Question

Find the point of intersection (if any) of the following pairs of lines. $$ \text { a. } \begin{array}{ll} x=3+t & x=4+2 s \\ & y=1-2 t & y=6+3 s \\ & z=3+3 t & z=1+s \\ & x=1-t & x=2 s \end{array} $$ b. \(\quad y=2+2 t \quad y=1+s\) $$ \begin{array}{c} z=-1+3 t \quad z=3 \\ \text { c. }\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} 3 \\ -1 \\ 2 \end{array}\right]+t\left[\begin{array}{r} 1 \\ 1 \\ -1 \end{array}\right] \\ {\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} 1 \\ 1 \\ -2 \end{array}\right]+s\left[\begin{array}{l} 2 \\ 0 \\ 3 \end{array}\right]} \end{array} $$ d. \(\left[\begin{array}{l}x \\ y \\\ z\end{array}\right]=\left[\begin{array}{r}4 \\ -1 \\\ 5\end{array}\right]+t\left[\begin{array}{l}1 \\ 0 \\ 1\end{array}\right]\) \(\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{r}2 \\\ -7 \\ 12\end{array}\right]+s\left[\begin{array}{r}0 \\ -2 \\\ 3\end{array}\right]\)

Short answer

(a) (2, 3, 0), (b) (0, \(\frac{17}{3}, 3)\), (c) No intersection, (d) (2, -1, 3).

Step-by-step solution

Equating Parametric Equations - Part a

For part (a), we have two sets of parametric equations for each of the x, y, and z coordinates. \[ x = 3 + t = 4 + 2s, \ y = 1 - 2t = 6 + 3s, \ z = 3 + 3t = 1 + s \] Solve these systems of equations to find if there exists a common \(t\) and \(s\).

Solving x-equations - Part a

From the equation \(x = 3 + t = 4 + 2s\), solve for \(t\) and \(s\). First set the equations equal: \(3 + t = 4 + 2s\). Simplifying gives \(t = 1 + 2s\).

Substituting into y-equation - Part a

Substitute \(t = 1 + 2s\) into the y equation \(y = 1 - 2t = 6 + 3s\), giving \(1 - 2(1 + 2s) = 6 + 3s\), which simplifies to \(-1 - 4s = 6 + 3s\), yielding \(s = -1\).

Finding t from s - Part a

Replace \(s = -1\) back in \(t = 1 + 2s\) to find \(t = 1 + 2(-1) = -1\).

Verify with the z-equation - Part a

Check using the z-equation \(z = 3 + 3t = 1 + s\). Substituting \(t = -1\) and \(s = -1\) gives \(3 + 3(-1) = 1 + (-1)\). Both give \(-2\), confirming solution.

Final Solution for Part a

The point of intersection is \((3 + t, 1 - 2t, 3 + 3t)\) when \(t = -1\) or \(s = -1\), resulting in \((2, 3, 0)\).

Equating Parametric Equations - Part b

For part (b), solve \(y = 2 + 2t = 1 + s\), \(z = -1 + 3t = 3\). From \(y = 2 + 2t = 1 + s\) get \(s = 2t + 1\). Substitute into \(z = -1 + 3t = 3\).

Solve z-equation - Part b

Set \(-1 + 3t = 3\) to find \(t\). Solving gives \(3t = 4\) and \(t = \frac{4}{3}\). Substitute into \(s = 2t + 1\) to get \(s = \frac{11}{3}\).

Verify Calculation - Part b

Verify by plugging \(t = \frac{4}{3}\) and \(s = \frac{11}{3}\) into both equations. Both should link each component back to the matching line coordinate. The point is \((x, y, z) = (0, \frac{17}{3}, 3)\).

Equating Parametric Equations - Part c

For part (c), solve for \(t\) and \(s\) in the vector equations: \[\begin{array}{l}x: 3+ t = 1 + 2s, \ y: -1 + t = 1 + 0s, \ z: 2 - t = -2 + 3s \end{array}\]

Solve Equations for x and y - Part c

Solve the first two equations: \(t = 2s - 2\) from \(3 + t = 1 + 2s\), and \(t = 2\) from \(-1 + t = 1\). They do not align, meaning these vectors don't intersect.

Equating Parametric Equations - Part d

Solve for \(t\) and \(s\) in the vector equations: \[\begin{array}{l}x: 4+ t = 2, \ y: -1 = -7 - 2s, \ z: 5 + t = 12 + 3s \end{array}\]

Solve for x, y, and z - Part d

For \(x\) solve \(4 + t = 2\) giving \(t = -2\). For \(y\) solve \(-1 = -7 - 2s\) giving \(s = -3\). For \(z\) solve \(5 + t = 12 + 3s\). If values match across t and s for both lines, then intersect.

Verify Intersecting Point Part d

Replace \(t = -2\) and \(s = -3\) into all three equations; check validity. If results are consistent, lines intersect there. Coordinates are \((2, -1, 3)\).

Compile and Conclude

From solving each system, pair (a) intersects at \((2, 3, 0)\), pair (b) intersects at \((0, \frac{17}{3}, 3)\), pair (c) does not intersect, and pair (d) intersects at \((2, -1, 3)\).

Key concepts

Line Intersection

Finding the point where two lines intersect is a common problem when dealing with parametric equations. In 3D geometry, lines are expressed in terms of parameters, often noted as \( t \) and \( s \). Each coordinate (x, y, z) of the line's point is defined by a separate equation depending on the parameter.
To find the intersection, you equate these equations for each coordinate: x, y, and z.

• Equate the equations for x-coordinates and solve for one of the parameters, say \( t \).
• Substitute the parameter value into the y-equation to solve for the other parameter, \( s \).
• Verify this solution in the z equation, ensuring consistency.

If all solutions fit consistently into their respective equations, then the lines intersect at a specific point. If not, the lines do not intersect or could be parallel.

Vector Analysis

Vector analysis is an essential skill in understanding lines in space. When parametric equations represent lines, they often use vector components to describe direction and position. A vector equation for a line in three-dimensional space can be written as \( \mathbf{r} = \mathbf{a} + t\mathbf{b} \), where \( \mathbf{a} \) is a fixed point on the line, \( \mathbf{b} \) is the direction vector, and \( t \) is the parameter.
This format helps in analyzing how lines interact with one another.

• For two lines to intersect, there must be a common point where their direction and position vectors meet.
• If the direction vectors are scalar multiples, the lines might be parallel but not necessarily intersecting.

Understanding how to manipulate these vectors and parameters lets you solve for interaction points effectively, as seen in the vector equations presented in our steps.

System of Equations

When solving for line intersection in parametric terms, you're essentially solving a system of equations. Each coordinate equation (x, y, z) gives rise to a distinct system when equated with its counterpart from the second line.
Here's how to tackle these:

• Start with one coordinate, solving for one parameter using simple algebra.
• Substitute any findings into the remaining equations to uncover more unknowns.
• Verify all results to ensure there's no contradiction between any of the equations.

Systems of equations in line intersection problem-solving are particularly straightforward because each parameter corresponds neatly to a single coordinate, simplifying the solution process.