vector question EM?

Question

The vector from the origin to the point A is given as (6,−2,−4), and the unit
vector directed from the origin toward point B is (2,−2, 1)/3. If points A and
B are ten units apart, find the coordinates of point B.

Answer 1

The vector AB = A + tB

AB = <6,-2,-4> + t<2,-2,1>

| AB | = √[(2t - 6)² + (-2t + 2)² + (t + 4)²] = 10

(2t - 6)² + (-2t + 2)² + (t + 4)² = 100
4t² - 24t + 36 + 4t² - 8t + 4 + t² + 8t + 16 = 100
9t² - 24t - 44 = 0

t = {24 ± √(576 + 4*9*44)} / 18
t = (24 ± √2160) / 18
t = (24 ± 12√15) / 18
t = (4 ± 2√15) / 3
t ≈ -1.2486556, 3.9153222

There are two possible locations for point B.

Edit: I see that I used 3 times the vector pointed toward B, but a multiple of that vector is still pointed toward B, so the caculation should still work.

Answer 2

The vector pointing towards B can be written as k(2,-2,1)/3 where k is the module of this vector.

The squared distance between points A and B is 100 and can be found like this:

(6 - 2k/3)^2 + (-2 + 2k/3)^2 + (-4 - k/3)^2 = 100

From this point on is just a matter of solving a quadratic equation, find the value of k (remember, is a module, so it must be positive) and with it calculate the coordinates of B