Solving 2-D vector along three directions (say at 0,60 and 120 degree)

[Mohmmad Maaitah]

Homework Statement

1-Can you resolve a 2-D vector along two directions which are not at 90 degrees to each other?
A- Yes, but not uniquely B- No. C-Yes uniquely


2-Can you resolve a 2-D vector along three directions (say at 0, 60 and 120 degrees)?
A-Yes, but not uniquely B-No. C-Yes uniquely.

Relevant Equations

None

So for 1 I know it's Yes you can, but I don't understand what uniquely means here so I can't say if it's uniquely or not.
for 2 I've never seen a 2-D vector broken into 3 reference axies so I guess No?
What really confuse me is the answers which goes 1-C and 4-A

 

[BvU]

Hi,

What you do when resolving a vector is to write that vector as a sum of the constituents, for example:$$\vec x = a\,\vec u + b\,\vec v$$where $a$ and $b$ are scalars and $\vec u$ and $\vec v$ are the constituents ('basis vectors'). Two vectors $\vec u$ and $\vec v$ in 2-D form a basis if they are independent, i.e. $\vec 0 = a\,\vec u + b\,\vec v \Rightarrow a=0 \ \& \ b=0$. The decomposition is unique; basically () you have two equations with two unknowns when you solve $\vec x = a\,\vec u + b\,\vec v \$.

And, going to exercise 4: once you have resolved $\vec x$ as $\vec x = a\,\vec u + b\,\vec v$, you have $$ \vec 0 = a\,\vec u + b\,\vec v -\vec x,$$ in other words: the decomposition of $\vec y$ as a sum of $d\,\vec u + e\,\vec v + f\, \vec x$ is no longer unique: you can always add a multiple of (a,b,-1) to (d,e,f) and still get the same $\vec y$.

The equation equivalent is two equations with three unknowns - there is one degree of freedom left over.

Does this answer your question ?

$\$

 

[jbriggs444]

So for 1 I know it's Yes you can, but I don't understand what uniquely means here so I can't say if it's uniquely or not.

For 1, you have a vector and you are asked about expressing it as a "linear combination" of two other vectors.
That is, you are asked about expressing the resultant vector $\vec{R}$ as some multiple $a$ of vector $\vec{x}$ plus some other multiple $b$ of vector $\vec{y}$. That is what "resolve" mean here.

You seem to realize that as long as $\vec{x}$ and $\vec{y}$ are not parallel then it is possible to express any vector $\vec{R}$ as a linear combination of $\vec{x}$ and $\vec{y}$:$$\vec{R} = a \vec{x} + b \vec{y}$$But you are concerned about the word "uniquely". What does it mean in this context?

Unique means that if you can find values for $a$ and $b$ that work, those are the only values that work. Any other pair of multiples will give you a result that differs from $\vec{R}$.

for 2 I've never seen a 2-D vector broken into 3 reference axies so I guess No?

For 2, it is the same idea. But this time you are asked about values for $a$, $b$ and $c$ that satisfy:$$\vec{R} = a \vec{x} + b \vec{y} + c \vec{z}$$Well, if you can express $\vec{R}$ as some multiple of $\vec{x}$ plus some multiple of $\vec{y}$ then adding in a multiple of $\vec{z}$ is no problem. Just set $c$ to zero.

But given values for $a$, $b$ and $c$ that work this time, are they unique? Will any other set of three values work? What are your thoughts on that?

 

[Mohmmad Maaitah]

Hi,

What you do when resolving a vector is to write that vector as a sum of the constituents, for example:$$\vec x = a\,\vec u + b\,\vec v$$where $a$ and $b$ are scalars and $\vec u$ and $\vec v$ are the constituents ('basis vectors'). Two vectors $\vec u$ and $\vec v$ in 2-D form a basis if they are independent, i.e. $\vec 0 = a\,\vec u + b\,\vec v \Rightarrow a=0 \ \& \ b=0$. The decomposition is unique; basically () you have two equations with two unknowns when you solve $\vec x = a\,\vec u + b\,\vec v \$.

And, going to exercise 4: once you have resolved $\vec x$ as $\vec x = a\,\vec u + b\,\vec v$, you have $$ \vec 0 = a\,\vec u + b\,\vec v -\vec x,$$ in other words: the decomposition of $\vec y$ as a sum of $d\,\vec u + e\,\vec v + f\, \vec x$ is no longer unique: you can always add a multiple of (a,b,-1) to (d,e,f) and still get the same $\vec y$.

The equation equivalent is two equations with three unknowns - there is one degree of freedom left over.

Does this answer your question ?

$\$

Thanks for your time, but I'm totally sure this is more complicated than what I thought it would be :O
I think you missed my simple question, what does unique mean and can I break a 2D vector into 3 components? (If I got the question right)

 

[Mohmmad Maaitah]

For 1, you have a vector and you are asked about expressing it as a "linear combination" of two other vectors.
That is, you are asked about expressing the resultant vector $\vec{R}$ as some multiple $a$ of vector $\vec{x}$ plus some other multiple $b$ of vector $\vec{y}$. That is what "resolve" mean here.

You seem to realize that as long as $\vec{x}$ and $\vec{y}$ are not parallel then it is possible to express any vector $\vec{R}$ as a linear combination of $\vec{x}$ and $\vec{y}$:$$\vec{R} = a \vec{x} + b \vec{y}$$But you are concerned about the word "uniquely". What does it mean in this context?

Unique means that if you can find values for $a$ and $b$ that work, those are the only values that work. Any other pair of multiples will give you a result that differs from $\vec{R}$.For 2, it is the same idea. But this time you are asked about values for $a$, $b$ and $c$ that satisfy:$$\vec{R} = a \vec{x} + b \vec{y} + c \vec{z}$$Well, if you can express $\vec{R}$ as some multiple of $\vec{x}$ plus some multiple of $\vec{y}$ then adding in a multiple of $\vec{z}$ is no problem. Just set $c$ to zero.

But given values for $a$, $b$ and $c$ that work this time, are they unique? Will any other set of three values work? What are your thoughts on that?

Oh thank you, this helps!
Okay so I now think it must be YES for both of 1 and 2 and both must be uniquely right?
Because it only works only one way for each question!
Please correct me if I'm wrong!

 

[BvU]

Not that complicated: unique means $\vec x = a\,\vec u + b\,\vec v$ has one and only one solution $(a,b)$.
(or: $\vec x = a\,\vec u + b\,\vec v \quad \&\ \vec x = c\,\vec u + d\,\vec v \Rightarrow a=c \ \&\ b=d$).

And yes, you can always write a 2-D vector as the sum of 3 components (as I've shown in post #2!), provided these component are not all three multiples of each other.

$\$

 

[jbriggs444]

Okay so I now think it must be YES for both of 1 and 2 and both must be uniquely right?
Because it only works only one way for each question!

The answer to 1 is indeed "Yes, uniquely".

But for number 2... Works only one way? Let us examine that assertion.

Say that we are trying to resolve $\vec{R}$ as a linear combination of $\vec{x}$, $\vec{y}$ and $\vec{z}$:$$\vec{r} = a\vec{x} + b\vec{y} + c\vec{z}$$

Let $c$ be whatever value you please.

We know from the answer to 1 that we can express $\vec{r} - c\vec{z}$ as a linear combination of $\vec{x}$ and $\vec{y}$.

Can you see how that gives us a lot of freedom to choose a value for $c$. What does that say about uniqueness?

 

[Mohmmad Maaitah]

But C must only be 0 to express the 2D vector!

 

[Mohmmad Maaitah]

Can you see how that gives us a lot of freedom to choose a value for $c$. What does that say about uniqueness?

I am sorry, I can't see how we are free to use any value for C expect 0 when we talk about 2D vector.

 

[Mohmmad Maaitah]

(or: $\vec x = a\,\vec u + b\,\vec v \quad \&\ \vec x = c\,\vec u + d\,\vec v \Rightarrow a=c \ \&\ b=d$).$\$

This confuses me a lot!

 

[BvU]

This confuses me a lot!

In what way ? Uniqueness means there is no more than one solution (one decomposition). Does the math to express that confuse you ? Can you try to formulate/define uniqueness in your own words ?

 

[Steve4Physics]

@Mohmmad Maaitah, considering this visually may help.

Suppose we want to resolve a vector into 3 components where the required directions of the components are at angles of 0º, 30º and 60º to the horizontal.

Here a 2 different ways:

Since there is more than one way, the resolution is not unique.

(I hope you can see that there are not just 2, but an infinite number of ways to do the resolution.)

 

[Mohmmad Maaitah]

In what way ? Uniqueness means there is no more than one solution (one decomposition). Does the math to express that confuse you ? Can you try to formulate/define uniqueness in your own words ?

Yeah it's the math that still confuse me can't get hold of it.
the post #12 helped me get it anyway!

 

[Mohmmad Maaitah]

I would love to see more simple math expression.
Thank you this helped a lot.

View attachment 333573ways:
Since there is more than one way, the resolution is not unique.

(I hope you can see that there are not just 2, but an infinite number of ways to do the resolution.)