Question

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 k-coefficients 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).

Answer 1

You have two vectors:

<b>a</b> = (7,11,3)
<b>b</b> = (x,y,z)

The vector from a to b can be found by

<b>a</b> + <b>d</b> = <b>b</b>

so, <b>d</b> = <b>b</b> - <b>a</b>
= (x-7,y-11,z-3)

so the distance is |<b>d</b>| in the direction of <b>d</b>.