Solution (make a sketch yourself): The constants in front of x and y give the coordinates of a normal vector to the line:
~nl =
 
2
3
 
.
This is then a direction vector for k (with k as in Method 1.11), and
since we know that k goes through Q, we can write k in parameter form:
k :
 
x
y
 
=
 
1
1
 
+ t
 
2
3
 
=
 
1 + 2h
1 + 3h
 
, t ∈ R.
To find the intersection, this is inserted for x and y in the equation for
the line:
2(1 + 2t) + 3(1 + 3t) = 6 ⇒ 4t + 9t = 1
so t =
1
13 and that gives the point:
N = (1 + 2 1
13 , 1 + 3 1
13 ) = ( 15
13 ,
16
13 ).


Let's break down the steps to solve this problem, which involves finding the intersection of a line and a normal vector to it.

### Step-by-Step Solution

1. **Normal Vector to the Line**:
   We're given the normal vector \(\vec{n}_L = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\), which is perpendicular to the line. The normal vector provides the direction of a line perpendicular to the given line, which we'll call \( k \).

2. **Parametric Form of Line \( k \)**:
   The line \( k \) passes through point \( Q = (1, 1) \) and has a direction vector equal to the normal vector \( \vec{n}_L = \begin{pmatrix} 2 \\ 3 \end{pmatrix} \). Therefore, we can write the parametric form of \( k \) as:

   \[
   k: \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 3 \end{pmatrix}
   \]

   Simplifying, we get:

   \[
   x = 1 + 2t, \quad y = 1 + 3t, \quad t \in \mathbb{R}
   \]

3. **Substitute \( x \) and \( y \) into the Line Equation**:
   We're given that the line equation is \( 2x + 3y = 6 \). To find the intersection, we substitute the parametric expressions for \( x \) and \( y \) into this equation:

   \[
   2(1 + 2t) + 3(1 + 3t) = 6
   \]

   Expanding both sides:

   \[
   2 + 4t + 3 + 9t = 6
   \]

   Simplifying:

   \[
   5 + 13t = 6
   \]

   Solving for \( t \):

   \[
   13t = 1 \quad \Rightarrow \quad t = \frac{1}{13}
   \]

4. **Find the Intersection Point \( N \)**:
   Now that we know \( t = \frac{1}{13} \), we substitute it back into the parametric equations for \( x \) and \( y \) to find the coordinates of the intersection point \( N \):

   \[
   x = 1 + 2 \cdot \frac{1}{13} = 1 + \frac{2}{13} = \frac{15}{13}
   \]
   \[
   y = 1 + 3 \cdot \frac{1}{13} = 1 + \frac{3}{13} = \frac{16}{13}
   \]

   Thus, the intersection point is:

   \[
   N = \left( \frac{15}{13}, \frac{16}{13} \right)
   \]

### Sketch of the Problem

The sketch of this problem would involve:
1. A line represented by the equation \( 2x + 3y = 6 \).
2. A point \( Q = (1, 1) \).
3. A line \( k \) passing through \( Q \) with a direction vector \( \vec{n}_L = \begin{pmatrix} 2 \\ 3 \end{pmatrix} \), intersecting the original line at the point \( N = \left( \frac{15}{13}, \frac{16}{13} \right) \).

Let me know if you'd like a sketch!

---

### Related Questions:
1. How is a normal vector to a line related to the slope of the line?
2. What is the general method to find the intersection of two lines in parametric form?
3. How would the solution change if the point \( Q \) were different?
4. Can we generalize this method to 3-dimensional lines and planes?
5. What happens if the lines are parallel, and how would that affect the intersection point?

### Tip:
Always check if the lines are parallel before solving, as parallel lines do not intersect unless they are coincident.

The constants in front of x and y give the coordinates of a normal vector to the line: ~nl = 2 3 . This is then a direction vector for k (with k as in Method 1.11), and since we know that k goes through Q, we can write k in parameter form: k : x y = 1 1 + t 2 3 = 1 + 2h 1 + 3h , t ∈ R. To find the intersection, this is inserted for x and y in the equation for the line: 2(1 + 2t) + 3(1 + 3t) = 6 ⇒ 4t + 9t = 1 so t = 1 13 and that gives the point: N = (1 + 2 1 13 , 1 + 3 1 13 ) = ( 15 13 , 16 13 ).

Explore how to find the intersection of a parametric line passing through point Q (1, 1) with a direction vector (2, 3) and a standard line 2x + 3y = 6. This problem involves parametric equations, vector geometry, and substitution methods to solve for the intersection point, suitable for undergraduate students or advanced high school learners.

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