# How to describe the intersection of two sets?

The following is a homework problem:

Let \begin{align*} W_1 &= \{(a_1, a_2, a_3) \in\mathbb{R}^3 \mid a_1 = 3 a_2\text{ and }a_3 = -a_2\}\\ W_2 &= \{(a_1, a_2, a_3) \in \mathbb{R}^3 \mid a_1 - 4 a_2 - a_3 = 0\} \end{align*} Describe the intersection of W1 and W2 and observe that it is a subspace.

I realize that the intersection is $$\{ (a_1, a_2, a_3) \in \mathbb{R}^3 \mid a_1 = 3 a_2\text{ and }a_3 = -a_2\text{ and }a_1 - 4 a_2 - a_3 = 0 \}.$$ Further, I've observed that this seems to just be the set, $W_1$, based on the conditions. I just kind of haphazardly messed with the formulas to try to figure stuff out with no real methodology.

What is the method to solving these kinds of problems? How do I solve this and similar problems?

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Assume that $(a_1,a_2,a_3)\in W_1\cap W_2$. As you note, you must have $a_1=3a_2$ and $a_3=-a_2$, in order for the point to be in $W_1$. And you must have $a_1-4a_2-a_3=0$ in order to be in $W_2$.
That means that the point must satisfy the following three equations: \begin{align*} a_1 -3a_2 &= 0\\ a_2 + a_3 &= 0\\ a_1 - 4a_2 - a_3 &= 0 \end{align*} So the intersection consists precisely of all solutions to this system of linear equations.
Here, we obtain that if $a_1 = 3a_2$ and $a_2=-a_3$, then necessarily $a_1-4a_2 - a_3 = 0$, so that the third equation is a consequence of the first two, which is what you observed: it turns out that $W_1\subseteq W_2$, so $W_1\cap W_2 = W_1$.
In general, if your sets are given as solution sets of equations, then finding the intersection is equivalent to solving a system of equations; whether this is easy or not will depend on the equations. And for even more generality, it may be extremely hard to say what the intersection is explicitly; for instance, trying to figure out exactly what integers are in $$\{ p\in\mathbb{Z}\mid p\text{ is prime}\}\cap \{n\in\mathbb{Z}\mid n=p+2\text{ for some prime }p\}$$ is the same as trying to figure out exactly what the twin primes are, and it is still unknown whether there are infinitely many of them or not....