D1: x=4 t, y=-4-t, z=6 2t, d2: x-5/2=y-11/4=z-5/2

General
1

d1: x=4 t, y=-4-t, z=6 2t, d2: x-5/2=y-11/4=z-5/2

2

Your question appears to involve two lines ( D_1 ) and ( D_2 ) in a 3D Cartesian coordinate system. First, let’s break down the parametric equations for each line and analyze them.

Line ( D_1 ):

The parametric equation for ( D_1 ) is given as:

[
x = 4t, \quad y = -4 - t, \quad z = 6 + 2t
]

This can be rewritten in vector form as:

[
\mathbf{r}_1(t) = \begin{bmatrix} x \ y \ z \end{bmatrix}
= \begin{bmatrix} 0 \ -4 \ 6 \end{bmatrix}

  • t \begin{bmatrix} 4 \ -1 \ 2 \end{bmatrix}
    ]

Here:

  • The point ( \mathbf{P}_1(0, -4, 6) ) is a point on the line.
  • The direction vector for ( D_1 ) is ( \mathbf{v}_1 = \begin{bmatrix} 4 \ -1 \ 2 \end{bmatrix} ).

Line ( D_2 ):

The symmetric equations for ( D_2 ) are given as:

[
\frac{x - 5}{2} = \frac{y - 11}{4} = \frac{z - 5}{2}
]

Introduce a parameter ( s ) (for simplicity) to rewrite this as parametric equations:

[
x = 5 + 2s, \quad y = 11 + 4s, \quad z = 5 + 2s
]

In vector form:

[
\mathbf{r}_2(s) = \begin{bmatrix} x \ y \ z \end{bmatrix}
= \begin{bmatrix} 5 \ 11 \ 5 \end{bmatrix}

  • s \begin{bmatrix} 2 \ 4 \ 2 \end{bmatrix}
    ]

Here:

  • The point ( \mathbf{P}_2(5, 11, 5) ) is a point on the line.
  • The direction vector for ( D_2 ) is ( \mathbf{v}_2 = \begin{bmatrix} 2 \ 4 \ 2 \end{bmatrix} ).

Task: Relationship Between ( D_1 ) and ( D_2 )

Now, let’s examine the relationship between the two lines ( D_1 ) and ( D_2 ). The possibilities include:

  1. The lines are parallel.
  2. The lines intersect at a point.
  3. The lines are skew (neither parallel nor intersecting).

1. Are the lines parallel?

To check if the lines are parallel, the direction vectors ( \mathbf{v}_1 = \begin{bmatrix} 4 \ -1 \ 2 \end{bmatrix} ) and ( \mathbf{v}_2 = \begin{bmatrix} 2 \ 4 \ 2 \end{bmatrix} ) should be scalar multiples of each other. That is, there should exist a scalar ( k ) such that:

[
\mathbf{v}_1 = k \mathbf{v}_2
]

Equating components:

[
\frac{4}{2} = \frac{-1}{4} = \frac{2}{2}
]

Checking these ratios:

  • ( \frac{4}{2} = 2 ),
  • ( \frac{-1}{4} = -0.25 ),
  • ( \frac{2}{2} = 1 ).

Since the ratios are not equal (( 2 \neq -0.25 \neq 1 )), the lines are not parallel.

2. Do the lines intersect?

To check for intersection, set the parametric equations of the two lines equal to each other. We need to solve for ( t ) and ( s ) such that:

[
\mathbf{r}_1(t) = \mathbf{r}_2(s)
]

That is:

[
\begin{cases}
4t = 5 + 2s \
-4 - t = 11 + 4s \
6 + 2t = 5 + 2s
\end{cases}
]

Solve these equations one by one.

From the first equation:
[
4t = 5 + 2s \implies 2t = 5/2 + s \implies t = \frac{5}{4} + \frac{s}{2}
]

From the second equation:
[
-4 - t = 11 + 4s
]

Substitute ( t = \frac{5}{4} + \frac{s}{2} ):

[
-4 - \left(\frac{5}{4} + \frac{s}{2}\right) = 11 + 4s
]

Simplify:

[
-4 - \frac{5}{4} - \frac{s}{2} = 11 + 4s
]
[
-\frac{16}{4} - \frac{5}{4} - \frac{s}{2} = 11 + 4s
]
[
-\frac{21}{4} - \frac{s}{2} = 11 + 4s
]

Multiply through by 4 to eliminate fractions:

[
-21 - 2s = 44 + 16s
]

Simplify:

[
-21 = 44 + 18s \implies -65 = 18s \implies s = -\frac{65}{18}
]

Substitute ( s = -\frac{65}{18} ) into ( t = \frac{5}{4} + \frac{s}{2} ):

[
t = \frac{5}{4} + \frac{-\frac{65}{18}}{2}
]
[
t = \frac{5}{4} - \frac{65}{36}
]

Find a common denominator for ( \frac{5}{4} ) and ( \frac{65}{36} ). The least common denominator is 36:

[
\frac{5}{4} = \frac{45}{36}, \quad \frac{65}{36} = \frac{65}{36}
]

[
t = \frac{45}{36} - \frac{65}{36} = -\frac{20}{36} = -\frac{5}{9}
]

Verification:

Now check if the third equation is satisfied when ( t = -\frac{5}{9} ) and ( s = -\frac{65}{18} ). Substituting:

Third equation: ( 6 + 2t = 5 + 2s )

[
6 + 2\left(-\frac{5}{9}\right) = 5 + 2\left(-\frac{65}{18}\right)
]
[
6 - \frac{10}{9} = 5 - \frac{130}{18}
]
[
\frac{54}{9} - \frac{10}{9} = \frac{90}{18} - \frac{130}{18}
]
[
\frac{44}{9} = \frac{-40}{18}
]

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