Fdtxjr

I know that for this proof I need to show that $A$ $subseteq$ $B$ and $B$ $subseteq$ $A$. Starting with $A$ $subseteq$ $B$, I started by setting the equations equal to each other and solving for $x$.

$2x$ - $y$ + $7z$ = $x$ - $y$ + $5z$ gives, $x$ = $-2z$. Plugging $x$ back into one of the equations, you get $(-2z)$ -$y$ + $7z$ = $0$. Simplifying, I get $y$ = $3z$. So, for all ($x$, $y$, $z$) $in$ $A$, ($x$, $y$, $z$) = ($-2z$, $3z$, $z$). That is, ($x$, $y$, $z$) = ($-2c$, $3c$, $c$) since we took ($x$, $y$, $z$) to be arbitrary. So ($x$, $y$, $z$) $in$ $B$ and $A$ $subseteq$ $B$.

Now, I'm having trouble going in the other direction and showing that $B$ $subseteq$ $A$. I have the following thus far: Assume ($x$, $y$, $z$) $in$ $B$. Thus, ($x$, $y$, $z$) = ($-2c$, $3c$, $c$). I'm not sure how to get from here to showing that $B$ $subseteq$ $A$.

Take any $(x,y,z)in B$, then $x= -2c$, $y=3c$ and $z=c$, so $$x-y+5z = -2c-3c+5c =0$$ and $$2x-y+7z = -4c-3c +7c =0$$ so $(x,y,z)in A$ and thus $Bsubseteq A$.

$A$ is a line: it is an intersection of two non-parallel planes.

$B$ is a line, explicitly parameterized by $c$.

Therefore:

To show $B subset A$, all you need is two points from $B$ that lie in $A$. Direct inspection shows that the origin ($c=0$) does. This is one point.

To get another, take, for example, $c = 1$, so the corresponding point in $B$ is:
$$
(x, y, z) = (-2, 3, 1).
$$
Plug it into the two equations that define $A$ and see that they are satisfied.

Try not to get too caught up in set concepts.

$A subset B$ means $x-y+5z = 0; 2x -y +7z = 0 implies exists c; x=-2c;y=3c; z = c$.

And $B subset A$ means if we let $z=c; y=3c; x=-2c implies x-y+5z = 0; 2x -y +7z = 0$

And $A = B$ means precisely $x-y+5z = 0; 2x -y +7z = 0 iff exists c; x=-2c;y=3c; z = c$

...

To be honest $B subset A$ is the easier way:

If $x = -2c; y = 3c; z = c$ then $x - y + 5z = (-2c) -3c + 5c =-5c + 5c= 0$ and $2(-2c)-3c +7c = -4c-3c + 7 = -7c + 7c=0$ and we are done for that direction.

That was it.

$Asubset B$ is a matter of showing:

$x - y + 5z = 0; 2x -y + 7z = 0$ is two equations with three unknowns. If we solve in terms of $z$ we get (one way or another[1]) $y = 3z$ and $x = -2z$ and that's it. Set $z=c$ and we are done.

[1] One way:

$y = x+ 5z = 2x + 7z$

$2x - x = 5z - 7z$

$x = -2z$

$y = -2z + 5z = 2(-2z) + 7z = 3z$.

Set $A$: Intersection of $2$ planes passing through the origin determines a straight line passing through the origin.

Ansatz:

$vec r = t(a,b,c)$, $t in mathbbR$.

Determine the direction vector $(a,b,c)$.

1) $ta -tb +t5c=0$;

2) $2ta-tb+7tc=0$;

1)$a-b+5c=0;$

2)$2a-b+7c=0$;

$a+2c=0$; $a =-2c;$

$b= -2c+5c=3c$;

$(a,b,c)= (-2c,3c,c)$.

Hence: $vec r = t (-2,3,1).$

Finally :

$A=$

$(x,y,z)=t(-2,3,1), t in mathbbR$

$=B.$