Taking the Inner Product with the Zero Vector

Note that an inner product is conjugate-linear in the second component (Details at http://en.wikipedia.org/wiki/Inner_product_space#Elementary_properties).

Thus, since 2 is its own conjugate, we have $2<v,0>=<v,2\cdot 0>=<v,0>$, hence $<v,0>=0$. So yes, it's true.

Think of it simply like this.

$\langle \mathbf{v} , \mathbf0 \rangle = \langle \mathbf{v}, \mathbf{0}*\mathbf{0} \rangle$

Here let one of the zeroes be the scalar, then the RHS will look like

$\langle \mathbf{v} , \mathbf0 \rangle = \langle \mathbf{v} , \mathbf{0}*\mathbf{0} \rangle = 0\cdot\langle \mathbf{v} , \mathbf{0} \rangle$

which is $0$.