Let L1 be the line passing through the points Q13 1 5 and Q2

Let L₁ be the line passing through the points Q₁=(3, 1, 5) and Q₂=(1, 5, 3) and let L₂ be the line passing through the point P₁=(6, 10, 2) with direction vector d=[3, 1, 2]^T. Determine whether L₁ and L₂ intersect. If so, find the point of intersection Q.

Solution

L1 passes through Q1(-3,1,5) and Q2(-1,-5,3)

we can find the direction vector of L1 as (Q1 - Q2)

direction vector = <-2, 6, 2>

parametric form of L1 =

x = -3 + (-2)*s                                             //  point + s * direction vector

y = 6 + 6*s

z = 5 + 2*s

we get

x = -3 -2s                         (1)

y = 1 + 6s                        (2)

z = 5 + 2s                        (3)

parametric form of L2 =

direction vector = <3,1,2>

x = -6 + 3t                         (4)        // point + t * direction vector

y = -10 + t                         (5)

z= -2 + 2t                          (6)

Lines intersect if we can find values for s and t such that

-3 -2s = -6 + 3t                              (7)        // from (1) and (4)  

1 + 6s = -10 + t                              (8)       // from (2) and (5)

5 + 2s = -2 +2t                                (9)        // from (3) and (6)

from (7) and (8) we need to solve for \"s\" and \"t\"

we get s= -3/2 and t = 2

now we subtituite these values in (9) if this equation saties then these lines intersect each other

5 + 2*(-3/2) = -2 +2*(2)

5 - 3 = -2 + 4

2 = 2           hence the lines will intersect

lines will intersect when s = -3/2 and t=2

L1 =

x = -3 - 2s = > -3 -2*(-3/2) = > 0

y = 1 + 6s => 1 + 6*(-3/2) => -8

z= 5+2s => 5+2*(-3/2) = 2

L2 =

x= -6+3t => -6 + 3*2 = >0

y= -10 + t => -10 + 2 = -8

z= -2 +2t = -2 +2*2 = 2

so the point of intersection is (0,-8,2)