- #1
yayscience
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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?