how do you find the coordinates of the point (x, y, z) on the plane z = 1 x + 1 y + 2 which is closest to the origin
the closest point is surely on a line from (0,0,0) meeting the plane at right angles.
but the normal to the given plane is (1,1,-1)
so the line must have direction (1,1,-1)
it's parametric equations would be:
x = k
y = k
z = -k
sub that in to the plane to find the value of k, sub back in the parametrics
and voila, there is your point.
To find the coordinates of the point (x, y, z) on the plane z = 1x + 1y + 2 that is closest to the origin, you can use the concept of the distance formula and minimize the distance between the origin and the point on the plane.
The distance between two points (x1, y1, z1) and (x2, y2, z2) in three-dimensional space can be calculated using the Euclidean distance formula:
d = √((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)
In this case, we want to minimize the distance between the origin (0, 0, 0) and the point (x, y, z) on the plane z = 1x + 1y + 2.
Let's substitute the values of x and y from the equation of the plane into the distance formula, considering z = 1x + 1y + 2:
d = √((x - 0)^2 + (y - 0)^2 + (1x + 1y + 2 - 0)^2)
Simplifying this equation yields:
d = √(x^2 + y^2 + (x + y + 2)^2)
To find the closest point to the origin, we need to minimize this distance. One way to do this is by finding the minimum value of the distance function.
Taking the derivative of the distance function with respect to x and y, and setting the derivatives equal to zero, will provide us with the x and y values that minimize the distance function.
By solving these equations, we can find the values of x and y, and then substitute them into the equation of the plane to calculate the value of z.
Note that if the point (x, y, z) lies on the plane z = 1x + 1y + 2, it must satisfy the equation for the plane.
By solving these equations simultaneously, you can determine the coordinates of the point (x, y, z) on the plane z = 1x + 1y + 2 that is closest to the origin.