Finding the angle between two 3 dimensional vectors 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

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