Basic Lin Alg - Angle Between Vectors

[sunmaz94]

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.

 

[magicarpet512]

Your <u,v> must be a vector, not a scaler.

 

[spamiam]

$$\cos\theta = \frac{-3}{\sqrt{6}\sqrt{6}} = ?$$

:P

 

[sunmaz94]

Your <u,v> must be a vector, not a scaler.

No...dot products always yield scalars. Thanks anyways.

 

[spamiam]

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 $\vec{u} \cdot \vec{v}$.

 

[sunmaz94]

$$\cos\theta = \frac{-3}{\sqrt{6}\sqrt{6}} = ?$$

:P

I did sqrt(6)^2 = 36 instead of 6! Wow! Stupid. Thanks!

 

[sunmaz94]

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]

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 $$\theta$$?

 

[sunmaz94]

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.