Determine whether the lines L1 and L2 are parallel, skew, or intersecting. If they intersect, find the point of intersection.L1: x / 1 = y - 1 / -1 = z - 2 / 3L2: x - 2 / 2 = y - 3 / -2 = z / 7

The lines are skew

To make things clearer, rewrite the given lines as follows;

[tex]L_1 : \frac{x}{1} = \frac{y-1}{-1} = \frac{z-2}{3}[/tex]

[tex]L_2 : \frac{x-2}{2} = \frac{y-3}{-2} = \frac{z}{7}[/tex]

(i) Test if the lines are parallel.

Two lines are parallel if the cross product of their directional vectors is equal to the zero vector 0 (which is equal to <0, 0, 0>). For example, if their directional vectors are m and n, then;

m x n = 0

Where;

m, n and 0 are vectors

In the given lines, L₁ and L₂, their directional vectors are given by the denominators of their symmetric equation.

For L₁, the directional vector = <1, -1, 3>

For L₂, the directional vector = <2, -2, 7>

Let

m = <1, -1, 3> =  i - j + 3k

n = <2, -2, 7> = 2i - 2j + 7k

Now, find the cross product of the vectors;

m x n   =   |  i        j        k |

              |  1       -1       3 |

              |  2      -2      7 |

m x n   =  i(-7 + 6) -j(7 - 6) + k(-2 + 2)

m x n   =  i(-1) -j(1) + k(0)

m x n   =  -i -j + 0k

Since the cross product does not give a zero vector, then the lines are not parallel.

(ii) Test if the lines intersect

To do this:

(a) First, we express the lines in parametric form rather than the symmetric form. Therefore

[tex]L_1 : \frac{x}{1} = \frac{y-1}{-1} = \frac{z-2}{3}[/tex]  is equated to say a;

   and

[tex]L_2 : \frac{x-2}{2} = \frac{y-3}{-2} = \frac{z}{7}[/tex] is equated to say b;

We have;

[tex]L_1 : \frac{x}{1} = \frac{y-1}{-1} = \frac{z-2}{3} = a[/tex]

[tex]L_2 : \frac{x-2}{2} = \frac{y-3}{-2} = \frac{z}{7} = b[/tex]

Split the lines into system of three equations in terms of a and b

For L₁

=> [tex]\frac{x}{1}[/tex] = a  which gives x = a

=> [tex]\frac{y-1}{-1}[/tex] = a which gives y = 1 - a

=> [tex]\frac{z-2}{3}[/tex] = a which gives z  = 3a + 2

∴ L₁ has these three equations

x = a                           -------------------------(i)

y = 1 - a                      --------------------------(ii)

z = 3a + 2                  --------------------------(iii)

For L₂

=> [tex]\frac{x-2}{2}[/tex] = b which gives x = 2b + 2

=> [tex]\frac{y-3}{-2}[/tex] = b which gives y = 3 - 2b

=> [tex]\frac{z}{7}[/tex] = b which gives z = 7b

∴ L₂ has these three equations

x = 2b + 2                        --------------(iv)

y = 3 - 2b                         ---------------(v)

z = 7b                              ----------------(vi)

(b) Secondly, we combine the set of equations of the two lines.

Equations of x : (i) and (iv)

a = 2b + 2                  

=> a - 2b = 2                   --------------------------------------(vii)

Equations of y: (ii) and (v)

1 - a = 3 - 2b

=> - a + 2b = 2               --------------------------------------(viii)

Equations of z: (iii) and (vi)

3a + 2 = 7b

=> 3a - 7b = -2               -------------------------------------(ix)

(c) Thirdly, solve equations (vii), (viii) and (ix) simultaneously to find the values of a and b

Add equations (vii) and (viii)

    a - 2b = 2

+

   -a + 2b = 2  

   0  +  0  = 4    

Since 0 + 0 ≠ 4, then the values of s and t cannot be determined from the system of equations due to inconsistency. Therefore, the lines L₁ and L₂ do not intersect.

(iii) Test if the lines are skew

Two lines are said to be skew if they are neither parallel nor intersecting.

Since the lines L₁ and L₂ are neither parallel nor intersecting, the lines are skew.

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