# Thread: line in three dimensions

1. ## line in three dimensions

I have an exam in a few hours and I would like someone to solve 2 questions and tell me each step they did along the way, 1 question i ahve done and the other is similar but i can seem to get the correct answer:

Q1: L=(0,-1,0) M=6,5,3) P=(3,0,1)

P is not on the line LM, find a point on line LM, called N, which is perpendicular to P.
This one i can do and I get N=(2,1,1)

Q2: x=2t y=-t z=4t.

Find a point Q on this line closest to a point P, not on this line.

I get the wrong answer for this 1, I think its because i am finding line PQ instead of QP. Is this where i am going wrong?, also if it is can someone tell me why it matters that QP is used and not PQ?

2. For Q2, I think you're missing some info - mainly what point P is.

To find the shortest distance means find the point perpendicular on line Q to point P.
In this case, you need to solve:
PQ.u = 0
u = (2, -1, 4) % taken from x=2t, y=-t & z=4t. i.e make t=1, point u therefore lies on line Q.
PQ = P - Q
Q = (2t, -t, 4t)
P= ?? %need to have point P here. We can't use (x,y,z) as we already have unknown variable t.
Unless you mean use point P from Q1. In that case
P = (3, 0, 1)
PQ = (3-2t, t, 1-4t)
PQ.u = (6 - 4t) - t + (4 - 16t) = 0
=> -21t +10 = 0
=> t = 10/21
Put this back into Q and you get:
(20/21, -10/21, 40/21) %this is the closest point on line Q to point P.

good luck with the exam!
Were there any questions you found particularly difficult?