
 
  • #13
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I mistyped it actually, the answer I was getting is 2/√3, because the homework asked earlier to calculate the dot product a.b and the length of |b| (amongst other things).
Yes, that's the right value for |a1|.
JulienB said:
a.b = a1b1 + a2b2 + a3b3 = 1 -2 + 3 = 2
|b| = √(b1^2+b2^2+b3^2) = √(1 + 1 + 1) = √3

So in the equation |a1| = a.b/|b|, I just replace a.b by 2 and |b| by √3.

I will take a deep look again at what you wrote and try to figure it out. I'm trying to understand why the length |a1| should be multiplied by a unit vector.
To make a vector with the right length (2/√3) and pointing in the right direction. In other words, to get what you're calling a1.
JulienB said:
Thank you very much for your patience, I appreciate your efforts of explaining me the process. I'm working hard to soon not get stuck on such problems anymore. :)
Julien.
 
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  • #14
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Yes, that's the right value for |a1|.
Wow that's great, I was scratching my head over and over again because I thought that was wrong.

To make a vector with the right length (2/√3) and pointing in the right direction. In other words, to get what you're calling a1.
Okay, I think I understand:
the "suitable unit vector" u to obtain a1 is a vector that would have the same direction as b and a magnitude of 1. For the magnitude of u to be 1, I must divide each of the component of b by |b| (= √3).

The unit vector u in direction of b would then be u = (1/√3,1/√3,1/√3).

Am I correct so far? It kinda look strange.. Then next step is to obtain a1 is:

a1 = |a1|.u = (2/3, 2/3, 2/3)

I hope I'm on the right track :)
 
  • #15
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Wow that's great, I was scratching my head over and over again because I thought that was wrong.



Okay, I think I understand:
the "suitable unit vector" u to obtain a1 is a vector that would have the same direction as b and a magnitude of 1. For the magnitude of u to be 1, I must divide each of the component of b by |b| (= √3).

The unit vector u in direction of b would then be u = (1/√3,1/√3,1/√3).

Am I correct so far? It kinda look strange.. Then next step is to obtain a1 is:

a1 = |a1|.u = (2/3, 2/3, 2/3)

I hope I'm on the right track :)
Yes, everything looks good.
 
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  • #16
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Thanks a lot for your help, I appreciate it.


J.
 
  • #17
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Sorry to come back, I have a short additional question: to determine a2, can I just make a2 = a - a1? I'm wondering because I get very complicated results.
 
  • #18
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Sorry to come back, I have a short additional question: to determine a2, can I just make a2 = a - a1? I'm wondering because I get very complicated results.
Yes. Since a = a1 + a2, you can subtract a1 from both sides to get your equation.
 

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