Final answer:
Using vectors and the cross product, the Cartesian equation of the plane passing through the points Q(0,2,3), R(4,0,3), and S(0,0,-3) is 12x + 24y - 8z - 24 = 0.
Step-by-step explanation:
To determine the Cartesian equation of a plane that passes through three points, we can use vector methods. Let's find two vectors that are parallel to the plane by subtracting the coordinates of the points to get the direction vectors. We can use points Q(0,2,3), R(4,0,3), and S(0,0,-3). Vector QR can be found by subtracting the coordinates of Q from R: QR = R - Q = (4,0,3) - (0,2,3) = (4,-2,0). Similarly, vector QS = S - Q = (0,0,-3) - (0,2,3) = (0,-2,-6). Next, we need to find the normal vector to the plane, which is perpendicular to both QR and QS vectors. The cross product of QR and QS will give us the normal vector N = QR x QS.
Using the components of QR and QS, the cross product N is:
- Nx = QRyQSz - QRzQSy = (-2)*(-6) - (0)*(-2) = 12
- Ny = QRzQSx - QRxQSz = (0)*(0) - (4)*(-6) = 24
- Nz = QRxQSy - QRyQSx = (4)*(-2) - (-2)*(0) = -8
Now, with the normal vector N = (12, 24, -8), we can write the Cartesian equation of the plane using the point-normal form: 12(x - x0) + 24(y - y0) + (-8)(z - z0) = 0. We can choose any of the given points for (x0, y0, z0); let's use point Q (0,2,3). Substituting Q into the equation gives:
12(x - 0) + 24(y - 2) + (-8)(z - 3) = 0
Which simplifies to:
12x + 24y - 48 - 8z + 24 = 0
And further simplifies to the final Cartesian equation of the plane:
12x + 24y - 8z - 24 = 0