Find an equation of the plane which passes through the point (
5
, 
5
, 
1
), and contains the line x=(
2
+ 
2
*t), y=(
5
+ 
7
*t) and z=(
2
+ 
4
*t)
Leave your answer in the form A x + B y + C z - D , where A, B, C are integers.
Answer

We are asked to find the equation of a plane that passes through the point \( (5, 5, 1) \) and contains the line:

\[
x = 2 + 2t, \quad y = 5 + 7t, \quad z = 2 + 4t
\]

### Step 1: Determine two vectors in the plane.

The parametric equations of the line give the coordinates of any point on the line as:

\[
(x, y, z) = (2 + 2t, 5 + 7t, 2 + 4t)
\]

Let \( t = 0 \) and \( t = 1 \) to find two points on the line:

- When \( t = 0 \), the point on the line is \( (2, 5, 2) \).
- When \( t = 1 \), the point on the line is \( (4, 12, 6) \).

So, two points on the line are \( P_1(2, 5, 2) \) and \( P_2(4, 12, 6) \).

### Step 2: Form two direction vectors.

The vector from \( P_1(2, 5, 2) \) to \( P_2(4, 12, 6) \) is:

\[
\mathbf{v_1} = (4 - 2, 12 - 5, 6 - 2) = (2, 7, 4)
\]

Next, find the vector from the given point \( P_3(5, 5, 1) \) to the point \( P_1(2, 5, 2) \):

\[
\mathbf{v_2} = (2 - 5, 5 - 5, 2 - 1) = (-3, 0, 1)
\]

### Step 3: Compute the normal vector to the plane.

The normal vector \( \mathbf{n} \) to the plane can be found by taking the cross product of \( \mathbf{v_1} \) and \( \mathbf{v_2} \):

\[
\mathbf{n} = \mathbf{v_1} \times \mathbf{v_2} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 7 & 4 \\
-3 & 0 & 1
\end{vmatrix}
\]

\[
\mathbf{n} = \mathbf{i} \left( 7 \cdot 1 - 4 \cdot 0 \right) - \mathbf{j} \left( 2 \cdot 1 - 4 \cdot (-3) \right) + \mathbf{k} \left( 2 \cdot 0 - 7 \cdot (-3) \right)
\]

\[
\mathbf{n} = \mathbf{i} (7) - \mathbf{j} (2 + 12) + \mathbf{k} (21)
\]

\[
\mathbf{n} = (7, -14, 21)
\]

Thus, the normal vector is \( \mathbf{n} = (7, -14, 21) \).

### Step 4: Form the equation of the plane.

The equation of the plane is given by:

\[
A(x - x_0) + B(y - y_0) + C(z - z_0) = 0
\]

where \( (x_0, y_0, z_0) \) is a point on the plane (we use \( (5, 5, 1) \)) and \( (A, B, C) \) is the normal vector \( (7, -14, 21) \).

Substituting these values:

\[
7(x - 5) - 14(y - 5) + 21(z - 1) = 0
\]

Expanding:

\[
7x - 35 - 14y + 70 + 21z - 21 = 0
\]

\[
7x - 14y + 21z + 14 = 0
\]

Thus, the equation of the plane is:

\[
7x - 14y + 21z - 14 = 0
\]

### Final Answer:

\[
7x - 14y + 21z - 14 = 0
\]

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. How do you find the distance from a point to a plane?
2. How do you find the equation of a plane through three points?
3. How do you determine if two planes are parallel or perpendicular?
4. What is the geometric interpretation of the cross product?
5. How do you find the angle between two planes?

**Tip:** The cross product of two vectors is perpendicular to both, making it a key tool in finding the normal vector to a plane.

Find an equation of the plane which passes through the point (5, 5, 1), and contains the line x=(2 + 2t), y=(5 + 7t), and z=(2 + 4t). Leave your answer in the form A x + B y + C z - D, where A, B, C are integers.

Discover how to find the equation of a plane passing through a point and containing a parametric line. This problem involves key vector geometry concepts, including parametric equations, cross products, and the equation of a plane. Suitable for high school students in Grades 11-12.

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