A plane, π, has 3x-5z+3=0 as its Cartesian equation. Determine the Cartesian equation of a plane that is perpendicular to π and contains the point P(2,9,-3)

the normal of the given plane is

(3,0,-5) which is then the direction of your line

A vector equation of that line is
r = (x,y,z) = (2,9,-3) + t(3,0,-5) where t is your parameter

x = 2 + 3t ---> t = (x-2)/3
y = 9 + 0t ---> y = 9
z = -3 - 5t --> t = (z+3)/-5

so

(x-2)/3 = (z+3)/-5 , y = 9

To find the equation of a plane perpendicular to a given plane and passing through a given point, we can follow these steps:

Step 1: Determine the normal vector of the given plane π.
Step 2: Use the normal vector obtained in Step 1 and the given point to write the equation of the plane.

Step 1: Determine the normal vector of the given plane π.
The Cartesian equation of a plane in the form Ax + By + Cz + D = 0, where A, B, and C are the coefficients of x, y, and z, respectively. In this case, the equation of plane π is given as 3x - 5z + 3 = 0.

To find the normal vector, we extract the coefficients of x, y, and z. In this case, the coefficients are A = 3, B = 0, and C = -5.

Therefore, the normal vector of plane π is N = (3, 0, -5).

Step 2: Use the normal vector obtained in Step 1 and the given point to write the equation of the plane.
The equation of a plane passing through a given point P(x1, y1, z1) and perpendicular to a vector N = (A, B, C) is given by:

A(x - x1) + B(y - y1) + C(z - z1) = 0.

In this case, the given point P is (2, 9, -3), and the normal vector N is (3, 0, -5).

Substituting these values, we get:

3(x - 2) + 0(y - 9) + (-5)(z - (-3)) = 0
3x - 6 - 5z + 15 = 0
3x - 5z + 9 = 0.

Therefore, the Cartesian equation of the plane perpendicular to π and containing the point P(2, 9, -3) is 3x - 5z + 9 = 0.

To find the Cartesian equation of a plane that is perpendicular to the given plane π and contains the point P(2, 9, -3), we need to determine the normal vector of π and use it to find the equation of the perpendicular plane.

The normal vector of a plane is the vector that is perpendicular to every vector lying in the plane. In this case, the given plane has the equation 3x - 5z + 3 = 0. By rearranging this equation in terms of x and z, we get:

3x + 3 = 5z
(3/5)x + 3/5 = z

So, the normal vector of the given plane is (3/5, 0, -1).

Now, we can use this normal vector to find the equation of the perpendicular plane. The equation of a plane can be written as:

ax + by + cz + d = 0

Where (a, b, c) is the vector normal to the plane and (x, y, z) are coordinates of a point on the plane. We already have the normal vector (3/5, 0, -1), and the point P(2, 9, -3) lies on the plane we want to find.

Substituting these values into the equation, we get:

(3/5)(x - 2) + 0(y - 9) + (-1)(z + 3) = 0
(3/5)(x - 2) - (z + 3) = 0
(3/5)x - 6/5 - z - 3 = 0
(3/5)x - z - 21/5 = 0

Therefore, the Cartesian equation of a plane that is perpendicular to π and contains the point P(2, 9, -3) is:

(3/5)x - z - 21/5 = 0