- #1
UrbanXrisis
- 1,196
- 1
considering the surface [tex]25x^2+25y^2+4z^2=54[/tex]
The parametric equation for a line going thought point P=(1,1,1) is
[tex]x=1+50t[/tex]
[tex]y=1+50t[/tex]
[tex]z=1+8t[/tex]
A plane an equation for the tangent plane through P.
Here's what I know:
the equation for a plane needs a perpendicular vector to the plane and a point on the plane. I have a point on the place P=(1,1,1) but how do I find the perpendicular vector? I have a line ON the plane, what can I do to get an equation that is perpendicular to the plane?
The parametric equation for a line going thought point P=(1,1,1) is
[tex]x=1+50t[/tex]
[tex]y=1+50t[/tex]
[tex]z=1+8t[/tex]
A plane an equation for the tangent plane through P.
Here's what I know:
the equation for a plane needs a perpendicular vector to the plane and a point on the plane. I have a point on the place P=(1,1,1) but how do I find the perpendicular vector? I have a line ON the plane, what can I do to get an equation that is perpendicular to the plane?