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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 -

    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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