Three questions about geometric vectors I was going through some problems and was unable to solve these:
(1) Let there be two planes (in three rectangular dimensions) given by $2x-3y+4z=2$ and $2x-3y+4z=6$ and let there be a line defined by the equations $$\frac{x-5}{3}=\frac{y+3}{2}=\frac{z}{6}.$$ The problem is to show that the line intersects both planes, say at $P$ and $Q,$ and then find $|PQ|,$ all this without explicitly finding $P$ or $Q,$ which would make the problem trivial of course.
I was only able to show the first part; namely that the planes are intersected by the line. One simply notes that a normal vector to both planes (which are parallel) casts a non-vanishing shadow on the directing vector of the line, which shows that the line is not parallel to the planes. But I simply don't know how to do the second part. This is where I need help.
(2) Let $a$ and $b$ be nonzero space vectors. Show that if $a\cdot c=b\cdot c$ for every vector $c,$ then $a=b.$ One can prove this if one lets $c=a-b,$ but I don't see how then $c$ would be arbitrary. Please help.
(3) This is just another way suggested to prove the well-known trigonometric identity $\cos (x+y)=\cos x \cos y-\sin x \sin y,$ namely: consider the vectors $(\cos x,\sin x)$ and $(\cos y,-\sin y).$ Then by talking their dot product one obtains the RHS of the identity; however, how does one obtain the LHS? I suspect it would have something to do with the vector identity $a\cdot b=|a||b|\cos\phi,$ but I can't see how to show that the angle between those two unit vectors above is $x+y,$ in order to complete the proof. Many thanks.
 A: (1) If you want to get $PQ$ without getting the coordinates of $P$ and $Q$, then consider the perpendicular line to both planes passing through $P$, suppose that this line intersects the other plane in a point $A$.
Now lets consider the right triangle $PAQ$ (right at $A$). $PA$ is the distance between the 2 planes so $PA=\frac{4}{\sqrt{29}}$ (check the calculations), and you can get the angle between $\vec{PA}$ and $\vec{PQ}$ using the dot product between the normal vector of the plane and the directing vector of the line, and with some trignometry you get the hypotenuse $PQ$.
(2) You have that $a.c=b.c$ for any $c$, so any vector $c$ that you pick, will satisfy the equality, picking for example $c=\vec i$ and then $c=\vec j$ and then $c=\vec k$, will allow you to reach the fact that $a=b$ (because by doing this you'll be getting that both vectors $a$ and $b$ have the same $x$, $y$, and $z$ components, since the dot product of $a$ with $\vec i$ is simply the projection of $a$ on the $x$-axis). Also taking $c=a+b$ is another method for getting to the solution.
(3) I liked this idea actually, it is the first time I see something like this. Now you are right it have something to do with $a.b=|a|.|b|.\cos \phi$. Now it is obvious that $|a|=|b|=1$, so we just have to prove that $\phi=x+y$. The easiest way to think about this is imagining yhe vectors in a trignometric circle, you have the vector $a=(\cos x,\sin x)$ making an angle $x$ with the $x$-axis, and now what is the difference between the vector $(\cos y,\sin y)$ and the vector $b=(\cos y,-\sin y)$? The second vector is symmetric to the first vector with respect to the $x$-axis, so the angle from $b$ to the $x$-axis is also $y$. (Another way of thinking about this is that we can express $b$ as $b=(\cos y,-\sin y)=(\cos (-y),\sin (-y))$. Finally if the angle between $b$ and the $x$-axis is $y$ and the angle between the $x$-axis and $a$ is $x$, so we have that the angle between $b$ and $a$ is $\phi=(x+y)$. (To see it clearly and without loss of generality, take both angles to be in the first quadrant, i.e. take them between $0$ and $\frac{\pi}{2}$)
