# Angle between pair of vectors from two planes at an angle mutually.

Suppose I have 2 planes mutually at an angle to each other. Then two vectors, one from each, would make same angle between them. Is that so?

To ease the problem, assume that one of the planes ($P$) is $z=0$ and the hinge is $r:y=z=0$.

The equation of the plane that cuts $P$ at $r$ with an angle $\alpha$ is $Q:y\sin\alpha+z\cos\alpha=0$. We can assume that $0<\alpha\le\pi/2$.

Now, a generic vector in $P$ is $(u,v,0)$ and a generic vector of the same length in $Q$ is $(u,v\cos \alpha, v\sin\alpha)$, so the cosine of the acute angle $\beta$ between them is $$\cos\beta=\frac{u^2+v^2\cos \alpha}{u^2+v^2}\ge\frac{u^2\cos\alpha+v^2\cos \alpha}{u^2+v^2}=\cos\alpha$$

So $\beta\le \alpha$.

That is, the acute angle between two vectors is at most the acute angle between two planes that contain them.

• @Sufyan The correct sign is $\ge$. Note that $u^2\ge u^2\cos\alpha$ and that the cosine is decreasing from $0$ to $\pi/2$. – ajotatxe Jan 5 '17 at 14:44
• The two vectors that you’ve chosen aren’t exactly “generic” since they have the same $x$-coordinate. Your summary at the end, while correct for such pairs, is incorrect in general. – amd Jan 5 '17 at 20:56

If I've understood your question, call the two planes you described plane A and plane B. Then consider a vector in plane A, but that is also completely contained in the 'hinge line'. Also consider a vector in plane B, but that is also completely contained in the 'hinge line'. Clearly these two vectors do not have an angle between them of 30 degrees.

The (acute) angle between pairs of vectors can be anything from $0$ to $180°$. Consider a pair of vectors that are both in the hinge—pointing in the same direction along the line of intersection of the two planes. Clearly, the angle between them is zero. Now, rotate one of these vectors in its plane. The angle between them will change continuously until it reaches a maximum of $180°$ when the vector is back in the hinge, but pointing in the opposite direction.

• Is your argument not contrary to what ajotatxe said in last. He says the acute angle between vectors $\beta$ can be no more than the acute angle between planes $\alpha$ but it may be equal or smaller. Or maybe you are telling something different and I am mixing it because I haven't got it yet? – Sufyan Naeem Jan 5 '17 at 20:19
• Unfortunately, ajotatxe didn’t consider all possibilities. He only looked at pairs of vectors with the same $x$-coordinate, for which the dihedral angle is indeed maximal. Now that you mention it, however, the acute angle between a pair of vectors at the hinge can be anything. I’ve modified my answer accordingly. – amd Jan 5 '17 at 21:02
• Actually my book tells me that the magnetic force on a charge moving with velocity v in a magnetic field B is given by, $F_B = |q|vB\sin{\theta}$ and it further says that $\theta$ is the angle between the plane of field and the plane of motion. However as far as I know, $\theta$ is the angle between directions of velocity and magnetic field and not their planes. Now I suppose, or maybe I am sure that my book is wrong. If you know something to tell me, I would be really grateful. I could ask this on Phy.SE but I think the sentence in book is mathematically wrong so I should ask this here first. – Sufyan Naeem Jan 5 '17 at 23:38
• @SufyanNaeem You’ll have to tell me what “the plane of field” and “the plane of motion“ mean in this context. The force, as you know, is proportional to the cross product of $\mathbf B$ and $\mathbf{\dot r}$, so perhaps the text is referring to planes for which these two vectors are normals? – amd Jan 6 '17 at 8:26
• @SufyanNaeem The force is always perpendicular to both the velocity and $\mathbf B$. If $\mathbf B$ is perpendicular to the page, then the resulting force will always be parallel to the page. Being perpendicular to $\mathbf{\dot r}$, the force is a pure deflection, but the particle is only going to move along a circular arc if $\mathbf B$ is constant. (Well, there is an “infinitesimal” circle in general, but I’m not going to go there. :)) – amd Jan 6 '17 at 18:03