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Coplanar Vector QuestionPlease help!

  • Thread starter spoc21
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  • #1
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Homework Statement


Determine if, [tex]\vec{w}[/tex]1 = [tex]\vec{2v}[/tex]1 + [tex]\vec{3v}[/tex]2 , [tex]\vec{w}[/tex]2 = [tex]\vec{v}[/tex]2 + [tex]\vec{2v}[/tex]3 and [tex]\vec{w}[/tex]3 = [tex]\vec{-v}[/tex]1 - [tex]\vec{3v}[/tex]3are coplanar.

⃗w1 = ⃗2v1 + ⃗3v2 , ⃗w2 = ⃗v2 + ⃗2v 3 and ⃗w3 = ⃗-v1 - ⃗3v3


The Attempt at a Solution



I tried to multiply it by scalars, k and l, but am very confused with the result...I would really appreciate any help!
 
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Answers and Replies

  • #2
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Homework Statement


Determine if, ⃗w1 = ⃗2v1 + ⃗3v2 , ⃗w2 = ⃗v2 + ⃗2v 3 and ⃗w3 = ⃗-v1 - ⃗3v3 are coplanar.



The Attempt at a Solution



I tried to multiply it by scalars, k and l, but am very confused with the result...I would really appreciate any help!
I am very confused with the question. Many of the characters you used render as square boxes.
 
  • #3
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I am very confused with the question. Many of the characters you used render as square boxes.
theyre vectors, like w1 is supposed to have an arrow above it[tex]\vec{w1}[/tex] sorry about that, Im new to these forums..
 
  • #4
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What information are you given about v1, v2, and v3?
 
  • #5
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What information are you given about v1, v2, and v3?

They are just variables, I'm supposed to determine if they're coplanar..I assumed that multiplying each by scalar, say k and l would work, but it got kind of confusing.
 
  • #6
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v1, v2, and v3 are vectors, so if they all happen to be coplanar, then so are w1, w2, and w3.

Are you sure there isn't some other information given about the B]v[/B]ivectors?
 
  • #7
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this is what I got

w1 = 2v1 + 3v2
w2 = v2 + 2v3
w3 = -v1 - 3v3


so what I did was:

solve for v2

v2 = w2 - 23

now I substituted v2 into eq 1

w1 = 2v1 + 3 (w2 -23)


w1 = 2v1 + 3w2 -63)


Now I found the value o w2 into our new equation

w1 = 2v1 + 3(v2 + 2v3) -63

w1 = 2v1 + 3v2 + 6v3 -63 (6 and 6 cancel out)


so we have :
w1 = 2v1 + 3v2


we know from our equations that w1 = 2v1 + 3v2


so therefore:

2v1 + 3v2 = 2v1 + 3v2


= 0

Since the sum = 0, we know that these vectors are linearly dependent, and hence they are coplanar..


Can you please tell if my work is correct..

Thank You!
 
  • #8
33,070
4,771
this is what I got

w1 = 2v1 + 3v2
w2 = v2 + 2v3
w3 = -v1 - 3v3


so what I did was:

solve for v2

v2 = w2 - 23
Typo in line above.
now I substituted v2 into eq 1

w1 = 2v1 + 3 (w2 -23)


w1 = 2v1 + 3w2 -63)


Now I found the value o w2 into our new equation

w1 = 2v1 + 3(v2 + 2v3) -63

w1 = 2v1 + 3v2 + 6v3 -63 (6 and 6 cancel out)


so we have :
w1 = 2v1 + 3v2
This (above) was given.
we know from our equations that w1 = 2v1 + 3v2
We also know that the equation above was given. There's no need to derive given information.
so therefore:

2v1 + 3v2 = 2v1 + 3v2
Well, duh. It's always true that something equals itself. Am I missing something here?
= 0
How can you say this? I can always say x = x, but I can't then come along and say that x = 0 out of the blue.
Since the sum = 0, we know that these vectors are linearly dependent, and hence they are coplanar..
If w1, w2, and w3 are linearly dependent, it must be that one of them is either a multiple of another, or that one of them is a linear combination of the others.
Can you please tell if my work is correct..

Thank You!
Assuming that vectors v1, v2, and v3 are linearly independent, I found w1, w2, and w3 to be linearly independent as well, meaning that they are NOT coplanar.
 
  • #9
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Assuming that vectors v1, v2, and v3 are linearly independent, I found w1, w2, and w3 to be linearly independent as well, meaning that they are NOT coplanar.
Would you mind sharing how you came to this conclusion..
 
  • #10
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4,771
No, not at all.

Set c1 w1 + c2 w2 + c3w3 = 0. (1)

Clearly equation 1 above has at least one solution (c1 = c2 = c3 = 0), and if that's the only solution then the w's are linearly independent. If there are other solutions, then the w's are linearly dependent.

In the equation above, replace w1 by 2v1 + 3v2, and replace w2 and w3 using the given information.

Equation 1 now becomes
c1(2v1 + 3v2) + c2(v2 + 2v3) + c3(-v1 - 3v3) = 0

<==> (2c1 - c3)v1 + (3c1 + c2)v2 + (2c2 - 3c3)v3 = 0

We're assuming that v1, v2, and v3 are linearly independent, so that means that the equation above has exactly one solution; namely the one where the coefficients of the v's are all 0.

Can you take it from there? Also, I am tacitly assuming the v's are linearly independent, and you seem to be doing so, also. Is it given that these vectors are linearly independent?
 
  • #11
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Also, I am tacitly assuming the v's are linearly independent, and you seem to be doing so, also. Is it given that these vectors are linearly independent?
Actually, the question only says to determine if they are coplanar, nothing about linear dependence or independence. However, I'm assuming that if these vectors are linearly dependent, they must be coplanar, so that's why I mentioned it here.
 
  • #12
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4,771
Actually, the question only says to determine if they are coplanar, nothing about linear dependence or independence. However, I'm assuming that if these vectors are linearly dependent, they must be coplanar, so that's why I mentioned it here.
Try to be more precise in what you say. When you say "if these vectors are linearly dependent, they must be coplanar" does "they" refer to the v vectors or the w vectors? If you were referring to the w vectors, then you're not being clear in what you wrote.

Also, if you don't have any information about the v vectors, there's not really much you can say about the w vectors.
 
  • #13
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This is what I ended up with:

[tex]\vec{w}[/tex]1 = s [tex]\vec{w}[/tex]2 + t [tex]\vec{w}[/tex]3


2v1 + 3v2 + 03 = sv2 + 2sv3 - tv1 - 3tv3

so we gather the like terms:

2v1 = -tv 1 ......(1)
3v2 = sv2.........(2)
2sv3 - 3tv3 = 0.......(3)

now we take out the common variables in each, v1 in eq (1), and v2 in eq (2), finding the value of the scalars s, and t.

so therefore t = -2
and s = 3

We substitute these values into eq (3), which is 2sv3 - 3tv3 = 0, and see if they satisfy the equation:

2(3) - 3(-2)
= 12


The values of s and t don't satisfy the equation illustrating that they are not coplanar..

Am I right?

Thanks!
 
  • #14
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Bump...anyone?
 
  • #15
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4,771
This is what I ended up with:

[tex]\vec{w}[/tex]1 = s [tex]\vec{w}[/tex]2 + t [tex]\vec{w}[/tex]3
How did you end up with this? Instead it looks like you're starting by assuming this, which means that you're assuming that the vectors are coplanar. This is valud to do if you suspect that the vectors are not coplanar, and you're trying for a proof by contradiction. I don't think this is what you are trying to do, so I would advise against making this assumption.

Go back to what I said in post 10, in the part with this equation:
(2c1 - c3)v1 + (3c1 + c2)v2 + (2c2 - 3c3)v3 = 0

For this equation to be true for arbitrary vectors v1, v2, and v3, the coefficients of these vectors must be zero, so solve for the constants c1, c2, and c3. If you get one solution (the zero solution) the vectors w1, w2, and w3 are linearly independent, hence not coplanar. If you get more than one solution, the w vectors are linearly dependent, hence coplanar.

2v1 + 3v2 + 03 = sv2 + 2sv3 - tv1 - 3tv3

so we gather the like terms:

2v1 = -tv 1 ......(1)
3v2 = sv2.........(2)
2sv3 - 3tv3 = 0.......(3)

now we take out the common variables in each, v1 in eq (1), and v2 in eq (2), finding the value of the scalars s, and t.

so therefore t = -2
and s = 3

We substitute these values into eq (3), which is 2sv3 - 3tv3 = 0, and see if they satisfy the equation:

2(3) - 3(-2)
= 12


The values of s and t don't satisfy the equation illustrating that they are not coplanar..

Am I right?

Thanks!
 
  • #16
87
0
ok thank you!
 

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