- #1

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## Homework Statement

Find the angle between

[tex]\begin{align*}

\vec{A} = 10\hat{y} + 2\hat{z} \\

and \\

\vec{B} = -4\hat{y}+0.5\hat{z}

\end{align*}[/tex]

using the cross product.

The answer is given to be 161.5 degrees.

## Homework Equations

[tex]

\left| \vec{A} \times \vec{B} \right| = \left| \vec{A} \right| \left| \vec{B} \right|sin(\theta)

[/tex]

## The Attempt at a Solution

[tex]

\left| \vec{A} \times \vec{B} \right| = [/tex] [tex]\left|

\begin{array}{ccc}

\hat{x} & \hat{y} & \hat{z} \\

0 & 10 & 2 \\

0 & -4 & 0.5

\end{array} \right| = \left| 13\hat{x} \right| = 13 [/tex]

The magnitude of A cross B is 13.

Next we find the magnitude of vectors A and B:

[tex] \left| \vec{A} \right| = \sqrt{10^2+2^2} = \sqrt{104} = 10.198039 [/tex]

and

[tex] \left| \vec{B} \right| = \sqrt{(-4)^2+(\frac{1}{2})^2} = \sqrt{16.25} = 4.0311289 [/tex]

multiplying the previous two answers we get:

41.109609

So now we should have:

[tex] \frac{13}{41.109609} = sin(\theta) [/tex]

Solving for theta, we get:

18.434951 degrees.

This is frustrating: 180-18.434951 = the correct answer. I'm not quite sure where I'm going wrong here.

I must be making the same mistake repeatedly. Another problem was the same thing, but with the numbers changed, and I also got the 180-{the answer I was getting} = {the correct answer}, but when I tried the example using the SAME methodology, I got the correct answer.

Can someone please share some relevant wisdom in my direction?