Coplanar Vector QuestionPlease help

[spoc21]

Homework Statement

Determine if, $$\vec{w}$$₁ = $$\vec{2v}$$₁ + $$\vec{3v}$$₂ , $$\vec{w}$$₂ = $$\vec{v}$$₂ + $$\vec{2v}$$₃ and $$\vec{w}$$₃ = $$\vec{-v}$$₁ - $$\vec{3v}$$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!

 

[Mark44]

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.

 

[spoc21]

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$$\vec{w1}$$ sorry about that, I am new to these forums..

 

[Mark44]

What information are you given about v₁, v₂, and v₃?

 

[spoc21]

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.

 

[Mark44]

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?

 

[spoc21]

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!

 

[Mark44]

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.

 

[spoc21]

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..

 

[Mark44]

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?

 

[spoc21]

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.

 

[Mark44]

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.

 

[spoc21]

This is what I ended up with:

$$\vec{w}$$₁ = s $$\vec{w}$$₂ + t $$\vec{w}$$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!

 

[spoc21]

Bump...anyone?

 

[Mark44]

This is what I ended up with:

$$\vec{w}$$₁ = s $$\vec{w}$$₂ + t $$\vec{w}$$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!

 

[spoc21]

ok thank you!