Posted by **Grace** on Saturday, January 7, 2012 at 9:34pm.

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).

- Precalc - incomplete -
**Steve**, Sunday, January 8, 2012 at 3:42pm
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**

= (x-7,y-11,z-3)

so the distance is |**d**| in the direction of **d**.

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