Vectors in R3: What am I doing wrong?

Find the equation of the plane through the points A (2, -1, 1), B (-1, 3, -1) and C (17/7, -5/7, 6/7).





The line through AB has direction (-3, 4, -2).


The line through AC has direction (3/7, 2/7, -1/7).





The vector product of these will be a vector which is perpendicular to both and therefore a normal to the plane.





I believe the vector product of (-3, 4, -2) and (3/7, 2/7, -1/7) is (0, -9/7, -18/7)... plz check.





Call this the normal vector, "n".





So, the equation of the plane is the locus of all points "r" such that the vector from r to A (or B or C) is perpendicular to "n".





Hence, we need: (r-A).n = 0, i.e. r.n = A.n ("." denotes scalar product)





In other words (x,y,z).((0, -9/7, -18/7) = (-3, 4, -2).(0, -9/7, -18/7)





So working out the scalar products above, the equation is:





(-9/7)y - (18/7)z = 0





BUT when I check this by substituting in the three points A, B and C to the equation above, they don't fit the equation. So something is wrong. What is it?

Vectors in R3: What am I doing wrong?
Instead of substituting A in the equation


r.n = A.n


you put (x,y,z).(0,-1,-2) = (-3,4,-2).(0,-1.-2)


Note that I've simplified by dividing through by 9/7; when concerned with direction only, we can multiply or divide by a scalar.





It should be (x,y,z).(0,-1,-2) = (2,-1,1).(0,-1,-2)


i.e. -y - 2z = -1


or


y + 2z = 1





In general, once you have the normal vector (a, b, c), the equation of the plane is


ax + by + cz = constant, and you find the constant by substituting the coordinates of any of the three given points.
Reply:i didn't check the vector product but I think there is the mistake, the other things are correct


But my advice is to get rid of 1/7 in the direction of AC.


You can multiply with 7, and work with a parallel vector





edit: oh yes, Hy is right


you subsituted the direction, not the point A, that was the mistake

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