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The problem asks us to find the equation of a plane with the following given dataThe problem asks us to find the equation of a plane with the following given 

- Normal vector \(\mathbf{n} = 9\mathbf{j}\), which means the normal vector is \( \mathbf{n} = \langle 0, 9, 0 \rangle \) (only in the \(y\)-direction).
- The plane passes through the point \(P(11, 3, 6)\).

The general equation of a plane is given by:

\[
ax + by + cz = d
\]

where \(\mathbf{n} = \langle a, b, c \rangle \) is the normal vector, and \((x_0, y_0, z_0)\) is a point on the plane.

### Step 1: Find the coefficients \(a\), \(b\), and \(c\)
The normal vector \(\mathbf{n} = \langle 0, 9, 0 \rangle\) gives us the values of \(a\), \(b\), and \(c\):
\[
a = 0, \quad b = 9, \quad c = 0
\]

### Step 2: Substitute the point into the equation
Now, we substitute the coordinates of the point \(P(11, 3, 6)\) into the plane equation \(ax + by + cz = d\):

\[
0(11) + 9(3) + 0(6) = d
\]

\[
0 + 27 + 0 = d \implies d = 27
\]

### Step 3: Write the final equation of the plane
Thus, the equation of the plane is:

\[
9y = 27
\]

or, simplified:

\[
y = 3
\]

This means the plane is parallel to the \(xz\)-plane and lies at \(y = 3\).

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Would you like more details on how to visualize the plane or any further explanations?

Here are 5 related questions for deeper understanding:
1. How can you derive the general form of a plane equation from its normal vector and a point on it?
2. What does it mean geometrically when the normal vector has only a \(y\)-component?
3. How would the plane equation change if the normal vector had components in the \(x\) or \(z\)-directions?
4. Can you find the distance from a point not on the plane to this plane?
5. How would the equation of the plane change if the normal vector were multiplied by a scalar?

**Tip:** For a plane to be parallel to one of the coordinate planes (like the \(xz\)-plane), its normal vector will have only one non-zero component in the perpendicular direction.

Write the equation of the plane with normal vector n = 9j passing through the point (11, 3, 6) in scalar form ax + by + cz = d. (Express numbers in exact form. Use symbolic notation and fractions where needed.)

Find the equation of a plane with a normal vector of 9j and passing through the point (11, 3, 6). This solution involves using the equation of a plane formula and is suitable for high school students in Grades 11-12 studying analytic geometry.

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