# Find the point on the plane x 2y 5 3z closest to the point 4 4 7

## Introduction

In mathematics, the closest point problem is the problem of finding the point on a given set (usually a hypersurface) which is closest to a given point. The closest point problem arises in many practical contexts, such as in numerical analysis, where it can be used to find efficient ways to sample a function; in data mining, where it can be used to cluster data points; and in robotics, where it can be used to plan safe trajectories for robots.

## The point on the plane closest to the given point

To find the point on the plane closest to the given point, we need to find the point on the plane that has the smallest distance to the given point. We can do this by using the equation of a plane and the distance formula.

The equation of a plane is:

Ax + By + Cz = D

where A, B, and C are the coefficients of the plane and D is a constant.

The distance between a point (x1, y1, z1) and a plane with equation Ax + By + Cz = D is:

distance = |Ax1 + By1 + Cz1 – D| / sqrt(A^2 + B^2 + C^2)

So, to find the point on the plane closest to our given point (4, 4, 7), we need to plug in our values into the distance formula and take the derivative with respect to x. We set this derivative equal to 0 and solve for x. This will give us our x-coordinate for the closest point on the plane. Then we can plug this back into our original equation for the plane to find our y- and z-coordinates.

distance = |2(4) + 5(4) + 3(7) – 5| / sqrt(2^2 + 5^2 + 3^2)
= |36 – 5| / 14
= 31 / 14
= 31/14 ~ 2.21429
derivative wrt x = 0 => 2x + 5(4) + 3(7) – 5 = 0
=> 2x = -31 => x = -15.5
Plugging this back into our original equation, we get:
-15.5 + 2y + 5(3) = 0
=> 2y = 10.5 => y = 5.25
And finally:
z = 7 => z = 7

## The algebraic method

The algebraic method is the most straightforward way to find the closest point on a plane to a given point. We simply need to solve for the point on the plane that is closest to the given point.

First, we’ll set up a few variables. Let’s call the given point P, and the plane x 2y 5 3z. We’ll also call the unknown point on the plane Q. We can then set up the following equation:

PQ = (x-4)^2 + (y-4)^2 + (z-7)^2

Now, we need to minimize this equation. To do so, we’ll take the partial derivative with respect to x, y, and z, and set each equal to 0:

∂PQ/∂x = 2(x-4) = 0
∂PQ/∂y = 2(y-4) = 0
∂PQ/∂z = 2(z-7) = 0

This gives us three equations in three unknowns, which we can solve to find Q:

x-4=0 => x=4
y-4=0 => y=4
z-7=0 => z=7

Thus, Q is the point 4 4 7 on the plane x 2y 5 3z.

## The geometric method

The geometric method is the most straight-forward way to find the closest point on a plane to a given point. This method involves drawing a line from the given point to the plane, and then finding the point where this line intersects the plane. The intersection point will be the closest point on the plane to the given point.

## Conclusion

The point on the plane x 2y 5 3z closest to the point 4 4 7 is the point 4 4 7.