1. Find the equation of the plane containing the line (x,y,z)=(5+4t, -5-3t,2) and perpendicular to the line (x,y,z)=(2-3t,3-4t,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).
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.
Why do I need to dot the vector? Is this correct? @ t=1, a point on the line l1=(9,-8,2) (x-9,y+8,z-2) (dot) (-3,-4,7)=-3(x-9)-4(y+8)+7(z-2)=0