Question

Find the shortest distance between A = (1, 2, 1) and the plane 2x - y + 2z + 4 = 0. Determine the point on the plane. Use projection A on the plane to find the shortest distance.

Answer 1

To find the shortest distance between a point and a plane, use projection. Find the normal vector, project the point onto the plane, and calculate the shortest distance.

Step-by-step explanation:

Shortest Distance between a Point and a Plane

To find the shortest distance between a point and a plane, we can use the concept of projection. Given point A = (1, 2, 1) and the equation of the plane 2x - y + 2z + 4 = 0, we can determine the shortest distance as follows:

1. Find the unit vector normal to the plane. The coefficients of x, y, and z in the plane equation give us the normal vector, which is (2, -1, 2).
2. Project vector A onto the plane by finding the orthogonal projection of A onto the plane's normal vector. The projection vector represents the distance from A to the plane. We can use the dot product to find the scalar component of the projection vector by dividing the dot product of A and the normal vector by the magnitude of the normal vector squared.
3. Multiply the normal vector by the scalar component to obtain the projection vector. Add the projection vector to point A to find the point on the plane closest to A.
4. The shortest distance is the magnitude of the projection vector obtained in the previous step.

In this case, the point on the plane closest to A is (0, 1, -2) and the shortest distance is 3 units.