Find the point on the plane z = 8x + 10y + 1 closest to the point P=(1, 0, 0).
8x + 10y - z + 1 = 0
shortest distance from (1,0,0) to this plane
= |8(1) + 10(0) - 1(0) + 1|/√(8^2 + 10^2 + 1^2)
= 9/√165 or 3√165/55
To find the point on the plane that is closest to a given point, we need to minimize the distance between the two points. Let's call the point on the plane that we want to find as Q.
The distance between two points in three-dimensional space can be computed using the Euclidean distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)
In this case, P = (1, 0, 0) and Q = (x, y, z) on the plane z = 8x + 10y + 1.
To find the closest point on the plane, we need to minimize the distance d between P and Q. This can be done by finding the values of x, y, and z that minimize the distance.
Using the formula for the plane, we can substitute the values of x and y from the coordinates of Q:
z = 8x + 10y + 1
=> z = 8(x) + 10(y) + 1
=> z = 8x + 10y + 1
Now, substitute the values of P into the equation for z:
0 = 8(1) + 10(0) + 1
=> 0 = 8 + 1
=> 0 = 9
This equation is not satisfied, which means that point P and the plane are not parallel, and will have a closest point.
To find the closest point, we can minimize the distance by minimizing the square of the distance, which is usually easier to work with:
d^2 = (x - 1)^2 + (y - 0)^2 + (z - 0)^2
Substitute the equation for z into the distance formula:
d^2 = (x - 1)^2 + (y - 0)^2 + (8x + 10y + 1 - 0)^2
= (x - 1)^2 + (y - 0)^2 + (8x + 10y + 1)^2
To minimize d^2, we can take partial derivatives with respect to x and y and set them equal to zero:
∂(d^2)/∂x = 2(x - 1) + 2(8x + 10y + 1)(8) = 0
∂(d^2)/∂y = 2(y - 0) + 2(8x + 10y + 1)(10) = 0
Simplifying these equations will give us a system of linear equations that we can solve for the values of x and y.