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This should be a fairly simple problem, but I've gotten it wrong a dozen times.

I have two vectors, vector $\vec{N}$ in the x-z plane with a slope of $\frac{1}{8}$, and $\vec{E}$ in the y-z plane with a slope of $\frac{1}{4}$.

The angle between two vectors is defined by:

$$\theta=\arccos(\frac{\vec{N}\cdot\vec{E}}{\vert{N}\vert\vert{E}\vert})$$

I start by finding the component form of each vector for the $i,j,$ and $k$ directions.

$$\vec{N}=\langle{i,0,k_N}\rangle$$ since the slope is $\frac{1}{8}$, $k_N=\frac{i}{8}$, $$\vec{N}=\langle{i,0,\frac{i}{8}}\rangle$$ Same goes for $k_E$ of $\vec{E}$ with the slope of $\frac{1}{4}$. $$\vec{E}=\langle{0,j,\frac{j}{4}}\rangle$$ The dot product of the two is simple because the first two products are zero. $$\vec{N}\cdot\vec{E}=\frac{ij}{32}$$ Now I look for the magnitude of each vector. $$\vert{N}\vert=\sqrt{i^2+0^2+(\frac{i}{8})^2}$$ $$=i\frac{\sqrt{65}}{8}$$ Then the $\vec{E}$ vector comes out to $$j\frac{\sqrt{17}}{4}$$ so $$\vert{N}\vert\vert{E}\vert=\frac{ji\sqrt{1105}}{32}$$ This works really well for finding $\frac{\vec{N}\cdot\vec{E}}{\vert{N}\vert\vert{E}\vert}$ because $32, i,$ and $j$ cancel out. $$\frac{\vec{N}\cdot\vec{E}}{\vert{N}\vert\vert{E}\vert}=\frac{1}{\sqrt{1105}}$$ Finally, I solve for $\arccos(\frac{\vec{N}\cdot\vec{E}}{\vert{N}\vert\vert{E}\vert})$ to get $\theta$. $$\theta=\arccos(\frac{1}{\sqrt{1105}})$$ $$\approx{1.54}$$ in radians.

This is wrong. The answer is $\approx{1.6}$ radians. Where did I go wrong?

EDIT: corrected the planes, this won't have an effect on the answer

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    Other than your putting the two vectors in the wrong planes, it looks OK to me. I also get approx. 1.54. – 2017-02-18
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    Fixed that, although it doesn't matter because of the parallelogram rule. I'm starting to think the computer is wrong and not me. – 2017-02-18

2 Answers 2

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So $\vec{N}\neq $, likewise $\vec{E} \neq ⟨0,j,j/4 ⟩$. try $\vec{N}= <0,j,k/8>$ and $\vec{E}=⟨i,0,k/4 ⟩$.

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    If $k$ is rise and $i$ is run, and $\frac{1}{8}$ is rise over run, $\frac{k}{i} = \frac{1}{8}$. How is $k\neq\frac{i}{8}$? Can you explain your reasoning? – 2017-02-18
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    Right sorry the unit vectors aren't needed at all. you just need the vectors $\vec {N} =$ <8,0,1> or <0,1,1/8> since $\vec {N}$ is in the y-z plane and $\vec {E} = <4,0,1>$ or <1,0,1/4> ad $\vec {E}$ is in the x-z axis. I hope this helps. – 2017-02-18
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    since $j$ and $i$ cancel, it doesn't really change the answer. In fact, any scalar would also leave the answer unchanged. The error must be somewhere else. – 2017-02-18
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Turns out the computer was wrong. Multiple human sources have confirmed that the answer is in fact $\approx{1.54}$