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To solve this problem, we will use the formula to find the distance \( D \) from a point \( P(x_1, y_1, z_1) \) to a plane \( Ax + By + Cz + D = 0 \). The distance is given by:

\[
D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]

The equation of the plane is \( 4x + 3y - 4z = -7 \), which can be rewritten as:

\[
4x + 3y - 4z + 7 = 0
\]

Here:
- \( A = 4 \)
- \( B = 3 \)
- \( C = -4 \)
- \( D = 7 \)

The point is \( (4, 4, -5) \), which means:
- \( x_1 = 4 \)
- \( y_1 = 4 \)
- \( z_1 = -5 \)

Substitute the values into the formula and compute the distance:

\[
D = \frac{|4(4) + 3(4) - 4(-5) + 7|}{\sqrt{4^2 + 3^2 + (-4)^2}}
\]

Let me calculate that.The distance from the point \( (4, 4, -5) \) to the plane \( 4x + 3y - 4z = -7 \) is approximately \( 8.59 \) units.

Would you like more details or have any questions? 

Here are 5 related questions to deepen your understanding:
1. How is the distance formula derived from vector projection principles?
2. What changes if the plane equation has a different constant term?
3. How do we find the closest point on the plane from a given point?
4. Can the distance formula be extended to higher dimensions?
5. What happens if the given point lies on the plane?

**Tip:** Always ensure the plane equation is in the standard form \( Ax + By + Cz + D = 0 \) before applying the distance formula.

Find the distance from the point (4, 4, -5) to the plane 4x + 3y - 4z = -7.

Learn how to find the distance from a point (4, 4, -5) to the plane 4x + 3y - 4z = -7 using the distance formula in 3D coordinate geometry. This problem covers key concepts in analytic geometry and is suitable for Grades 11-12.

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