I just started learning calc 3 and I'm confused about the topic of direction angles of vectors. It all began when my textbook defined them as
The direction angles of a nonzero vector a are the angles $\alpha$, $\beta$, and $\gamma$ (in the interval $[0, \pi]$ that a makes with the positive $x$-, $y$-, and $z$-axes, respectively (see Figure 3).
That definition makes sense! The problem arises in Figure 3. Why do the arrows representing the angles look like that? Pretty much every other figure on the same topic does this thing where the little arc representing the angle starts on one axis and ends up on the vector!
The organic chemistry tutor drew them in a way that I would expect in this video. So I just assumed that everyone else was bad at drawing angles and went on to work on some exercises, and the one that made me make this post was the one that asked to find the direction cosines and direction angles of the vector $\langle c,c,c \rangle$ where $c>0$. My wrong answer was $\pi/4=45°$ for $\alpha$, $\beta$, and $\gamma$. Isn't the angle of the vector to each axes $45°$? The correct answer according to the textbook was approximately 57.4°. How does one get to this answer? I can derive it correctly algebraically using the direction cosines but I want to understand this topic geometrically, just so I know what I'm doing.
In other words, why is the answer not $45$ degrees, and does the strange way the figures are drawn have anything to do with this?
I'm with David K here. The depiction you praise in the video is exactly the same as the depiction in the figure you disparage, except for being more crudely drawn. I can only assume that this crudeness has somehow allowed you to mistake the drawing as matching a misconception you have. I can only take guesses as to what this misconception is, but a big clue is your belief that the angles for $(c,c,c)$ would be $45^\circ$.
It is hard to draw these angles clearly, because they exist in three dimensions, but the drawing is only in two dimensions. This flattening of the real situation disguises the angles involved. To get a better understanding, it might help if you have a physical cube that you can look at. Say, a Rubik's cube or a 6-sided die. Consider the origin of the coordinate system as being the center of the cube, and the three coordinate axes as emerging from the cube in the centers of three faces. The vector $(c,c,c)$ points in the direction of the vertex where those three faces meet.
Now, if you are looking at only two dimensions, with only an $x$ and $y$-axis, the vector $(c,c)$ points exactly halfway between those two axes, which meet in a $90^\circ$ angle, So the angle between $(c,c)$ and each of the axes will be half of that $90^\circ$ angle, and thus is $45^\circ$. This is all within one plane. In three dimensions, this is the plane $z = 0$. The $x$ and $y$ coordinate axes already have $z=0$ so they are in that plane. But the vector lying halfway between them is $(c,c,0)$, not $(c,c,c)$. That is the vector with direction angles to the $x, y$ axes of $45^\circ$. Its direction angle to the $z$-axis is $90^\circ$.
On the cube, $(c,c,0)$ points to the midpoint of the edge between the faces for the $x$ and $y$ axes. So $45^\circ$ is the angle between the vector pointing to the center of a face and the vector pointing to the middle of an edge of that face. But the direction angle for $(c,c,c)$ is the angle between the middle of a face and a vertex of that face. If you look at the cube, it should be obvious that the center of the face has a bigger angle to a vertex than it does to the middle of an edge. The direction angles for $(c,c,c)$ are larger than the $45^\circ$ angles of $(c,c,0)$.
As Will Jagy has said, the direction angles $\theta_x, \theta_y, \theta_z$ satisfy the formula $$\cos^2 \theta_x + \cos^2 \theta_y + \cos^2 \theta_z = 1$$
For the point $(c,c,c)$, symmetry demands that $\theta_x = \theta_y = \theta_z$, so calling the common value $\theta$, we have $$3\cos^2 \theta = 1\\\cos \theta = \dfrac 1{\sqrt 3}\\\theta \approx 54.7^\circ$$