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Find a nonzero vector normal to the plane 



To find a nonzero vector normal to the plane described by the equation \( z - 5(x - 2) = -4(1 - y) \), we first need to rewrite this equation in a standard plane form:

### Step 1: Expand both sides
Starting with the given equation:
\[
z - 5(x - 2) = -4(1 - y)
\]

Expand both sides:
\[
z - 5x + 10 = -4 + 4y
\]

### Step 2: Rearrange the equation
Move all terms to one side of the equation to get it in a form similar to \(Ax + By + Cz = D\):
\[
z - 5x + 10 = -4 + 4y
\]
\[
z - 5x - 4y + 10 + 4 = 0
\]
\[
-5x + 4y + z + 14 = 0
\]

Thus, the plane equation becomes:
\[
-5x + 4y + z + 14 = 0
\]

### Step 3: Identify the normal vector
For a plane in the form \(Ax + By + Cz = D\), the normal vector to the plane is given by \(\langle A, B, C \rangle\).

From the equation:
\[
-5x + 4y + z = -14
\]
The coefficients of \(x\), \(y\), and \(z\) are \(-5\), \(4\), and \(1\), respectively. Therefore, a nonzero normal vector to the plane is:

\[
\mathbf{n} = \langle -5, 4, 1 \rangle
\]

Would you like further details on how to verify this result or apply this normal vector?

Here are 5 related questions:
1. How can we use the normal vector to find the angle between two planes?
2. What is the significance of the normal vector in 3D geometry?
3. How do we find the distance between a point and the plane?
4. How can we calculate the area of a triangle formed by three points in space using vectors?
5. How can we convert this plane equation into parametric form?

**Tip**: The direction of the normal vector is crucial for determining the orientation of the plane. It also helps in calculating projections onto the plane.

Find a nonzero vector normal to the plane described by the equation z - 5(x - 2) = -4(1 - y).

This solution explains how to find a nonzero vector normal to the plane described by the equation z - 5(x - 2) = -4(1 - y). The solution includes the steps to rearrange the equation, identify key coefficients, and calculate the normal vector. Ideal for students in Grades 10-12.

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