If angular velocity & angular acceleration are vectors, why not angular displacement?

Set your copy of Halliday, Walker & Resnick on the table so the front cover is parallel to the table and visible to you. Now your right hand flat on the book with your thumb and forefinger forming a ninety degree angle. Orient your hand so your thumb is parallel to the spine of the book and pointing toward the top edge. Your forefinger should be parallel to the lines of text on the cover, pointing to the right.

This makes a nice basis for a book-based coordinate system. Your thumb points along the x-hat axis, your forefinger along the y-hat axis. To complete a right-hand system, the z-hat axis points into the book.

Pick the book up and make a +90 degree rotation about the x-hat axis. The spine of the book should be horizontal and facing up. Now make a +90 degree rotation about the y-hat axis (as rotated by that first rotation). You should be looking at the front cover of the book but oriented vertically with text flowing toward the ground.

If you start all over again but reverse the order of operations, +90 degree rotation about the y-hat axis followed by a +90 degree rotation about the x-hat axis, you should be looking at the spine of the book rather than the front cover. The front cover is oriented vertically, but with text running parallel to the ground. You can put your book back in the bookcase.

Rotation in three dimensional space and higher is not commutative (rotation A followed by rotation B is not necessarily the same as rotation B followed by rotation A).

Another hint that there's something different between rotation and translation is the number of parameters needed to describe the two in some N-dimensional space. Translation in N dimensional space obviously needs N parameters. Lines have one degree of freedom, planes, two, three dimensional space three, and so on. Lines don't rotate. There are no rotational degrees of freedom in one dimensional space. Rotation does make sense in two dimensional space, where a single scalar (one rotational degree of freedom) completely describes rotations. Three dimensional space has three degrees of freedom. Four dimensional space? It has six. Three dimensional space is the only space for which the number of rotational degrees of freedom and number of translational degrees of freedom are equal to one another.

This unique characteristic of three dimensional space is why you can treat angular velocity as a vector. The vector cross product (something else that is unique to three dimensional space) means introductory students can be taught about rotations without having to learn about Lie theory or abstract algebras. That wouldn't be the case for students who live in a universe with four spatial dimensions.

The wikipedia link does an adequate job of explaining this.

You can find a very mathematical discussion in this partial duplicate question, but here is my simple take on it.

The angular displacement in three dimensions does have a vector nature in the sense of having both a magnitude (the angle through which you turn something) and a direction (the axis about which it is turned).

However, for a true vector quantity ${\bf A} + {\bf B} = {\bf B} + {\bf A}$, and this is not true for angular displacement.

Hold a pencil with its point vertically upwards. Now rotate it through 90 degrees so the point is directly away from you (i.e. about a horiontal axis), followed by a 90 degree rotation about the vertical axis defined by the pencil originally. Note the position of the pencil point.

Now go back to the original position. Rotate the pencil through 90 degrees around its own vertical axis, followed by rotating it 90 degrees about a horizontal axis. The point ends up in a different place right?

Angular displacement is define by change in angle i.e. ${\Delta \theta}=({\theta}_1 -{\theta}_2)$, where $\theta$ is taken as positive in anti-clock wise.
Hence angular displacement ${\Delta \theta}$ has both magnitude and direction. But let when a particle rotate from a point A in $\theta$ angle in anti-clock wise to point B and then back through $\theta$, it comes its initial point A but when a particle starts rotating from B through an angle $\theta$ and after reaching to A it backs to B. Since in two case the end point is different, we conclude that angular displacement is not commutative. That's why it is not a vector quantity.

Now we know that when a particle rotating with a angular velocity $\vec{\omega}$ it linear velocity $\vec{v}$ is always directed in tangential direction of the path. And angular velocity is defined by: $\vec{\omega}={\frac{{\vec{r}} X {\vec{v}}}{{|{\vec{r}}|}^2}}$. Hence angular velocity is perpendicular to the plane containing ${\vec{r}}$ and ${\vec{\omega}}$ and also obey vector commutative law. That's why angular velocity is a vector quantity.