Parametric vector form of the line

[ronho1234]

The parametric vector form of the line 1
 is given as r1 = u1 + rv1 (r element of real field)
where u1 is the position vector of P1 = (1,1,−3) and v1=vectorP1P2

where
P2 = (3,3,−2) .
The parametric vector form of the line 2
 is given as r2 = u2 + sv2 (s element of real field)
where 2 u is the position vector of P3 = (−2,0,2) and
v2= −j− k .
(a) Give the parametric scalar equations of the lines l1
 and l2
 .
(b) Find the unit vector ˆn with negative i component which is perpendicular to
both l1
 and l2

(c) The shortest distance between two lines is the length of a vector that
connects the two lines and is perpendicular to both lines. For l1
 and l2

this is expressed in the vector equation 2 1
r −r = tnˆ where t element of real field is a
parameter. Write this equation as 3 scalar equations and hence obtain a
system of three linear equations for the three parameters r, s and t .
(d) Solve this system of equations for r, s and t and hence find the shortest
distance between the two lines 1
 and 2
 .
(e) Find the point Q on line 1
 which is closest to line 2

I've done bits and pieces of the question bu I'm especially stuck on parts c d and e. please help thank you

 

[LightHero]

:) a) Parametric scalar equations of the lines l1 and l2 are: l1: x = 1 + r, y = 1 + 2r, z = -3 - rl2: x = -2 + s, y = 0 - s, z = 2 + s b) The unit vector ˆn with negative i component which is perpendicular to both l1 and l2 is ˆn = (-1, 0, 1).c) The vector equation for the shortest distance between two lines is 1x - 2(-2 + s) = t(-1),1y - 2(0 - s) = t(0),1z - 2(2 + s) = t(1).d) Substituting the parametric equations of the lines into the vector equation, we get:1 + r - 4 + 2s = -t,2r - s = 0, -r + s = t.Solving this system of equations for r, s and t yields:r = s/2,t = 3/2,s = 3/2.The shortest distance between the two lines is d = √((1+r-4+2s)^2+(2r-s)^2+(-r+s)^2) = √(9/4).e) The point Q on line 1 which is closest to line 2 is given by the parametric vector equation r1 = u1 + rv1. Substituting the values of r and s obtained in part d, we get: Q = (1 + r, 1 + 2r, -3 - r) = (7/2, 2, -5/2).