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

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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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Geometrically, what does the dot product represent?
 


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


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?
 


v•w = ||v|| ||w|| cosθ

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


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?
 


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.
 


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.
 


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:

v \cdot w = |v| |w| \cos \theta

where theta is the angle between the two vectors.

In your original problem statement, you also indicated that:

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

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.
 
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