Question

Use Lagrange multipliers to find the point on the plane

x−2y+3z=6 that is closest to the point (0, 1, 3).
a) (1, 2, 1)
b) (0, 0, 2)
c) (3, 1, 2)
d) (2, 1, 3)

Answer 1

To find the point on the plane closest to (0, 1, 3), we can use Lagrange multipliers. After solving the system of equations, we find that the point on the plane closest to (0, 1, 3) is (1, 2, 1).

Step-by-step explanation:

To find the point on the plane that is closest to the point (0, 1, 3), we can use Lagrange multipliers. The equation of the plane is x - 2y + 3z = 6. We want to minimize the distance between the point and the plane, which is the square root of the sum of the squares of the differences in the x, y, and z coordinates.

We set up the Lagrange function L = (x - 0)² + (y - 1)² + (z - 3)² + λ(x - 2y + 3z - 6), where λ is the Lagrange multiplier. Taking partial derivatives with respect to x, y, z, and λ and setting them equal to zero, we can solve for x, y, z, and λ.

After solving the system of equations, we find that the point on the plane closest to (0, 1, 3) is (1, 2, 1). Therefore, the correct answer is option a) (1, 2, 1).