I have two questions:
$1)$ How can I find the line of intersection between the planes
$$x+ 2y +z =4 \\ \mathrm{and} \\ 2x+y-z=5$$
$2)$ How do I find an equation for a line that goes through $A = (2,0,2)$ and $B = (4,1,0)$ that is parallel to the answer in the first question?
I find the intuition on these problems really difficult, so if you describe how you do it, I would be very grateful.
Best Answer
For the first one, you want to solve the system $$\begin{cases} x+ 2y +z =4 \\ 2x+y-z=5\end{cases}$$ for $(x,y,z)$. For example, summing I am getting that $x+y=3$. Then I am getting that $z=1-y$, so that $$(x,y,z)=(3-y,y,1-y)=(3,0,1)+y(-1,1,-1)$$ Thus, the intersection is a line $$\mathscr L:(3,0,1)+t(-1,1,-1)\;\;;\; t\in\Bbb R$$
For the second one, suppose we can write your line as $$\mathscr L':w+tv\;\;;\;t\in\Bbb R$$ for some vectors $v,w$. Saying that $\mathscr L'$ is parallel to the line we found above means their directions are parallel. Any line is fully determined by a point and a direction, you should be able to obtain what $\mathscr L'$ is. It seems the exercise wants two lines, one that goes through $A$ and one through $B$.
Recall that if we have a line $v+tw$ and we already know what $w$ is, and know that $v'$ is a point in it, we may let say $t=1$ so that $v+w=v'\implies w=v'-v$, that is, we may let $w$ be $v'$ plus any scalar multiple of the direction we want.