Describe $15/(4,3,2,1)$ as a vector

Question

I was doing some exercises on multivariable calculus (MIT OCW multivariable calculus with Herbert Gross, lesson 3 part III) and I ran into the following:
Describe the algebraic linear equation for which $\frac {15}{(4,3,2,1)}$ is a solution set.
Describe $\frac {15}{(4,3,2,1)}$ as a vector.
Unfortunately I could not afford the book that the course requires (Thomas' Calculus, Chapter 5, Supplementary notes), but based on the answers I think I can infer what is going on.
They say the solution set is $$\{ \bar x: \bar x \cdot (4,3,2,1)=15\}$$ Expanding the dot product gives the equation (I understand that). From this I guess they divide by $(4,3,2,1)$ to "solve" for $x$. This in and of itself is difficult to swallow because I had always thought dividing by vectors is not defined. What is the intuition or the rigor behind this?
Then the text says (to answer its second problem) that any vector in the direction of $(4,3,2,1)$ is simply a scaled version of that vector i.e. $(4t,3t,2t,t)$. In order for it to be in the solution set it must satisfy the linear equation provided above. Solving $(4,3,2,1) \cdot (4t,3t,2t,t) =15$ gives $t=1/2$. The text claims that $\frac {15}{(4,3,2,1)}=(2,3/2,1,1/2)$ after substituting in for $t=1/2$. My main questions are: why must the scaled-by-$t$-vector be in the same direction as $(4,3,2,1)$? And: why can we solve the system this way?
Here is a link to the course website if anyone wants to check the problem out: http://ocw.mit.edu/resources/res-18-007-calculus-revisited-multivariable-calculus-fall-2011/study-materials/

Answer 1

It looks as if this non-standard notation is asking to find the vector $x$ of the smallest Euclidean norm so that $x\cdot (4,3,2,1) = 15$. This is the same as making an orthogonal projection (do you have knowledge about such projections?) of the origin onto the plane ${\cal P}$ given by that equation. The normal to the plane is precisely $n=(4,3,2,1)$ and to get from the origin to the nearest point on ${\cal P}$ the most economical is to go in the direction of the normal until you hit the plane, i.e. finding $t$ so that $t n$ belongs to ${\cal P}$ (the solution you sketch). Using Pythagoras it is not difficult to see that any other point on the plane will have a larger distance to the origin.