https://file.aiquickdraw.com/mathbot/file/17280373720319pcqfoan.jpg

Let's break down the three parts of the problem based on the image you've uploaded:

### Part i) 
We are asked to turn the two parametric lines back into linear equations of the form \( ax_1 + bx_2 = c \) and solve the system. The given lines are:

\[
\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} + \alpha \begin{pmatrix} 1 \\ 1 \end{pmatrix}
\]
\[
\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix} + \beta \begin{pmatrix} 2 \\ 1 \end{pmatrix}
\]

We need to convert these parametric lines into linear equations by eliminating the parameters \( \alpha \) and \( \beta \).

For the first line:
\[
x_1 = 2 + \alpha \quad \text{and} \quad x_2 = 0 + \alpha = \alpha
\]
Thus, from the first line, \( x_2 = x_1 - 2 \).

For the second line:
\[
x_1 = 1 + 2\beta \quad \text{and} \quad x_2 = 3 + \beta
\]
Thus, from the second line, \( x_2 = \frac{x_1 - 1}{2} + 3 \).

Now we have the system:
1. \( x_2 = x_1 - 2 \)
2. \( x_2 = \frac{x_1 - 1}{2} + 3 \)

### Part ii)
To solve for the interception directly, we equate the two expressions for \( x_2 \) from the system above:

\[
x_1 - 2 = \frac{x_1 - 1}{2} + 3
\]

First, multiply both sides of the equation by 2 to eliminate the fraction:
\[
2(x_1 - 2) = (x_1 - 1) + 6
\]
\[
2x_1 - 4 = x_1 + 5
\]
Now subtract \( x_1 \) from both sides:
\[
x_1 - 4 = 5
\]
Solving for \( x_1 \):
\[
x_1 = 9
\]

Substitute \( x_1 = 9 \) into the first equation \( x_2 = x_1 - 2 \):
\[
x_2 = 9 - 2 = 7
\]

Thus, the two lines intersect at the point \( (9, 7) \).

### Part iii)
In the third part, we are given two new parametric lines in 3D space and asked whether they still intersect:

\[
\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} + \alpha \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}
\]
\[
\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} + \beta \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}
\]

The system becomes:

\[
\begin{aligned}
x_1 &= 2 + \alpha \quad \text{and} \quad x_1 = 1 + 2\beta \\
x_2 &= 0 + \alpha \quad \text{and} \quad x_2 = 3 + \beta \\
x_3 &= 1 + \alpha \quad \text{and} \quad x_3 = 1 + \beta
\end{aligned}
\]

We now solve for \( \alpha \) and \( \beta \) by equating each of the components \( x_1, x_2, x_3 \):

From \( x_3 = 1 + \alpha = 1 + \beta \), we get \( \alpha = \beta \).

Substitute \( \alpha = \beta \) into the equations for \( x_1 \) and \( x_2 \):

For \( x_1 \):
\[
2 + \alpha = 1 + 2\alpha
\]
\[
2 = 1 + \alpha \quad \Rightarrow \quad \alpha = 1
\]

For \( x_2 \):
\[
0 + \alpha = 3 + \alpha
\]
This leads to a contradiction, \( 0 = 3 \), meaning the two lines **do not intersect** in 3D space.

---

Would you like further clarifications or more details on any part of this solution? 

Here are 5 questions to explore this topic further:
1. How would you verify if two lines are parallel or skew in 3D space?
2. Can you explain geometrically why two lines in 2D with different slopes must intersect?
3. How would you generalize the process of finding intersections in higher dimensions?
4. How can the cross product help determine if two lines in 3D space are parallel?
5. What role does the direction vector play in determining whether lines in parametric form intersect?

**Tip:** In 3D, two lines might not intersect even if they are not parallel—they could be *skew* lines. Always check for consistency in all components when solving for intersections.

Given two lines in parametric form, where alpha, beta are real numbers, solve for their intersection and determine if they intersect in 3D space.

Learn how to find the intersection of two lines in parametric form, first in 2D and then in 3D. Explore the system of linear equations used to eliminate parameters and check if lines intersect. Understand key concepts in linear algebra and geometry, such as slopes, parametric equations, and the differences between parallel, intersecting, and skew lines.

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