Given points A(1, 3, −2), P1(2, 0, −1) and
P2(4, −2, −1)
Find the point P on the line through P1 and P2 that is closest to A
direction vector of P1P2 = <2 , -2, 0>
so the equation of the plane which had <2, -2, 0> as its normal is
2x - 2y + 0z = c
but A(1,3,-2) lies on it, so
2(1) - 2(3) + 0 = c
c = -4
plane equation: 2x - 2y = -4 or
x - y = -2
equation of line through P1P2:
x = 2 + 2t
y = 0- 2t
z = -1
sub that into the equation of the plane
2+2t - (0-2t) = -2
4t = -4
t = -1
the point is (0, 2, -1)
check my arithmetic