Find the veetor and Cartesian equation of the plane passing through the points 4(2,3,1), 8(4,-5,3) and parallel to X-axis

To find the vector and Cartesian equations of a plane passing through the points \( A(2,3,1) \), \( B(4,-5,3) \), and parallel to the X-axis, we can follow these steps:

### Step 1: Find two vectors on the plane
Since the plane passes through points \( A \) and \( B \), we can create two vectors on the plane by subtracting one point from the other.

- **Vector \( \vec{AB} \):**
  \[
  \vec{AB} = \vec{B} - \vec{A} = (4, -5, 3) - (2, 3, 1) = (4-2, -5-3, 3-1) = (2, -8, 2)
  \]
  So, \( \vec{AB} = (2, -8, 2) \).

- **Vector parallel to the X-axis**: Since the plane is parallel to the X-axis, it must contain a vector that is parallel to the X-axis. A vector parallel to the X-axis is simply \( (1, 0, 0) \).

### Step 2: Find the normal vector to the plane
The normal vector to the plane is perpendicular to any two vectors that lie on the plane. So, we take the cross product of \( \vec{AB} = (2, -8, 2) \) and the vector parallel to the X-axis \( \vec{i} = (1, 0, 0) \).

The cross product \( \vec{N} = \vec{AB} \times \vec{i} \) is calculated as follows:
\[
\vec{N} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & -8 & 2 \\
1 & 0 & 0
\end{vmatrix}
\]
Expanding the determinant:
\[
\vec{N} = \hat{i} \begin{vmatrix} -8 & 2 \\ 0 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 2 \\ 1 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -8 \\ 1 & 0 \end{vmatrix}
\]
\[
\vec{N} = \hat{i} (0 - 0) - \hat{j} (0 - 2) + \hat{k} (0 + 8)
\]
\[
\vec{N} = 0 \hat{i} + 2 \hat{j} + 8 \hat{k}
\]
Thus, the normal vector is:
\[
\vec{N} = (0, 2, 8)
\]

### Step 3: Find the Cartesian equation of the plane
The general equation of a plane is given by:
\[
Ax + By + Cz = D
\]
Where \( (A, B, C) \) is the normal vector to the plane. So, we substitute the normal vector \( (0, 2, 8) \) into this equation:
\[
0x + 2y + 8z = D
\]
Now, to find \( D \), substitute the coordinates of one of the points on the plane. Let's use point \( A(2, 3, 1) \):
\[
0(2) + 2(3) + 8(1) = D
\]
\[
D = 6 + 8 = 14
\]
Thus, the Cartesian equation of the plane is:
\[
2y + 8z = 14
\]

### Step 4: Find the vector equation of the plane
The vector equation of a plane is given by:
\[
\vec{r} = \vec{r_0} + s\vec{v_1} + t\vec{v_2}
\]
Where \( \vec{r_0} \) is the position vector of a point on the plane (e.g., point \( A(2, 3, 1) \)), and \( \vec{v_1} \) and \( \vec{v_2} \) are two non-parallel vectors lying on the plane (e.g., \( \vec{AB} \) and the X-axis vector).

Thus, the vector equation of the plane is:
\[
\vec{r} = (2, 3, 1) + s(2, -8, 2) + t(1, 0, 0)
\]
This represents the plane passing through \( A(2, 3, 1) \), parallel to the X-axis.

### Final Results:
1. **Vector Equation of the Plane**: 
   \[
   \vec{r} = (2, 3, 1) + s(2, -8, 2) + t(1, 0, 0)
   \]
   
2. **Cartesian Equation of the Plane**:
   \[
   2y + 8z = 14
   \]

Let me know if you would like further clarification or have additional questions!

---

**Next Steps to Explore:**
1. How to determine the distance from a point to a plane.
2. Deriving the vector equation of a plane from three non-collinear points.
3. How the normal vector affects the orientation of the plane.
4. Applications of planes in 3D geometry and physics.
5. How to find the intersection of two planes.

**Tip:** Always check if the given plane is parallel to any coordinate axis, as this simplifies the calculation by giving you easy-to-spot parallel vectors!

Find the vector and Cartesian equation of the plane passing through the points 4(2,3,1), 8(4,-5,3) and parallel to X-axis

Learn how to find the vector and Cartesian equations of a plane passing through the points (2, 3, 1) and (4, -5, 3), and parallel to the X-axis. This step-by-step guide involves concepts such as cross product, vector equations, and the Cartesian equation of a plane. Suitable for high school students in Grades 10-12.

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