Precalc
posted by Grace .
At the point were a line intersects a plane [with the equation (24x+32y+40z=480) and three points of A(5, 10, 1), B(6, 3, 6), and C(12, 1, 4)], the vector i, vector j, and vector kcoefficients of the line equal the corresponding values of x, y, and z for the plane. By substituting these coefficients, for x, y, and z in the equation for the plane, find the directed distance, d, from the known point (7, 11, 3) to the intersection point (x, y, z).

You have two vectors:
a = (7,11,3)
b = (x,y,z)
The vector from a to b can be found by
a + d = b
so, d = b  a
= (x7,y11,z3)
so the distance is d in the direction of d.