Calculating the Equation of a Plane Perpendicular to a Given Line

In summary, the equation for the plane that passes through the point (2,-1,3) and is perpendicular to the line v=(1,-2,2)+t(3,-2,4) can be found by using the direction vector of the line as the normal vector for the plane and calculating the equation using a point in the plane.
  • #1
daynat18
2
0
Equation for plane that passes through the point (2,-1,3) and is perpendicular to the linev=(1,-2,2)+t(3,-2,4)?

I'm not exactly sure where I'm supposed to go since the only examples in my book show the plane perpendicular to a vector written in a different form. Not in this one
 
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  • #2
If a line is given in the form (x0, y0, z0[/sub])+ t(A, B, C) then Ai+ Bj+ Ck is a vector pointing in the direction of the line. Is that enough?
 
  • #3
So is it supposed to be that I can find the plane based on simply that vector since any of the normal vectors would be orthogonal to the plane?
 
  • #4
daynat18 said:
So is it supposed to be that I can find the plane based on simply that vector since any of the normal vectors would be orthogonal to the plane?

Well the normal will be perpendicular to the plane, but so is the line v. So wouldn't the normal to the plane be given by the direction of the line v? Given the normal to the plane and a point in the plane, you should be able to calculate the equation of the plane.
 

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