Question

Let L be the line with parametric equations

x = 2+3t
y = 3−2t
z = 2+t

Find the shortest distance d from the point P0=(−5, 1, −4) to L, and the point Q on L that is closest to P0. Use the square root symbol '√' where needed to give an exact value for your answer.
d = _
Q = (_, _, _)


I know we have to deal with them in vectors. I think we should be making a vector that is LQ but otherwise idk what to do. If someone could help that would be great.

Answer 1

let Q(x,y,z) be that closest point.
then vector PQ = <x+5 , y-1 , z+4 >
we also know the given line has direction vector <3, -2, 1>

The shortest distance is when PQ is a perpendicular
which means that the dot product between the two vectors is zero
<x+5, y-1, z+4> dot <3, -2, 1> = 0
3x + 15 -2y + 2 + z + 4 = 0
3x - 2y + z = -21

subbing in our values for x, y , and z
3(2+3t) - 2(3-2t) + (2+t) = -21

6+9t-6+4t+2+t = -21
14t = -23
t = -23/14 , not very nice

Using :
x = 2+3t
y = 3−2t
z = 2+t

the point Q is

Q(2 - 69/14 , 3 + 46/14 , 2 - 23/14)
or
Q( -41/14 , 88/14 , 5/14)

PQ = √[ (-5+41/14)^2 + (1-88/14)^2 + (-4-5/14)^2 ]
= √ [ 841/196 + 5476/196 + 3721/196 ]
= √(10038/196)
= (√10038)/14

I was expecting "nicer" numbers. I did not write the solution down on paper first, but did this on screen, where the chances of arithmetic error are greater.
Better check my arithmetic, I know my method is correct.

Answer 2

87

Answer 3

This is correct! Thank you for helping!

Answer 4

How do we get the d value though?