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 \( AB \):

\[ \vec{AB} = (4 - 0, 5 - (-1), 1 - (-1)) = (4, 6, 2). \]

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 \( CD \):

\[ \vec{CD} = (-4 - 3, 4 - 9, 4 - 4) = (-7, -5, 0). \]

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

Equate the parametric equations to find a common point:

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:

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

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

\[ 4 \cdot \frac{5}{2} = 10 \implies 3 - 7t = 10 \implies -7t = 7 \implies t = -1. \]

Verify with the y-coordinate:

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

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

The equation holds, so \( s = \frac{5}{2} \), \( t = -1 \) is consistent.

Step 4: Find the Intersection Point

For line \( AB \), using \( s = \frac{5}{2} \):

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

Point: \( (10, 14, 4) \).

For line \( CD \), using \( t = -1 \):

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

Point: \( (10, 14, 4) \).

The points match, confirming intersection at \( (10, 14, 4) \).

Step 5: Check for Skew or Coplanar Lines

Direction vectors: \( AB: (4, 6, 2) \), \( CD: (-7, -5, 0) \).

Check if parallel by comparing ratios:

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

The lines are not parallel and share a point, so they intersect.

Final Answer

The lines intersect at:

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