Question

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

can someone plz help me?

Answer 2

The direction of vector AB is (9,8,12)
So a plane perpendicular to vector AB will be
9x + 8y + 12z = c, where c is a constant.
Let's find the plane such that C(6,2,3) lies on it
9(6) + 8(2) + 12(3) = c
c = 106
so the plane containing C and having AB as a normal is
9x + 8y + 12z = 106

the line AB has parametric equation of
x = 0+9t
y=0+8t
z = 0 + 12t
so to find where AB cuts the plane:
9(9t) + 8(8t) + 12(12t) = 106
289t = 106
t = 106/289
So the point of intersection of the plane and the line is
x = 9(106/289) = 954/289
y = 8(106/289) = 848/289
z = 12(106/289) = 1272/289

now use the distance formula to find the shortest distance:
distance
= √(6 - 954/289)^2 + (2 - 848/289)^2 + (3-1272/289)^2 )
= √( (780/289)^2 + (-270/289)^2 + (-450/289)^2)
= (1/289)√883800
= (10/289)√8838 = appr 3.253

check my arithmetic, I was expecting nicer numbers.

Answer 3

I got 2.109 instead of 3.253 but thanks for helping me! It must've taken you a long time to type all of that.

Answer 4

Math is intriguing because there are many ways to solve the same problem.
Here's another way by minimizing the distance between the line and the point.

1. find the parametric form of the line AB, which is relatively easy because the line passes through the origin, so the parametric form is
AB: <9t,8t,12t>
(note: <...> represents a vector)
2. find the function that represents the square of the distance of (6,2,3) from AB:
d(t)=(9t-6)^2+(8t-2)^2+(12t-3)^2
3. find the value of t that minimizes the distance d(t):
d(d(t))/dt=24(12t-3)+18(9t-6)+16(8t-2)=0
which simplifies to 2(289t-106)=0
or
t=106/289
4. Substitute t in AB to get
(954/289, 848/289, 1272/289)
5. The distance between C and AB is then the square-root of d(106/289)
=sqrt(2925/289)=3.181

Answer 5

frog