Your $v_1$ and $v_2$ need to be orthonormal. To expand on learnmore's answer, essentially, the reason you need orthogonality for this to work is that if your $v_1$ and $v_2$ are not orthogonal, then they will have a non-zero dot-product $v_1\cdot v_2$. This means that $v_2$ carries some weight "in the direction" $v_1$. Your intuition that $c = a\cdot v_1$ is the "amount of $a$ in the direction $v_1$" is correct - keep that intuition! Similarly, $d=a\cdot v_2$ is the amount of $a$ in the direction of $v_2$.
However - since $v_1$ and $v_2$ are not perpendicular, the number $c$ has "piece" of $v_2$ in it, and the number $d$ has a "piece" of $v_1$ in it. So, when you try to expand $a$ in the basis $\{v_1,v_2\}$, you would need an extra term to compensate for the "non-orthogonal mixing" between $v_1$ and $v_2$.
The technical details are as follows. Since $v_1$ and $v_2$ are linearly independent, we can write
$$
a = \alpha v_1+\beta v_2
$$ for some scalars $\alpha, \beta$. Now, take the dot product of $a$ with $v_1$ and expand it out:
$$
a\cdot v_1 = (\alpha v_1+\beta v_2)\cdot v_1 = \alpha v_1\cdot v_1 + \beta v_1\cdot v_2 = \alpha + \beta v_1\cdot v_2
$$ similarly, expand out $a\cdot v_2$:
$$
a\cdot v_2 = \alpha v_1\cdot v_2 + \beta
$$
Those extra terms ($\beta v_1\cdot v_2$ and $\alpha v_1\cdot v_2$) express the non-orthogonality. Written another way, we have
$$
\alpha = a\cdot v_1 - \beta v_1\cdot v_2
$$ and
$$
\beta = a\cdot v_2 - \alpha v_1\cdot v_2
$$ which shows clearly that the correct expansion coefficients have $a\cdot v_j$, but also another piece compensating for the non-orthogonality. I could go on - you can use matrices and such, but hopefully this is enough to convince you.
Alright let's take a general vector:
$$
\vec v=(x,y,z)
$$
now let's make it unitary, dividing it by it's norm:
$$
\hat v=\left(\frac{x}{\sqrt{x^2+y^2+z^2}},\frac{y}{\sqrt{x^2+y^2+z^2}},\frac{z}{\sqrt{x^2+y^2+z^2}}\right)
$$
now let's make it orthogonal to your other vector $\vec a=(-1,3,5)$
the ortogonality condition is ofcourse:
$$
\vec a \cdot \hat v=0 \iff -x+3y-5z=0
$$
solve it for x:
$$
x=3y-5z
$$
now substitute this result in $\hat v$:
$$
\hat v=\left(\frac{3y-5z}{\sqrt{y^2 + (3 y - 5 z)^2 + z^2}},\frac{y}{\sqrt{y^2 + (3 y - 5 z)^2 + z^2}},\frac{z}{\sqrt{y^2 + (3 y - 5 z)^2 + z^2}}\right)
$$
this is a unitary vector orthogonal to $\vec a$. Since it still depends on 2 variables you can conclude that there are infinitely many vectors with those 2 properties.
You could have answered it without any calculation just by imagining that on the tip on vector $\vec a$ you can put another vector orthogonal to it of length 1 and you can rotate that vector using $\vec a$ as an axis of rotation still keeping it of length 1 and orthogonal in infinitely many ways.
moreover one could also demonstrate that it really only depends on one parameter.
using the substitution:
$$
\begin{cases}
y=r \cos t\\
z=r \sin t
\end{cases}
$$
do the substitution and you'll see that the $r$ simplifies and the resulting vector will only depend on $t$.
Best Answer
Apart of the other anwers, there is indeed a relation between Euclidean distance $d(X,Y)$ and the inner product $\langle X, Y \rangle$ if the vectors $X, Y \in \mathbb{R}^N$ are normalized, that is $\langle X, X \rangle=\langle Y, Y \rangle = 1$, in particular $$\frac{d(X,Y)^2}{2} = 1 - \langle X, Y \rangle.$$ This can be shown as $$d(X,Y)^2 = \langle X - Y, X - Y \rangle = \langle X, X \rangle + \langle Y, Y \rangle - 2 \langle X, Y \rangle = 2 (1 - \langle X, Y \rangle)$$