Questions: ASSIGNMENT Assignment 2 Items Details Items - Due Oct 18, 2024 11:59 PM. - Late work is permitted until Oct 21, 2024 11:59 PM. It will be turned in automatically after the late work period ends. Submit answer Get help Practice similar Find the distance from the point (-1,2,5) to the plane -4x+3y-5z=5. Submit answer Next item Answers Answer Points - B 1 Report technical issue · Email instructor

Solution Steps

To find the distance from a point to a plane, we use the formula for the distance \( D \) from a point \((x_1, y_1, z_1)\) to a plane \(Ax + By + Cz + D = 0\), which is given by:

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

In this problem, the point is \((-1, 2, 5)\) and the plane is \(-4x + 3y - 5z = 5\). We need to rewrite the plane equation in the form \(Ax + By + Cz + D = 0\) by moving the constant to the left side, resulting in \(-4x + 3y - 5z - 5 = 0\).

1. Identify the coefficients \(A\), \(B\), \(C\), and \(D\) from the plane equation.
2. Substitute the point coordinates and the coefficients into the distance formula.
3. Calculate the distance using Python.

The given plane equation is

\[ -4x + 3y - 5z = 5 \]

We can rewrite it in the standard form \(Ax + By + Cz + D = 0\) by moving the constant to the left side:

\[ -4x + 3y - 5z - 5 = 0 \]

Thus, we have \(A = -4\), \(B = 3\), \(C = -5\), and \(D = -5\).

The point from which we want to find the distance is \((-1, 2, 5)\). We will substitute these coordinates into the distance formula:

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

Substituting the values:

\[ D = \frac{|-4(-1) + 3(2) - 5(5) - 5|}{\sqrt{(-4)^2 + 3^2 + (-5)^2}} \]

Calculating the numerator:

\[ |-4(-1) + 3(2) - 5(5) - 5| = |4 + 6 - 25 - 5| = |4 + 6 - 30| = |10 - 30| = |20| \]

Calculating the denominator:

\[ \sqrt{(-4)^2 + 3^2 + (-5)^2} = \sqrt{16 + 9 + 25} = \sqrt{50} = 5\sqrt{2} \]

Now substituting the values back into the distance formula:

\[ D = \frac{20}{5\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \]

Final Answer