How can I solve this normed inner product space problem without a prefix?

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In summary, the conversation discusses the dot product and its geometric representation, as well as the relationship between the dot product and vector lengths and angles. The problem given involves finding the length of a vector, and the conversation provides hints and steps for solving it using properties of inner products.
  • #1
Sheldinoh
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This question was on my math calc and analytical geometry final and i couldn't figure out how to do this. Its probably something pretty simple but i couldn't figure it out

1. Homework Statement [/b]
||v|| = 1 ||w|| = 2 v•w= (-1/2)||v|| ||w||
||3v+w|| = ?

Thanks in advance for the help
 
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  • #2


Geometrically, what does the dot product represent?
 
  • #3


im not really sure. is it that the dot product geographically represents the length of one vector over another vector
 
  • #4


Geometrically, the lengths of the vectors are involved in the dot product, but so is the angle between the two vectors.

Do you know what this relationship is?
 
  • #5


v•w = ||v|| ||w|| cosθ

but how would i get to ||3v+w|| from knowing this
 
  • #6


Sheldinoh said:
v•w = ||v|| ||w|| cosθ

but how would i get to ||3v+w|| from knowing this

For inner products (including the dot product), we have a few definitions that we use.

The first one is that we define the norm of a vector through its inner product to be
|v| = SQRT(<v,v>) >= 0.

We also define the inner product to be distributive ie <(v + w),x> to be <v,x> + <w,x>

Given these definitions, can you now solve the problem?
 
  • #7


would it be:
||3v+w||= ( 16x2+(3vy + wy)2 ) 1/2 .

If so, i was kinda under the impression that the professor wanted an numerical answer.
 
  • #8


Sheldinoh said:
would it be:
||3v+w||= ( 16x2+(3vy + wy)2 ) 1/2 .

If so, i was kinda under the impression that the professor wanted an numerical answer.

I'll give you a hint:

||3v + w|| = SQRT(<3v + w,3v + w>) (using the fact that |v| = SQRT(<v,v>)).

Think about the other properties of the dot (and in general inner) products.
 
  • #9


Sheldinoh said:
im not really sure. is it that the dot product geographically represents the length of one vector over another vector
I think you are trying to say it is the length of the projection of one vector on the other.
 
  • #10


Sheldinoh said:
v•w = ||v|| ||w|| cosθ

but how would i get to ||3v+w|| from knowing this
Yes, correct. With inner products, we have:

[tex]v \cdot w = |v| |w| \cos \theta [/tex]

where theta is the angle between the two vectors.

In your original problem statement, you also indicated that:

[tex]v \cdot w = \left(-\frac{1}{2}\right) |v| |w|[/tex]

Now, as a next step, using these two facts, are you able to deduce what the cosine of the angle between the two vectors is? And from there, deduce what the actual angle is?
 
  • #11


Sheldinoh said:
would it be:
||3v+w||= ( 16x2+(3vy + wy)2 ) 1/2 .

If so, i was kinda under the impression that the professor wanted an numerical answer.

Building on my first hint, we can use the property of distributivity:

||3v + w|| = SQRT(<3v + w, 3v +w>) = SQRT(<3v + w,3v> + <3v + w,w>)
= SQRT(<3v,3v> + <3v,w> + <3v,w> + <w,w>)
= SQRT(9<v,v> + <w,w> + 6<v,w>)

Given what you know, you should be able to find the numeric answer to your question.
 
  • #12


chiro said:
Building on my first hint, we can use the property of distributivity:

||3v + w|| = SQRT(<3v + w, 3v +w>) = SQRT(<3v + w,3v> + <3v + w,w>)
= SQRT(<3v,3v> + <3v,w> + <3v,w> + <w,w>)
= SQRT(9<v,v> + <w,w> + 6<v,w>)

Given what you know, you should be able to find the numeric answer to your question.
This is the quick and, in some ways, "simple" way to get the answer.

But I'm also thinking that most people don't see normed inner product spaces until they get to a course on real analysis, which might be a bit ahead of where the OP is at.

But yes, you've basically laid out the right answer here..
 
  • #13


psholtz said:
This is the quick and, in some ways, "simple" way to get the answer.

But I'm also thinking that most people don't see normed inner product spaces until they get to a course on real analysis, which might be a bit ahead of where the OP is at.

But yes, you've basically laid out the right answer here..

Ohhh ok. In my math degree we were introduced to inner product spaces in a second year one semester linear algebra course and all the definitions like distributivity, linearity, positive definiteness and so on were done in about 10 minutes.

To the OP, if you haven't done things in this way and you've done linear algebra, get out a book and learn about inner product spaces and the criteria for an inner product to have. The inner product spaces build on vector spaces.
 

1. What are norms of vectors?

Norms of vectors are mathematical quantities used to measure the size or magnitude of a vector. They are often represented by the symbol ||x|| and can be calculated using different formulas depending on the type of norm being used.

2. Why are norms of vectors important?

Norms of vectors are important because they provide a way to quantify the magnitude of a vector, which can be useful in a variety of applications including physics, engineering, and data analysis. They also allow for comparisons between vectors and can be used to determine the distance between two vectors.

3. What are some common types of norms used in vector problems?

Some common types of norms used in vector problems include the Euclidean norm (also known as the 2-norm), the Manhattan norm (also known as the 1-norm), and the maximum norm (also known as the infinity norm). Each of these norms has its own formula for calculating the magnitude of a vector.

4. How do you calculate the Euclidean norm of a vector?

The Euclidean norm, also known as the 2-norm, is calculated by taking the square root of the sum of the squared components of a vector. In other words, the Euclidean norm of a vector x = (x1, x2, x3) is calculated as ||x|| = √(x1^2 + x2^2 + x3^2).

5. Can norms of vectors be negative?

No, norms of vectors cannot be negative. They are always positive values because they represent the magnitude or size of a vector, which cannot be negative. However, in some cases, the result of a norm calculation may be zero, indicating that the vector has no magnitude or is a zero vector.

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