Find equation of a plane c) Containing the line I / x = - 2 + 3t y = 4 + 2t z = 3 - 1 and perpendicular to the plane pi / x - 2y + z = 5

To find the equation of a plane that contains the line and is perpendicular to another plane, we need to follow a systematic process.

### Given:
1. The parametric form of the line is:
   \[
   \frac{x + 2}{3} = t, \quad y = 4 + 2t, \quad z = 3 - t
   \]
   or equivalently:
   \[
   x = -2 + 3t, \quad y = 4 + 2t, \quad z = 3 - t
   \]
   
2. The plane \(\pi\) has the equation:
   \[
   x - 2y + z = 5
   \]

We need to find the equation of the plane that contains the given line and is perpendicular to \(\pi\).

### Step 1: Find the direction vector of the line
From the parametric equations of the line:
\[
x = -2 + 3t, \quad y = 4 + 2t, \quad z = 3 - t
\]
The direction vector \(\vec{d}\) of the line is:
\[
\vec{d} = \langle 3, 2, -1 \rangle
\]

### Step 2: Find the normal vector of the given plane \(\pi\)
The equation of the plane is \(x - 2y + z = 5\), so its normal vector is:
\[
\vec{n}_{\pi} = \langle 1, -2, 1 \rangle
\]

### Step 3: The plane we are finding is perpendicular to \(\pi\)
Since the desired plane is perpendicular to \(\pi\), its normal vector \(\vec{n}_{\text{plane}}\) must lie in the direction of both the normal vector \(\vec{n}_{\pi}\) of \(\pi\) and the direction vector \(\vec{d}\) of the line.

The normal vector of the required plane can be found by taking the cross product of \(\vec{d}\) and \(\vec{n}_{\pi}\):
\[
\vec{n}_{\text{plane}} = \vec{d} \times \vec{n}_{\pi}
\]
\[
\vec{n}_{\text{plane}} = \langle 3, 2, -1 \rangle \times \langle 1, -2, 1 \rangle
\]

### Step 4: Compute the cross product
We use the determinant formula for the cross product:
\[
\vec{n}_{\text{plane}} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
3 & 2 & -1 \\
1 & -2 & 1
\end{vmatrix}
\]
\[
\vec{n}_{\text{plane}} = \hat{i}((2)(1) - (-1)(-2)) - \hat{j}((3)(1) - (-1)(1)) + \hat{k}((3)(-2) - (2)(1))
\]
\[
\vec{n}_{\text{plane}} = \hat{i}(2 - 2) - \hat{j}(3 + 1) + \hat{k}(-6 - 2)
\]
\[
\vec{n}_{\text{plane}} = \hat{i}(0) - \hat{j}(4) + \hat{k}(-8)
\]
\[
\vec{n}_{\text{plane}} = \langle 0, -4, -8 \rangle
\]
Thus, the normal vector of the required plane is \(\langle 0, -4, -8 \rangle\), or equivalently \(\langle 0, 1, 2 \rangle\) after dividing by \(-4\).

### Step 5: Find a point on the plane
We can use a point from the line. Let \(t = 0\), then the point on the line is:
\[
P( x = -2, y = 4, z = 3 )
\]

### Step 6: Write the equation of the plane
The equation of a plane with normal vector \(\langle A, B, C \rangle\) passing through a point \(P(x_0, y_0, z_0)\) is:
\[
A(x - x_0) + B(y - y_0) + C(z - z_0) = 0
\]
Substituting \(A = 0\), \(B = 1\), \(C = 2\), and the point \(P(-2, 4, 3)\):
\[
0(x + 2) + 1(y - 4) + 2(z - 3) = 0
\]
\[
y - 4 + 2(z - 3) = 0
\]
\[
y - 4 + 2z - 6 = 0
\]
\[
y + 2z = 10
\]

### Final Answer:
The equation of the plane is:
\[
y + 2z = 10
\]

Would you like more details or have any questions?

### Related questions:
1. How do you find the equation of a plane using three points?
2. What is the significance of the cross product in geometry?
3. How do you verify if a line lies in a plane?
4. What are other ways to define a plane besides normal vectors?
5. Can two planes be both perpendicular and parallel to the same line?

### Tip:
The cross product of two vectors gives a vector perpendicular to both, which is very useful in geometry, especially for finding normals to planes.

Find the equation of a plane containing the line x = -2 + 3t, y = 4 + 2t, z = 3 - t, and perpendicular to the plane x - 2y + z = 5.

This solution explains how to find the equation of a plane that contains the line x = -2 + 3t, y = 4 + 2t, z = 3 - t, and is perpendicular to the plane x - 2y + z = 5. The problem involves cross products and parametric equations, making it suitable for high school students in Grades 11-12.

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