Determine whether the lines l1 and l2 are parallel, skew, or intersecting. If they intersect, find the point of intersection.
l1: x=3-2t, y=7+4t, z=-3+8t
l2: x=-1-u, y=18+3u, z=7+2u
The Correct Answer and Explanation is :
Let’s analyze the given parametric equations for lines ( l_1 ) and ( l_2 ), and determine whether they are parallel, skew, or intersecting.
Parametric Equations:
For line ( l_1 ):
[
\begin{aligned}
x_1 &= 3 – 2t \
y_1 &= 7 + 4t \
z_1 &= -3 + 8t
\end{aligned}
]
For line ( l_2 ):
[
\begin{aligned}
x_2 &= -1 – u \
y_2 &= 18 + 3u \
z_2 &= 7 + 2u
\end{aligned}
]
Step 1: Direction Vectors
First, we extract the direction vectors of the lines. The direction vector of ( l_1 ) is the coefficient of the parameter ( t ) in the equations of ( l_1 ):
[
\mathbf{d_1} = (-2, 4, 8)
]
Similarly, for ( l_2 ), the direction vector is the coefficient of ( u ) in the equations of ( l_2 ):
[
\mathbf{d_2} = (-1, 3, 2)
]
Step 2: Check if the Lines Are Parallel
Two lines are parallel if their direction vectors are proportional (i.e., one is a scalar multiple of the other). To check this, we compare the components of ( \mathbf{d_1} ) and ( \mathbf{d_2} ).
[
\frac{-2}{-1} = 2, \quad \frac{4}{3} \neq 2, \quad \frac{8}{2} = 4
]
Since the components are not proportional, the lines are not parallel.
Step 3: Check if the Lines Intersect
To determine if the lines intersect, we set the parametric equations of ( l_1 ) equal to the parametric equations of ( l_2 ):
[
\begin{aligned}
3 – 2t &= -1 – u \
7 + 4t &= 18 + 3u \
-3 + 8t &= 7 + 2u
\end{aligned}
]
Solving this system of equations, we first rearrange them:
[
\begin{aligned}
3 – 2t &= -1 – u \quad \Rightarrow \quad 2t – u = 4 \
7 + 4t &= 18 + 3u \quad \Rightarrow \quad 4t – 3u = 11 \
-3 + 8t &= 7 + 2u \quad \Rightarrow \quad 8t – 2u = 10
\end{aligned}
]
We now solve the system of equations:
- ( 2t – u = 4 )
- ( 4t – 3u = 11 )
- ( 8t – 2u = 10 )
From equation (1), solve for ( u ):
[
u = 2t – 4
]
Substitute ( u = 2t – 4 ) into equations (2) and (3):
Substitute into equation (2):
[
4t – 3(2t – 4) = 11 \quad \Rightarrow \quad 4t – 6t + 12 = 11 \quad \Rightarrow \quad -2t = -1 \quad \Rightarrow \quad t = \frac{1}{2}
]
Now substitute ( t = \frac{1}{2} ) into ( u = 2t – 4 ):
[
u = 2\left(\frac{1}{2}\right) – 4 = 1 – 4 = -3
]
Step 4: Check the Point of Intersection
Now, we check the values of ( t = \frac{1}{2} ) and ( u = -3 ) in the parametric equations of both lines.
For ( l_1 ), with ( t = \frac{1}{2} ):
[
\begin{aligned}
x_1 &= 3 – 2\left(\frac{1}{2}\right) = 3 – 1 = 2 \
y_1 &= 7 + 4\left(\frac{1}{2}\right) = 7 + 2 = 9 \
z_1 &= -3 + 8\left(\frac{1}{2}\right) = -3 + 4 = 1
\end{aligned}
]
For ( l_2 ), with ( u = -3 ):
[
\begin{aligned}
x_2 &= -1 – (-3) = -1 + 3 = 2 \
y_2 &= 18 + 3(-3) = 18 – 9 = 9 \
z_2 &= 7 + 2(-3) = 7 – 6 = 1
\end{aligned}
]
Since the coordinates of both lines match at ( (2, 9, 1) ), the lines intersect at this point.
Conclusion
The lines ( l_1 ) and ( l_2 ) are intersecting and the point of intersection is ( (2, 9, 1) ).