In my lectures, the relationship $\underline{\hat{e_k}} \cdot \underline{\hat{ e_j}}$ is given to be equal to $\delta_{kj}$ - how is this so?
From my understanding this is equal to $ \left( \begin{array}\\ 0\\ 0\\ 1 \end{array} \right) \cdot \left( \begin{array}\\ 0\\ 1\\ 0 \end{array} \right) $, which would give $\left( \begin{array}\\ 0\\ 0\\ 0 \end{array} \right)$; not the kroenecker delta.
For context, this is discussed whilst going over an introduction to the moment of inertia tensor.
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$\begingroup$The dot product is a scalar, not a vector, so you are obvious wrong.
And then you seem to attach specific meanings to $k$ and $j$ when there are just used as variables in that statement.
$\underline{\hat{e_k}}$ (btw: that's an awful lot of indications it's a vector) just means the $k$'th unit vector, with that in mind try to calculate the dot product again.
$\endgroup$ 1 $\begingroup$The dot products is measuring angles between vectors. Especially, if $a$ and $b$ are unit vectors, then $a\cdot b=\cos\angle(a,b)$, where $\angle(a,b)$ is the angle between $a$ and $b$.
So the angle between the unit vector $e_i$ and $e_i$ is $0^\circ$ (because it's the same vector twice). Note that $\cos(0^\circ)=1$. Therefore
$$e_i\cdot e_i=\cos\angle(e_i,e_i)=\cos0^\circ=1=\delta_{ii}.$$
On the other hand, for $i\not=j$, the unit vectors $e_i$ and $e_j$ are perpendicular to each other (by definition). This means $\angle(e_i,e_j)=90^\circ$ and $\cos 90^\circ=0$. Therefore
$$e_i\cdot e_j=\cos\angle(e_i,e_j)=\cos 90^\circ=0=\delta_{ij}.$$
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