Find the point on the plane z = 6x + 9y + 1 closest to the point P = (1, 0, 0). Hint: Minimize the square of the distance. (x, y, z) =
I will ignore the hint and use a very straight-forward method
the shortest distance will of course be the normal to the plane going through (1,0,0)
plane:
6x + 9y - z + 1 = 0
a normal to the plane is (6,9,-1)
and passing through (1,0,0) would be the line
x = 6t + 1
y = 9t
z = -t
intersecting this with the plane:
6(6t+1) + 9(9t) -(-t) + 1 = 0
36t + 6 + 81t + t = -1
118t = -7
t = -7/118
point of contact of the normal with plane is
(-42/118 + 1, -63/118, 7/118) = (38/59 , -63/118, 7/118)
expected nicer numbers, but ....
distance between (1,0,0) and the above weird point
= √((-42/118+1 - 1)^2 + (-63/118)^2 + (7/118)^2 )
= √(5782)/118
= appr .6444
To find the point on the plane closest to the point P(1, 0, 0), we need to find the values of x, y, and z that satisfy the plane equation and minimize the square of the distance between the two points.
Given the equation of the plane: z = 6x + 9y + 1
We want to minimize the square of the distance, which is given by:
D^2 = (x - 1)^2 + (y - 0)^2 + (z - 0)^2
Let's substitute the value of z from the plane equation into the distance equation:
D^2 = (x - 1)^2 + (y - 0)^2 + (6x + 9y + 1 - 0)^2
D^2 = (x - 1)^2 + (y - 0)^2 + (6x + 9y + 1)^2
We can now take the partial derivatives of D^2 with respect to x and y, and set them equal to zero to find the critical points. We will minimize D^2 by finding the values of x and y that satisfy these equations.
Taking the partial derivative with respect to x:
∂D^2/∂x = 2(x - 1) + 2(6x + 9y + 1)(6)
Setting it equal to zero:
2(x - 1) + 12(6x + 9y + 1) = 0
2x - 2 + 72x + 108y + 12 = 0
74x + 108y + 10 = 0
74x = -108y - 10
Taking the partial derivative with respect to y:
∂D^2/∂y = 2(y - 0) + 2(6x + 9y + 1)(9)
Setting it equal to zero:
2(y - 0) + 18(6x + 9y + 1) = 0
2y + 108x + 162y + 18 = 0
270y + 108x + 18 = 0
270y = -108x - 18
Solving the two equations simultaneously, we have:
74x = -108y - 10
270y = -108x - 18
Simplifying these equations will give us the values of x and y that satisfy both equations and correspond to the point on the plane closest to P.
After simplification, we get:
37x + 54y + 5 = 0
135x + 135y + 15 = 0
Solving these equations, we find:
x = -3/5
y = -2/5
(z can be obtained by substituting x and y values into the plane equation)
Therefore, the point (x, y, z) closest to P(1, 0, 0) on the plane z = 6x + 9y + 1 is approximately (-3/5, -2/5, 7/5).
To find the point on the plane closest to point P, we need to minimize the square of the distance between P and an arbitrary point (x, y, z) on the plane.
First, let's write the equation of the plane:
z = 6x + 9y + 1
Now, let's write the equation for the distance between P and an arbitrary point (x, y, z) on the plane:
Distance^2 = (x - 1)^2 + y^2 + z^2
Since we want to minimize the distance, we need to minimize the square of the distance.
Now, substitute the equation of the plane into the equation for the distance:
Distance^2 = (x - 1)^2 + y^2 + (6x + 9y + 1)^2
To find the minimum distance, we need to find the values of x and y that minimize the above equation.
To do that, we can take the partial derivatives of Distance^2 with respect to x and y and set them equal to 0:
∂(Distance^2)/∂x = 2(x - 1) + 2(6x + 9y + 1)(6) = 0
∂(Distance^2)/∂y = 2y + 2(6x + 9y + 1)(9) = 0
Solve the above system of equations to find the values of x and y that minimize the distance. Once you have x and y, substitute them back into the equation of the plane (z = 6x + 9y + 1) to get the value of z. That will give you the coordinates (x, y, z) of the point on the plane closest to point P.