Help with dot product for vectors

[warfreak131]

Question
Let vectors \vec{A} =(2,1,-4), \vec{B}=(-3,0,1), \vec{C}=(-1,-1,2).

What is the angle (in radians) \theta_{AB} between \vec{A} and \vec{B}?Important equations
\vec{A} \cdot \vec{B} = \vert \vec{A} \vert \,\vert \vec{B}\vert \cos(\theta), where \theta is the angle between \vec{A} and \vec{B}.

My attempt

\vec{A}\cdot\vec{B} = (2*-3) + (1*0) + (-4*1) = -6 - 4 = -10

|\vec{A}||\vec{B}| = (2*3) + (1*0) + (4*1) = 10

therefore, -10 = 10*cos(\theta)

-1 = cos(\theta)
arccos(-1) = \theta
180 = \theta
180 = \pi radians

but it says \pi isn't the right answer

But I don't think that sounds right, because \vec{A} is at \arctan{\frac{1}{2}} (26.7) degrees, and \vec{B} runs along the negative X axis at 180 degrees, So that's a difference of 153.3 degrees. And if you convert that to radians, you get approximately \frac{17\pi}{20} radians, which isn't right either.

 

[mplayer]

You did your dot product right but the magnitude of the vectors is wrong. Recall that the magnitude of a vector is square root of the sum of its individual components squared...the length of the vector.

For example, the magnitude of vector A would be |A| = sqrt(2^2 + 1^2 + -4^2) = sqrt(21)

Hope that helps.

 

[HallsofIvy]

Question
Let vectors \vec{A} =(2,1,-4), \vec{B}=(-3,0,1), \vec{C}=(-1,-1,2).

What is the angle (in radians) \theta_{AB} between \vec{A} and \vec{B}?


Important equations
\vec{A} \cdot \vec{B} = \vert \vec{A} \vert \,\vert \vec{B}\vert \cos(\theta), where \theta is the angle between \vec{A} and \vec{B}.

My attempt

\vec{A}\cdot\vec{B} = (2*-3) + (1*0) + (-4*1) = -6 - 4 = -10

|\vec{A}||\vec{B}| = (2*3) + (1*0) + (4*1) = 10

This is wrong. Find |A| and |B| separately and then multiply!

|A|= \sqrt{2^2+ 1^2+ (-4)^2}

therefore, -10 = 10*cos(\theta)

-1 = cos(\theta)
arccos(-1) = \theta
180 = \theta
180 = \pi radians

but it says \pi isn't the right answer

But I don't think that sounds right, because \vec{A} is at \arctan{\frac{1}{2}} (26.7) degrees, and \vec{B} runs along the negative X axis at 180 degrees, So that's a difference of 153.3 degrees. And if you convert that to radians, you get approximately \frac{17\pi}{20} radians, which isn't right either.