Finding the Vector Eqn of a Line

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SUMMARY

The discussion focuses on finding the vector equation of a line that passes through the point P(2, -1, 3) and is perpendicular to the plane defined by the equation 3x - 2y - z = 0. The solution involves using the normal vector of the plane, which is determined to be the direction vector for the line. The normal vector is calculated as 3i - 2j - k, leading to the vector equation Line(x) = P + td, where d is the direction vector.

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Homework Statement



Find the Vector eqn of a line which passes through P(2, -1, 3) and is perpendicular to the plane 3x -2y -z =0.

Homework Equations



Line(x) = P + td where d is a direction vector

The Attempt at a Solution



My main problem here is trying to find the direction vector for the line. I have really no idea how to find that. Any help is greatly appreciated :)
 
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I think I figured it out...I can use the normal vector of the plane as my direction vector.
 
Yes, that's exactly right. And that, of course, is just 3i- 2j- k.
 

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