Question

Show that the line passing through the points A $(0, -1, -1)$ and B $(4, 5, 1)$ intersects the line joining points C $(3, 9, 4)$ and D $(-4, 4, 4)$.

Hint:

When solving for the intersection of two lines in space, equate the parametric equations of the lines and solve for the parameter. If the solution is inconsistent, the lines do not intersect.

Answer 1

Step 1: Parametric Equations of the Lines

Line \( AB \):

Points: \( A(0, -1, -1) \), \( B(4, 5, 1) \).

Direction vector of \( AB \):

\[ \vec{AB} = (4, 6, 2). \]

Parametric equations for line \( AB \) (using point \( A \) and parameter \( s \)):

\[ x = 4s, \quad y = -1 + 6s, \quad z = -1 + 2s. \]

Line \( CD \):

Points: \( C(3, 9, 4) \), \( D(-4, 4, 4) \).

Direction vector of \( CD \):

\[ \vec{CD} = (-7, -5, 0). \]

Parametric equations for line \( CD \) (using point \( C \) and parameter \( t \)):

\[ x = 3 - 7t, \quad y = 9 - 5t, \quad z = 4. \]

Step 2: Set Up Equations for Intersection

To find the intersection, we equate the corresponding coordinates:

1. x-coordinate: \( 4s = 3 - 7t \)
2. y-coordinate: \( -1 + 6s = 9 - 5t \)
3. z-coordinate: \( -1 + 2s = 4 \)

Step 3: Solve the Equations

From the z-coordinate equation:

\[ -1 + 2s = 4 \implies 2s = 5 \implies s = \frac{5}{2}. \]

Substitute \( s = \frac{5}{2} \) into the x-coordinate equation:

\[ 4 \left(\frac{5}{2}\right) = 3 - 7t \implies 10 = 3 - 7t \implies -7t = 7 \implies t = -1. \]

Verify these values using the y-coordinate equation:

Left side: \( -1 + 6 \left(\frac{5}{2}\right) = -1 + 15 = 14 \).

Right side: \( 9 - 5(-1) = 9 + 5 = 14 \).

The y-coordinate equation is satisfied, confirming \( s = \frac{5}{2} \) and \( t = -1 \) are consistent.

Step 4: Find the Intersection Point

Using \( s = \frac{5}{2} \) in the parametric equations for line \( AB \):

\[ x = 4 \left(\frac{5}{2}\right) = 10, \quad y = -1 + 6 \left(\frac{5}{2}\right) = 14, \quad z = -1 + 2 \left(\frac{5}{2}\right) = 4. \]

The intersection point is \( (10, 14, 4) \).

Using \( t = -1 \) in the parametric equations for line \( CD \):

\[ x = 3 - 7(-1) = 10, \quad y = 9 - 5(-1) = 14, \quad z = 4. \]

The intersection point is \( (10, 14, 4) \).

Both lines yield the same point, confirming they intersect.

Step 5: Check for Skew or Coplanar Lines

Direction vectors are \( \vec{AB} = (4, 6, 2) \) and \( \vec{CD} = (-7, -5, 0) \).

Check for parallelism by comparing the ratios of the components:

\[ \frac{4}{-7} eq \frac{6}{-5} eq \frac{2}{0}. \]

The lines are not parallel. Since they share a common point, they intersect.

Final Answer

The lines intersect at:

\[ \boxed{(10, 14, 4)} \]