Geometry with vectors 
Let $V = 25i  + 15j  - 30k$. What angles does this vector make with the $x$, $y$, and $z$ axes?

Can someone go all the way through and show me how this is done? Thanks!
 A: The angle that $\def\V{{\bf V}}$ makes with the $x$-axis is the same as the angle it makes with the vector $\def\I{{\bf i}}\I = (1,0,0)$, which points the same way.
$$\def\vdot{{\cdot}}\V\vdot \I = |\V||\I|\cos\theta$$
where $\theta$ is the angle you want.
But $\V\vdot \I$ is just $25\cdot 1 + 15 \cdot 0 + -30\cdot 0 = 25$.
So the angle here is $$\theta =\cos^{-1} \frac{25}{|\V||\I|}.$$
Where $|\V| = \sqrt{25^2 + 15^2 + (-30)^2}$ and $|\I| = \sqrt{1^2+0^2+0^2}$. 
Can you do the rest yourself?
A: The angle $\theta$ between the vectors $u,v$ can be found using $\cos \theta=\dfrac{\langle u,v\rangle}{\|u\|\|v\|}$ where $\langle .,.\rangle$ is the dot product. See this.
A: $V=5(5i+3j-6k)$, thus $V$ has the same direction as $5i+3j-6k$.
Unit (easy to handle) vectors with the directions of the axes are $1i+0j+0k$ for the $x$-axis, $0i+1j+0k$ for the $y$-axis, and $0i+0j+1k$ for the $z$-axis.
To find the angle between two vectors given their cartesian components, you can use the dot product.
$$a\cdot b=|a||b|\cos\theta.$$
In other words:
$$\cos^{-1}\left(\dfrac{a\cdot b}{|a||b|}\right)=\theta.$$
