# Thread: Point to a line 3-d

1. ## Point to a line 3-d

alright so I am having doubt in my professors answer to this question or maybe I am doing something wrong so before i go speak with him, i wanna ask some other professionals

Find the distance of the point P = (1,2,-3) to the line: (1,-2,1) + t(2,3,4)

The vector from (1,2,-3) to (1,-2,1) is (0,4, -4) = A
and the direction vector is (2,3,4) = B

so sinθ = |AxB|/ [|A|][|B|]

where |AxB| = 12
|A| = √32
|B| = √29

so |A|sinθ = 12/√29

however my profs answer is √(32 - 16/29)

sorry i din't latex it i was in a rush

2. I get the same answer as your prof, (by a different method from the one that you have used).
I think that your value for |AxB| is wrong.

3. Originally Posted by treetheta
alright so I am having doubt in my professors answer to this question or maybe I am doing something wrong so before i go speak with him, i wanna ask some other professionals

Find the distance of the point P = (1,2,-3) to the line: (1,-2,1) + t(2,3,4)

The vector from (1,2,-3) to (1,-2,1) is (0,4, -4) = A
and the direction vector is (2,3,4) = B

so sinθ = |AxB|/ [|A|][|B|]

where |AxB| = 12
|A| = √32
|B| = √29

so |A|sinθ = 12/√29

however my profs answer is √(32 - 16/29)
You should find that $\displaystyle A\times B = (28,-8,-8) = 4(7,-2,-2)$, from which $\displaystyle |A\times B| = 4\sqrt{57}$, giving the answer to the problem as $\displaystyle 4\sqrt{\tfrac{57}{29}}$.

The prof's method was probably to minimise the distance from the point (1+2t,–2+3t,1+4t) to the point (1,2,–3) by completing the square. That distance is $\displaystyle \sqrt{4t^2+(3t-4)^2+(4t+4)^2} = \sqrt{29(t+\tfrac4{29})^2 + 32-\tfrac{16}{29}}$, which also comes out as $\displaystyle 4\sqrt{\tfrac{57}{29}}$ when t has the minimising value –4/29.

Personally, I prefer the cross product method to the minimisation method (provided that it's done correctly).