Let A = (0,0,0), B = (9,8,12), and C = (6,2,3). Find coordinates for the point on line AB that is closest to C.

Answer 1

Let the perpendicular distance from C to AB meet AB at D
vector AB = [9,8,12]
equation of plane through C and D having AB as a normal:
9x + 8y + 12z = c
but (6,2,3) lies on it , so the plane is
9(6)+8(2)+12(3) = c = 106

plane equation: 9x+8y+12z = 106

equation of line AB:
x = 0+9t
y= 0+8t
z= 0+12t
where does AB cut the plane?
9(9t) + 8(8t) + 12(12t) = 106
289t = 106
t = 106/289
so point D is (954/289 , 848/289 , 1272/289)

check my arithmetic

Answer 2

GOOD GRIEF, Just realized I did this same question a week ago.
At least I ended up with the same answer.

http://www.jiskha.com/display.cgi?id=1484278204