What is wrong with the high-school definition of a vector? Why is the high-school definition of a vector as "a quantity with a magnitude and a direction" incomplete? For example, Griffiths Introduction to Electrodynamics book says:

The deﬁnition of a vector as "a quantity with a magnitude and direction" is not altogether satisfactory.
(section 1.1.5 "How vectors transform")

However, I am not very satisfied with his chain of arguments.
 A: The main problem with "magnitude and direction" is the follow-up question: "What's a direction?" It's something that can uniquely be defined by a unit vector. Not very satisfactory, but no worse than, "A vector is an element of a vector space".
The physics we're talking about is formulated in Euclidean space, aka ${\mathbb R}^3$, and physical laws are presented as relationships between geometric objects that represent the symmetries of that space. There are several objects that do that, the scalar for instance. Since it is unchanged by rotations in space, its representation is trivial.
The simplest non-trivial geometric object is the vector, in fact in the form of spherical vectors, which are eigenstates of rotations, it is the fundamental representation of the rotational symmetries of $SO(3)$, the group of rotations in ${\mathbb R}^3$. That's a bit much for high school, so we're introduced to them in their Cartesian form, $(\hat x, \hat y, \hat z)$, where they have the interpretation as orthogonal unit arrows from which any and all vectors can be constructed. Foregoing that as too mathematical, we're left to describe them as things with magnitude and direction, and that does capture their essence.
Magnitude is not a difficult concept: if you have a velocity, $\vec v$, then the idea of $2\vec v$ is intuitive. Every student should have an intuitive idea of "direction", and the fact that if you rotate 180 degrees, you're now in "the opposite" direction, or if you rotate a full 360 degrees, your direction is unchanged is indeed a defining property of vectors. Moreover, it's easy to see why there are 3 independent basis vectors.
Compared with other geometric objects that represent the symmetries of space, that's no so bad. How would you describe a natural 2nd rank tensor? It has an alignment, but not a direction, it has magnitude....it can also have bulge. Moreover, there are 5 basis tensors. Not very intuitive.
Meanwhile, spinors have a magnitude and a direction...and a sign, so it's more than just a vector....but there are only 2 basis spinors? How does that work out?
So "magnitude and direction" isn't so bad.
A: I don't have the book to read Griffiths arguments, but what causes confusion on that high school definition is when we start to use vectors in curved surfaces. For example, if a sailor keeps the same speed and the same route according to the compass (say $221^\circ$), its velocity vector apparently is always the same, because it has the same magnitude and direction. But it is not true, what is more evident close to one pole.
The notion of basis vectors, and how they can change point to point is also important for their concept. However, when first presented in cartesian coordinates, the notion of basis vectors seem superfluous.
A: $\def\A{{\bf A}}
\def\N{{\bf 0}}$Here is a simple-minded reason that this definition is not satisfactory (having nothing to do with transformations): the zero vector has no direction.
Note that if $\A\ne\N$, then
$\A = |\A|\hat \A$, where $|\A|$ is the magnitude of $\A$ and $\hat \A = \A/|\A|$ is the unit vector in the $\A$ direction, i.e., where $\hat \A$ is the direction of $\A$.
(Note that $|\hat\A| = 1$ by this definition.)
Thus, every nonzero vector is a quantity with magnitude and direction.
In fact, it is the product of these quantities.
Now note that since $|\N| = 0$, the direction of the zero vector is undefined.
A: It's almost perfectly satisfactory. First, there are three ways to introduce vectors:

*

*Geometric

*Algebraic

*Transformational

All three definitions are important and it's important to understand how they relate to each other. The first two ways are common in high school mathematics. The geometric way is magnitide with direction and the second is axiomatically as an element of a vector space.
Griffiths says that the geometric definition is unsatisfactory so as to introduce the transformational definition. This is important because this is how tensors are traditionally introduced. This is not often simple to see as at the same time tensors are introduced as tangent tensors, so there is additional structure that complicates the picture. But at this level the structure is not neccessary.
However, in Griffiths electrodynamics tensors are not used. So why does Griffith introduce the transformational definition? This is because he wants to introduce axial vectors otherwise also known as polar vectors or pseudo-vectors. The magnetic vector is not an ordinary vector but an axial vector. In fact, angular momentum and torque are also axial vectors.
To define an axial vector actually requires expanding the notion of a vector to include a representation of the vector space. The representation required for an axial vector is called the sign representation, the simplest representation.
So Griffiths is right: the definition of a vector - and not just the 'high-school definition' but all three definitions - is not sufficient, an expansion is neccesary to include a representation to properly model physical phenomena like angular velocity, angular momentum and the magnetic vector. This is often not made explicit and results in quite a bit of confusion.
A: Good question; many basic discussions of vectors do not clearly explain this, in my opinion. The definition you provide is for a "free vector"; specifically,  one that is dependent only on direction and magnitude, and not its absolute location relative to the origin of a coordinate system.  Engineering developments distinguish among free, bounded, and sliding vectors.  But, physics developments generally treat vectors as free; for example, in developing the equations of motion in a non-inertial frame that undergoes both translation and rotation relative to an inertial frame, the vectors are free vectors.  Of course the effect of a vector depends on its location; for example, for a force to act on a particle the force vector must be located at the position of the particle.  The text Introduction to Vector Analysis by Harry Davis has a good discussion of vectors in physics treated as free vectors.  Also the text Mechanics by Symon discusses free vectors used on physics.
A: 
Why is the high-school definition of a vector as "a quantity with a magnitude and a direction" incomplete?

The high school definition of vectors applies to Euclidean vectors, or, more generally, to vectors within an inner product space. However, the mathematician’s definition of vectors and vector spaces allows for vectors that possess neither magnitude nor direction.
A: What Griffiths is getting at is that a vector can be thought of as more than just a single list of coordinates that specifies a magnitude and direction. Rather, it is an object that exists independent of a chosen coordinate system. As such, it can be represented by an infinite number of different lists of coordinate values, depending on the chosen basis (I'm assuming we're working with vector spaces over infinite fields like the real numbers). Viewed this way, it is the knowledge of how these different coordinate lists transform into each other to describe the same object that defines the vector. This notion of coordinate transformations is not typically emphasized in the so-called high school description.
A: This was originally supposed to be a comment but I think in the context of the question, specifically Griffiths Introduction to Electrodynamics , it may well be the answer also.
Intuitively, yes a vector is "a quantity with a magnitude and a direction" but there's an issue rising due to the nature of the vector that i believe no one has pointed out so far— Vectors do not add up following algebraic rules. They have their own vector addition rules that needs to be followed.
I will present a counter-example to the "high-school definition", but before that lets take a classic "high school" example of vectors. Consider two forces acting on an object. Let one of them be 4N and the other 3N. If the forces are parallel, perpendicular and antiparallel then the Net force on the object would be 7N, 5N,1N respectively.
This example is important, because it illustrates the angle depends of the vectors. In other words the effect of a vector always changes of the angles is changed.
Now, let's take an example in electrodynamics that Griffith would approve. Let's consider a wire carrying current 5A from South to North and another wire carrying current 4A from East to West.
Now let the wires meet at a junction and then join together in the North-West direction (ill add a pic later).  What is the current carried by the resulting wire? Without a doubt it is always 9A. This is because current adds linearly as a result of Kirchoffs current rule rather than "vector"ly. Actually, regardless of where the resulting wire is pointing to, the resultant current will always be 9A. If it is not, then that would imply that the  conservation of Charge is violated! The conservation of charge is after all the basis of Kirchoffs first rule.
Does the current(s) taken here have a magnitude? Yes. Do they have a direction? Yes. Are they vectors? No, because this particular physical quantity does not depend on the angles as charge is a scalar and it should always be conserved. I suspect this is what Griffiths was coming to when he said the high school defintion is incorrect.
In conclusion, a vector following vector addition is important. It implies that the nature of the vector depends on the direction also. If the quantity does not follow vector addition, then it is not a vector even if we can assign a direction to it.
A: In addition to this answer , here is another simple argument why definition of vector is not satisfactory. Electric current in simple circuits has both magnitude (Amps in SI units) and direction (positive to negative terminal or from high to low potential).
But electric current does not satisfy either the triangle law of vector addition nor the parallelogram law of vector addition. Thus it cannot be regarded as a vector.
Similarly there are many quantities (Higher dimensional) which have magnitude and direction but don't follow the usual triangle law of vector addition.
A: The definition is not satisfactory because vector spaces do not come equipped with a notion of direction or magnitude. The matrix
$$
\begin{pmatrix}
1 & 1 \\ 0&1
\end{pmatrix}
$$
defines an automorphism of $\mathbb{R}^2$ as a vector space, but preserves neither magnitude nor direction (understood as the angle between vectors).
To have a notion of magnitude and direction, we need a metric (inner product), i.e. a non-degnerate symmetric bilinear form.
The confusion comes from always thinking of vector spaces as having a natural inner product. But often there is no natural or even meaningful inner product. For example, the set of pairs $(\text{\$ spent on apples}, \text{\$ spent on oranges})$ is a perfectly good and even useful vector space. What should the magnitude of such a vector be? There's no meaningful answer.
