The two lines are in symmetric equations. L1 is (x-1)/2=(y+1)/3=-z. L2 is (x+1)/2=(y-1)/3=3-z. Find the equation of the plane containing the lines.
The point on L1 is (1,-1,0) with slope of (2,3,-1)
The point on L2 is (-1,1,3) with slope of (2,3,-1)
Normal vector =V(P1P2)XSlope
The Attempt at a Solution
With same slope, you know the lines are parallel.
So the vector from P1 to P2 is (-2,2,3)
Cross product of vector P1 to P2 and the slope is (-11,-8,4)
My thought was that the normal vector has the value of (a,b,c) in the equation for plane. I plugged in the values from the normal vector along with P1 to get:
According to the solutions sheet, the answer is 11(x-1)-4(y+1)+10z=0.
I got the x0,y0,z0 correct but why are the a, b, and c values wrong?