Step 1: Write both kets in terms of basis vectors.
Take the standard basis $|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $|1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$. Then $|v_1\rangle = \frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle)$ and $|v_2\rangle = \frac{1}{\sqrt{2}}(|0\rangle - i|1\rangle)$.
Step 2: Multiply out the combination term by term.
$|v_1\rangle \otimes |v_2\rangle = \frac{1}{2}(|0\rangle + i|1\rangle) \otimes (|0\rangle - i|1\rangle)$. Expand this like a normal product of two binomials, keeping track of which ket comes from which particle:
\[ = \frac{1}{2}\Big[ |0\rangle\otimes|0\rangle - i|0\rangle\otimes|1\rangle + i|1\rangle\otimes|0\rangle - i^2|1\rangle\otimes|1\rangle \Big] \]
Step 3: Simplify using $i^2 = -1$.
The last term becomes $-i^2|1\rangle\otimes|1\rangle = +|1\rangle\otimes|1\rangle$. So
\[ |v_1\rangle \otimes |v_2\rangle = \frac{1}{2}\Big[ |00\rangle - i|01\rangle + i|10\rangle + |11\rangle \Big] \]
Step 4: Map the four basis kets to column positions.
In the standard 4-dimensional ordering $|00\rangle, |01\rangle, |10\rangle, |11\rangle$, the coefficients found above (1, -i, i, 1) become the four entries of the column vector directly. This confirms
\[ |v_1\rangle \otimes |v_2\rangle = \frac{1}{2}\begin{pmatrix} 1 \\ -i \\ i \\ 1 \end{pmatrix} \]
Step 5: Rule out the matrix-shaped options.
Options written as $2\times2$ matrices would be appropriate for an operator such as $|v_1\rangle\langle v_2|$, not for a tensor product of two kets, which must stay a single column with four entries. That rules out both matrix options directly, and comparing signs rules out the other column option.
Final Answer:
Both routes agree: the answer is the column $(1, -i, i, 1)$ scaled by one half, option (A).\[ \boxed{\frac{1}{2}\begin{pmatrix} 1 \\ -i \\ i \\ 1 \end{pmatrix}} \]