Find an equation for the plane through the points (4,3,5),(1,2,0),(1,0,2).(4,3,5),(1,2,0),(1,0,2).


To find the equation of a plane passing through three points, we need the following steps:

### Step 1: Define the general equation of a plane

The equation of a plane in 3D space is of the form:
\[
Ax + By + Cz + D = 0
\]
where \(A\), \(B\), and \(C\) are the components of the normal vector to the plane, and \(D\) is a constant.

### Step 2: Find two vectors in the plane

Given three points \(P_1(4,3,5)\), \(P_2(1,2,0)\), and \(P_3(1,0,2)\), we first compute two vectors that lie in the plane. These can be obtained by subtracting the coordinates of one point from the others:

1. Vector \(\vec{v_1} = P_2 - P_1 = (1-4, 2-3, 0-5) = (-3, -1, -5)\)
2. Vector \(\vec{v_2} = P_3 - P_1 = (1-4, 0-3, 2-5) = (-3, -3, -3)\)

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

The normal vector \(\vec{n} = (A, B, C)\) to the plane is perpendicular to both \(\vec{v_1}\) and \(\vec{v_2}\). We find \(\vec{n}\) by computing the cross product of \(\vec{v_1}\) and \(\vec{v_2}\):

\[
\vec{n} = \vec{v_1} \times \vec{v_2} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-3 & -1 & -5 \\
-3 & -3 & -3
\end{vmatrix}
\]

Expanding this determinant:

\[
\vec{n} = \mathbf{i} \left((-1)(-3) - (-5)(-3)\right) - \mathbf{j} \left((-3)(-3) - (-5)(-3)\right) + \mathbf{k} \left((-3)(-3) - (-3)(-1)\right)
\]
\[
\vec{n} = \mathbf{i} (3 - 15) - \mathbf{j} (9 - 15) + \mathbf{k} (9 - 3)
\]
\[
\vec{n} = \mathbf{i} (-12) - \mathbf{j} (-6) + \mathbf{k} (6)
\]
\[
\vec{n} = (-12, 6, 6)
\]

Thus, the normal vector to the plane is \(\vec{n} = (-12, 6, 6)\).

### Step 4: Write the equation of the plane

Now that we have the normal vector \((-12, 6, 6)\), we can substitute these values into the general plane equation \(Ax + By + Cz + D = 0\):

\[
-12(x - 4) + 6(y - 3) + 6(z - 5) = 0
\]

Expanding this equation:

\[
-12x + 48 + 6y - 18 + 6z - 30 = 0
\]
\[
-12x + 6y + 6z = 0
\]
Dividing the equation by 6 to simplify:

\[
-2x + y + z = 0
\]

Thus, the equation of the plane is:
\[
\boxed{-2x + y + z = 0}
\]

Let me know if you would like any further details or explanations!

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Here are 5 related questions to deepen your understanding:

1. How can you find the distance from a point to a plane using the plane equation?
2. What is the geometric interpretation of the cross product of two vectors?
3. How do you find the equation of a plane parallel to a given one?
4. How would the equation change if we translated the plane by a certain vector?
5. How do you check if a given point lies on the plane?

**Tip:** The cross product of two vectors provides a perpendicular vector, which is useful for finding planes and lines in 3D geometry.

Find an equation for the plane through the points (4,3,5), (1,2,0), (1,0,2).

Learn how to find the equation of a plane passing through three points in 3D space: (4, 3, 5), (1, 2, 0), and (1, 0, 2). This solution covers the use of vector subtraction, cross product, and plane equations, suitable for students in Grades 10-12.

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