‖PQ×u‖
D = --------
‖u‖
where P is the point P(2,-5,-2) and Q is any point on the line, say when t=0
x=4+2t, y=-4-3t, z=5+4t we have the point Q(4,-4,5),
and u = < 2, -3, 4>, the direction vector for the line.
We calculate the vector PQ = < 4-2, -4-(-5), 5-(-2) > = < 2, 1, 7 >
We find the cross product:
PQ×u =
= 25i + 6j - 8k
We find the norms:
‖PQ×u‖ = Ö25²+6²+(-8)² = Ö725 = 5Ö29
‖u‖ = Ö2²+(-3)²+4² = Ö29
‖PQ×u‖ 5Ö29
D = -------- = ------ = 5
‖u‖ Ö29
Edwin