Final answer:
To find the point on the plane x-2y+3z=6 that is closest to the point (0,1,1), we can use the concept of orthogonal projection.
Step-by-step explanation:
To find the point on the plane x-2y+3z=6 that is closest to the point (0,1,1), we can use the concept of orthogonal projection. The orthogonal projection of a point onto a plane is the point on the plane that is closest to the given point.
To find the orthogonal projection, we need to find the normal vector of the plane. The coefficients of x, y, and z in the equation of the plane give us the components of the normal vector. In this case, the normal vector is (1, -2, 3).
The projection of the point (0,1,1) onto the plane can be found using the formula:
Projection = Point - Distance * Normal Vector
Substituting the values, we have:
Projection = (0,1,1) - Distance * (1, -2, 3)
Since the projection is on the plane, it must satisfy the equation of the plane. Substituting the coordinates of the projection into the equation of the plane, we get:
0 - 2(1) + 3(1) = 6 - Distance
Simplifying the equation, we have:
5 = 6 - Distance
Distance = 6 - 5
Distance = 1
Substituting the value of Distance into the formula for the projection, we have:
Projection = (0,1,1) - 1 * (1, -2, 3)
Projection = (0,1,1) - (1, -2, 3)
Projection = (-1, 3, -2)
Therefore, the point on the plane x-2y+3z=6 that is closest to the point (0,1,1) is (-1, 3, -2).