Plane equation perpendicular to line
1. Find the equation of the plane containing the line (x,y,z)=(5+4t, 53t,2) and perpendicular to the line (x,y,z)=(23t,34t,5+7t).
2. Relevant equations Cross product? Dot product? Ax+By+Cz=D? 3. The attempt at a solution I'm really new with this material and any aid would be greatly appreciated. The only thing I can think of to do would be the cross product of (4,3,0) and (3, 4, 7). 
Re: Plane equation perpendicular to line
A normal to the plane is <3, 4, 7>, which I got from the equation of the perpendicular line. You're given that the plane contains the line (x, y, z) = (5 + 4t, 5  3t, 2), so it should be a simple matter to find a point on this line, which I will call (x0, y0, z0). Once you have a point on a plane and a normal to the plane, the equation of the plane can be gotten by dotting the vector (x  x0, y  y0, z  z0) with the plane's normal vector.

Re: Plane equation perpendicular to line
Why do I need to dot the vector? Is this correct?
@ t=1, a point on the line l1=(9,8,2) (x9,y+8,z2) (dot) (3,4,7)=3(x9)4(y+8)+7(z2)=0 
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