considering the surface [tex]25x^2+25y^2+4z^2=54[/tex](adsbygoogle = window.adsbygoogle || []).push({});

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 eqation 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?

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# Parametric equations for a line

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