Can someone help me with this problem? I bet it will be pretty easy for the most of you:
Through the line $p$ draw a plane that is parallel to the $x$-axis, where p is defined by: $x=5-2t, y=9t-1, z=-7t-2$.
Since it's parallel to the $x$-axis, I know that the equation of the plane will have this form: $By+Cz+D=0.$ I also know that there is a point that lies on the plane $P(5,2,0)$ and a vector $p(-2,1,3)$.
I don't know how to use this information, though. Any help?
OK, the line p has equations:
$x = -2t + 5 $
$y = t + 2 $
$z = 3t $
Let the equation of the plane be: $ax + by + cz = d$
Since it's parallel to the $x$ axis, $a=0$ and $d\ne0$ (you also found that).
Take any two points from your line e.g. $(5,2,0)$ and $(3,3,3)$.
Now put these two points into the plane equation:
$by + cz = d$
You get a system with 2 equations.
$b.2 + c.0 = d$
$b.3 + c.3 = d$
Solving it for $(b,c)$ you get $b=d/2$, $c=-d/6$.
So the equation of the plane becomes:
$(d/2) \cdot y - (d/6) \cdot z = d$
Multiply it by $6/d$ and you get the final equation: $3y-z=6$