Final answer:
To find the equation of a plane that passes through two given points and is perpendicular to another plane, we can use the cross product of the normal vectors of the two planes.
Step-by-step explanation:
To find the equation of a plane that passes through two given points and is perpendicular to another plane, we can use the cross product of the normal vectors of the two planes.
First, let's find the normal vector of the given plane 2z = 5x + 4y. The coefficients of x, y, and z in the equation represent the components of the normal vector. So, the normal vector is n1 = (5, 4, 2).
Next, we find the direction vector of the plane passing through the given points by subtracting the coordinates of the two points. Let's call this vector v. v = (0, -2, 5) - (-1, 3, 1) = (1, -5, 4).
Finally, we can find the equation of the plane by taking the cross product of the normal vector n1 and the direction vector v:
n1 × v = (5, 4, 2) × (1, -5, 4) = (-28, -18, -21)
So, the equation of the plane that passes through the points (0,−2,5) and (−1,3,1) and is perpendicular to the plane 2z = 5x 4y is -28x - 18y - 21z = 0.