Question

Let T be the plane −x−y = −5. Find the shortest distance d from the point P0=(4, −2, −3) to T, and the point Q in T that is closest to P0. Use the square root symbol '√' where needed to give an exact value for your answer.

Answer 1

To find the shortest distance from point P0 to the plane T, we need to find the distance between the point and a point on the plane that is closest to P0. This can be done by finding the perpendicular distance from P0 to the plane. The shortest distance d from P0 to the plane T is √(2/5), and the point Q in T that is closest to P0 has coordinates (7/5, -2/5, 0).

Step-by-step explanation:

To find the shortest distance from point P0 to the plane T, we need to find the distance between the point and a point on the plane that is closest to P0. This can be done by finding the perpendicular distance from P0 to the plane.

The equation of the plane T is -x - y = -5, which can also be written as x + y = 5. Since the plane is in the form of x + y + z = c, the normal vector to the plane is (1, 1, 0). The distance d from P0 to T can be found using the formula d = |(P0 - Q) * n| / |n|, where Q is a point on the plane, n is the normal vector, and | | denotes the magnitude of a vector.

By substituting the values into the formula, the shortest distance d from P0 to the plane T is √(2/5), and the point Q in T that is closest to P0 has coordinates (7/5, -2/5, 0).