Draw a plane through a line parallel to the $x$-axis

Question

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?

sorry i was to lazy to write it properly. i posted it like a picture now — CHaOS, Mar 07 '16 at 19:05

peter.petrov · Answer 1 · 2016-03-07T19:33:36.753

1

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: $3y-z=6$

edited Mar 07 '16 at 19:33

answered Mar 07 '16 at 19:19

peter.petrov

12,568

I think the OP changed the equations of the line p after I wrote this answer. – peter.petrov Mar 07 '16 at 19:27
2

no, they are correct. thanks a lot man, i owe you one :) – CHaOS Mar 07 '16 at 19:40

Draw a plane through a line parallel to the $x$-axis

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