There is a method of finding the equation by using this equation: r.(di+ej.fk)=1 and substituting the position vectors in turn for r. Then the simultaneous equations are solved for d, e and f. I don't know how it works. Why is the scalar product 1?
There is a method of finding the equation by using this equation: r.(di+ej.fk)=1 and substituting the position vectors in turn for r. Then the simultaneous equations are solved for d, e and f. I don't know how it works. Why is the scalar product 1?
You did not mention one important point: di+ ej+ fk is a normal vector, perpendicular to the plane. Now, if r is a vector lying in the plane, in particular if r= <x- a, y- b, z- a> where (a, b, c) is given as a specific point in the plane and (x, y, z) represents a variable point in the plane, then r is perpendicular to di+ ej+ fk and so their dot product is 0.
If you have three points in the plane, say, (a, b, c), (d, e, f), (g, h, i), then (a- d, b- e, c- f) and (a- g, b- h, c- i) are vectors lying in the plane and so you can take di+ ej+ fk to be the cross product of those two vectors.
Suppose that is a unit vector perpendicular to the plane and that is the position vector of some fixed point within the plane.
If now is the position vector of another point within the plane, then will be a vector in the plane and so perpendicular to
In that case,
so
or
This can now be written as
Three different position vectors for three different points within the plane produce three equations in which the unknowns are