Find the distance from the point (3, 1, −4) to the plane
4x − y − 3z = 1.

To find the distance \( d \) from the point \( P(3, 1, -4) \) to the plane defined by the equation \( 4x - y - 3z = 1 \), we can use the formula:

\[
d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]

Here, \( A = 4 \), \( B = -1 \), \( C = -3 \), and \( D = -1 \) (from rearranging the plane equation to the form \( Ax + By + Cz + D = 0 \)). The coordinates of the point are \( (x_0, y_0, z_0) = (3, 1, -4) \).

Now, substitute the values into the formula:

1. Calculate the numerator:
   \[
   |4(3) - 1(1) - 3(-4) - 1| = |12 - 1 + 12 - 1| = |22| = 22
   \]

2. Calculate the denominator:
   \[
   \sqrt{4^2 + (-1)^2 + (-3)^2} = \sqrt{16 + 1 + 9} = \sqrt{26}
   \]

Now, plug these into the distance formula:

\[
d = \frac{22}{\sqrt{26}} \approx \frac{22 \sqrt{26}}{26} = \frac{11 \sqrt{26}}{13}
\]

So, the distance from the point to the plane is:

\[
d = \frac{22}{\sqrt{26}} \approx 4.3
\]

Would you like further details or have any questions? Here are some related questions:

1. How do you derive the equation of a plane from three points?
2. Can you explain how the distance formula changes in different dimensions?
3. What are some applications of finding the distance from a point to a plane?
4. How would the distance change if the plane's coefficients were altered?
5. Can you find the distance from a point to a line in 3D? 

**Tip:** Always double-check your calculations by substituting back into the original equation.

Find the distance from the point (3, 1, −4) to the plane 4x − y − 3z = 1.

Explore the calculation of the distance from the point (3, 1, -4) to the plane defined by 4x - y - 3z = 1. This solution involves key concepts in geometry and coordinates, applying the distance formula to derive the result. Suitable for high school students.

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