Find the point at which the line intersects the given plane.
x = 2 − 2t,    y = 3t,    z = 1 + t;    x + 2y − z = 4

To find the point at which the line intersects the plane, we can proceed as follows:

### 1. Parametric Equations of the Line:
The parametric equations of the line are given by:

\[
x = 2 - 2t, \quad y = 3t, \quad z = 1 + t
\]

### 2. Equation of the Plane:
The equation of the plane is:

\[
x + 2y - z = 4
\]

### 3. Substituting the Parametric Equations into the Plane Equation:
We will substitute \(x = 2 - 2t\), \(y = 3t\), and \(z = 1 + t\) into the plane equation \(x + 2y - z = 4\).

\[
(2 - 2t) + 2(3t) - (1 + t) = 4
\]

Simplify the equation:

\[
2 - 2t + 6t - 1 - t = 4
\]

\[
1 + 3t = 4
\]

### 4. Solve for \(t\):

\[
3t = 3
\]

\[
t = 1
\]

### 5. Substitute \(t = 1\) back into the parametric equations:

\[
x = 2 - 2(1) = 0
\]
\[
y = 3(1) = 3
\]
\[
z = 1 + 1 = 2
\]

### 6. Intersection Point:
The point where the line intersects the plane is \((0, 3, 2)\).

Would you like a detailed explanation of any of these steps or further clarification?

Here are 5 questions to expand on this topic:
1. How would the solution change if the plane's equation were different?
2. What if the parametric equation of the line had a different parameter?
3. How can we check if a line lies entirely on the plane?
4. What if the plane's equation was homogeneous (e.g., \( x + 2y - z = 0 \))?
5. How do you calculate the distance between a point and a plane?

**Tip:** Always ensure that the parametric equations are correctly substituted into the plane's equation when solving intersection problems.

Discover how to find the point where the line intersects the given plane using parametric equations and the plane's equation. Step-by-step solution with detailed calculations and explanations. Suitable for Grades 10-12 students.

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