Find an equation of the plane.
the plane through the point 
(6, −3, 6)
 and perpendicular to the vector 
−i + 3j + 5k


To find the equation of a plane that passes through a given point and is perpendicular to a given vector, we use the **standard form of a plane** equation:

\[
a(x - x_0) + b(y - y_0) + c(z - z_0) = 0
\]

where \( (x_0, y_0, z_0) \) is a point on the plane, and \( \langle a, b, c \rangle \) is the **normal vector** of the plane.

### Given:
- Point on the plane: \( (6, -3, 6) \)
- Normal vector: \( \mathbf{n} = \langle -1, 3, 5 \rangle \) (from the vector \( -i + 3j + 5k \))

### Step-by-Step Solution:

1. **Substitute** the normal vector \( \langle a, b, c \rangle = \langle -1, 3, 5 \rangle \) into the plane equation:
   \[
   -1(x - 6) + 3(y + 3) + 5(z - 6) = 0
   \]

2. **Expand** the terms:
   \[
   -x + 6 + 3y + 9 + 5z - 30 = 0
   \]

3. **Simplify**:
   \[
   -x + 3y + 5z - 15 = 0
   \]

4. **Rearrange** to get the standard form of the plane equation:
   \[
   x - 3y - 5z = -15
   \]

### Final Answer:
The equation of the plane is:

\[
x - 3y - 5z = -15
\]

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Would you like further details or have any questions?

Here are 5 related questions for practice:
1. How do you find the equation of a plane passing through three points?
2. What is the meaning of a normal vector to a plane?
3. How do you find the distance between a point and a plane?
4. How do you determine if two planes are parallel?
5. How do you find the line of intersection between two planes?

**Tip:** Always verify your plane's equation by substituting the given point to ensure it satisfies the equation!

Find an equation of the plane. The plane passes through the point (6, −3, 6) and is perpendicular to the vector −i + 3j + 5k.

Learn how to find the equation of a plane that passes through the point (6, −3, 6) and is perpendicular to the vector −i + 3j + 5k. This problem covers key concepts in 3D analytic geometry and vector mathematics, suitable for Grades 10-12.

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