Respuesta :
Answer:
The lines are skew
Step-by-step explanation:
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.
