compute the angle between the vector u = [1, 2, 3] and the plane
2x + 3y - z = 5
this one is really stumping me, I am not even sure where to begin..
One way which I can think of doing this, is to consider instead the plane $\displaystyle 2x + 3y - z = 0$ ; the angle between this plane and your vector will be the same. The idea is that now, this plane is a subspace of $\displaystyle \mathbb{R}^3$. Find a basis for it and orthonormalize it using the Gram-Schmidt process. Then calculate the projection of the vector on the plane $\displaystyle 2x + 3y - z = 0$; the angle you seek will be the angle between $\displaystyle u$ and its projection on the plane, which you can find using the dot product.
Perhaps a bit messy but it shouldn't take more than 6 minutes if you are familiar with everything above. Otherwise there is probably another (shorter) way.
Well, a plane can be fully defined by its normal vector. Once you find the normal vector coordinates (derive from plane equation, very simple), calculate the angle between the 2 vectors. The angle that your vector does with the plane will be 90degrees minus the angle that it does with the normal vector of the plane. If for example you get an angle of 40deg with the normal vector, the plane angle is 50deg. If the angle is 170deg, the plane angle is 80deg (through the other side of the plane). Very simple.
There are many ways to calculate the angle between vectors of known coordinates, I'll let you do it.
cheers,
Nuke
The vector $\displaystyle N=<2,3,-1>$ is the normal to the plane,
The angle $\displaystyle \Theta = \arccos \left( {\frac{{\left| {N \cdot U} \right|}}{{\left\| N \right\|\left\| U \right\|}}} \right)$ is the acute angle between $\displaystyle N~\&~U$.
Thus $\displaystyle \frac{\pi}{2}-\Theta$ is the acute angle between $\displaystyle \text{the plane}~\&~U$.