Question

Let L be the line through the point Q = (−1, 11, 11) and direction vector →d=[−1, 2, 2]T.
Find the two distinct points R1 and R2 on L at distance
3√10
to the point P = (1, 4, 1).

R1 = (0, 0, 0)

R2 = (0, 0, 0)

Answer 1

A general point R on our line L is
R(-1-t, 11+2t, 11+2t)
We want RP = 3/√10

Let's use our distance between two points formula
P(1,4,1) and R(-1-t, 11+2t, 11+2t)

√( (1+1+t)^2 + (4-11-2t)^2 + (1-11-2t)^2 ) = 3/√10
√(2+t)^2 + (-7-2t)^2 + (-10-2t)^2 ) = 3/√10
square both sides and expand
4 + 4t + t^2 + 49 + 28t + 4t^2 + 100 + 40t + 4t^2 = 9/10
9t^2 + 72t + 153 = 9/10
90t^2 + 720t + 1530 - 9 = 0
90t^2 + 720t + 1521 = 0

Argghhhhh, supposed to get 2 real values of t,
getting complex roots, can't find my error!!

Can somebody find where my error is ?

Answer 2

Found out what the problem is.
It is in the question itself, not my calculations.

Using projection, I just calculated that the perpendicular distance from
(1,4,1) to the given line is 3 units, which would be the closest to the line
we could get from the given point.
We are asked to find where the distance between (1,4,1) and a point on
the line is 3/√10.
of course 3/√10 < 3, so there are no points that close.