Basic Lin Alg - Angle Between Vectors

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Homework Help Overview

The discussion revolves around finding the angle between two vectors, specifically the vectors [-1, 1, 2] and [2, 1, -1]. The problem involves using the cosine formula related to the dot product and the magnitudes of the vectors.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the dot product and the norms of the vectors, with some questioning the interpretation of the dot product as a scalar versus a vector. There are also attempts to clarify the correct application of the cosine formula.

Discussion Status

There is ongoing clarification regarding the dot product and its properties. Some participants have identified errors in calculations, while others are exploring how to derive the angle from the cosine value. The discussion reflects a mix of attempts to correct misunderstandings and to verify calculations.

Contextual Notes

Participants mention careless errors in calculations and the need for careful interpretation of mathematical terms, indicating a focus on precision in mathematical reasoning.

sunmaz94
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Homework Statement



Find the angle between the vectors [-1, 1, 2] and [2, 1, -1].

Homework Equations



cos(theta) = <u, v> / ||u||||v||

The Attempt at a Solution



I get <u, v> = -3 and both norms equal to the square root of six. Then cos(theta) equals (-1/12), but I don't think this is correct. Can someone find my careless error? Thanks.
 
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Your <u,v> must be a vector, not a scaler.
 
<br /> \cos\theta = \frac{-3}{\sqrt{6}\sqrt{6}} = ?<br />

:P
 
magicarpet512 said:
Your <u,v> must be a vector, not a scaler.

No...dot products always yield scalars. Thanks anyways.
 
magicarpet512 said:
Your <u,v> must be a vector, not a scaler.

No, \langle \cdot, \cdot \rangle is the inner product, the same thing as the dot product.

\langle \vec{u}, \vec{v} \rangle is the same as \vec{u} \cdot \vec{v}.
 
spamiam said:
<br /> \cos\theta = \frac{-3}{\sqrt{6}\sqrt{6}} = ?<br />

:P

I did sqrt(6)^2 = 36 instead of 6! Wow! Stupid. Thanks!
 
spamiam said:
No, \langle \cdot, \cdot \rangle is the inner product, the same thing as the dot product.

\langle \vec{u}, \vec{v} \rangle is the same as \vec{u} \cdot \vec{v}.

Exactly.
 
magicarpet512 said:
Your <u,v> must be a vector, not a scaler.

sorry, i didnt mean to say that!

As spamiam is hinting, how do you get \theta?
 
magicarpet512 said:
sorry, i didnt mean to say that!

As spamiam is hinting, how do you get \theta?

I've figured it out now...careless error...cos(theta) is -1/2 so theta is 120 degrees or 2pi/3 radians.
 

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