It is true that two vectors are dependent if they "point in the same (or opposite) direction", i.e. if they are aligned.
But that is not totally true for three vectors in $3$D or more.
In the sense that, when the three vectors are aligned, i.e. parallel, i.e. when they are scalar multiples of each other, they are for sure dependent.
But the definition of linear dependency of three vectors is wider than being parallel: it includes also the case in which they are co-planar, although not parallel.
If you want to see that geometrically, taking the three vectors as position vectors from the origin, if they define a full $3$D parallelepiped then they are independent, if instead the parallelepiped collapses into a flat figure or segment then the vectors are dependent.
Algebraically this translates into the fact whether the matrix formed by the three vectors has full rank ($3$) or less.
Similarly for $n$ vectors of $m$ dimensions.
Then from the theory of linear system you know that, in a homogeneous system, if the matrix has full rank then it has the only solution $(0,0, \cdots, 0)$ which corresponds to the combination coefficients to be all null.
In reply to your comment, in ${\mathbb R}^2$ if you have two non-aligned = independent vectors, then a third one will lie on their same plane (the $x,y$ plane).
In the geometric interpretation, the parallelepiped (the hull) will be flat, i.e. dimension 2, which is less than 3, the number of vectors.
In the algebraic interpretation, a matrix $3 \times 2$ cannot have a rank greater than two: so 3 (or more) 2D vectors are necessarily dependent.
final note (to clarify what might be the source of your confusion)
The (in)dependence of $n$ vectors in ${\mathbb R}^m$ is defined for the whole set of $n$ vectors: they might be dependent, notwithstanding that a few of them ($q<n, \; q\le m$) could be independent. Yet if one is dependent on another (or other two, etc.), then the whole set is dependent.
And in fact it is a common task, given $n$ vectors, to find which among them represent an independent subset: the minor in the matrix with non-null determinant, the larger giving the rank.
Best Answer
Well, if it were, then we would have a very curious situation. Try replacing $0$ with some fixed vector $v_0 \ne 0$. Then the set $\{ 0 \}$ is independent, but the set $\{ v_0 \}$ isn't!
To make things worse: if you had two vectors $a$ and $b$ such that $\{ a, b, v_0 \}$ was independent in the standard sense, then the set $\{ \lambda a + \mu b \mid \lambda, \mu \in \mathbb{R} \}$ is independent, despite being a whole subspace!
EDIT: Maybe a specific example will help. Say we define "independent" to mean "there is a linear combination that sums to $\langle 1,1,1 \rangle$. Then the set $\{ \langle 1, 0, 0 \rangle, \langle 0, 1, 0 \rangle, \langle 1, 1, 0 \rangle \}$ is independent (no matter what combination you take, the $z$-component is zero, not one, so you can never get $ \langle 1, 1, 1 \rangle $). But this is clearly silly, because one vector is the sum of the other two, and so whatever our definition is describing, it doesn't capture the notion of independence.