Nick's Calculus Problem: Find Closest Point on Plane

In summary, to find the point on the plane $x-2y+3z=6$ that is closest to the point $(0,1,1)$, we use Lagrange multipliers to minimize the square of the distance between the two points. This leads to a system of equations which can be solved to find the coordinates of the closest point.
  • #1
MarkFL
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Here is the question:

Need help with calculus problem? Optimization.?


Find the point on the plane x - 2y +3z = 6 that is closest to the point (0,1,1).

I can't figure this out. Does anyone know how to do this? Thanks.

I have posted a link there to this thread so the OP can see my work.
 
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  • #2
Hello Nick,

Let's let our objective function be the square of the distance from the point on the plane $(x,y,z)$ to the given point $(0,1,1)$:

\(\displaystyle f(x,y,z)=x^2+(y-1)^2+(z-1)^2\)

The point $(x,y,z)$ is constrained to be on the given plane, hence our constraint is:

\(\displaystyle g(x,y,z)=x-2y+3z-6=0\)

Using Lagrange multipliers, we obtain the system:

\(\displaystyle 2x=\lambda\)

\(\displaystyle 2(y-1)=-2\lambda\)

\(\displaystyle 2(z-1)=3\lambda\)

This implies:

\(\displaystyle \lambda=2x=1-y=\frac{2}{3}(z-1)\)

From this, we obtain:

\(\displaystyle y=1-2x,\,z=3x+1\)

Substituting for $y$ and $z$ into the constraint, we find:

\(\displaystyle x-2(1-2x)+3(3x+1)-6=0\)

Solving for $x$ (and using the values for $y$ and $z$ in terms of $x$) we find:

\(\displaystyle x=\frac{5}{14}\implies y=\frac{2}{7},\,z=\frac{29}{14}\)

Thus, the point on the plane $x-2y+3z=6$ that is closest to the point $(0,1,1)$ is:

\(\displaystyle \left(\frac{5}{14},\frac{2}{7},\frac{29}{14} \right)\)
 

FAQ: Nick's Calculus Problem: Find Closest Point on Plane

What is "Nick's Calculus Problem: Find Closest Point on Plane"?

"Nick's Calculus Problem: Find Closest Point on Plane" is a mathematical problem that involves using calculus to determine the point on a plane that is closest to a given point. This problem is often used in the field of engineering and computer graphics.

What is the formula for finding the closest point on a plane?

The formula for finding the closest point on a plane involves using the distance formula and the equation of the plane to set up a system of equations. This system can then be solved using calculus techniques such as partial derivatives to find the coordinates of the closest point.

What are the applications of "Nick's Calculus Problem: Find Closest Point on Plane"?

This problem has many practical applications, including determining the shortest distance between a point and a plane in engineering and finding the optimal placement of objects in computer graphics. It can also be used in real-world scenarios such as calculating the closest distance between an airplane and a storm cloud.

Is there a specific method or approach to solving this problem?

Yes, there are specific steps and methods to solving "Nick's Calculus Problem: Find Closest Point on Plane". These include setting up the system of equations, finding the partial derivatives, and using optimization techniques to solve for the coordinates of the closest point.

Are there any special cases or exceptions to this problem?

Yes, there are some special cases and exceptions to consider when solving "Nick's Calculus Problem: Find Closest Point on Plane". These can include situations where the given point is already on the plane, or when the plane is parallel to the coordinate axes. In these cases, the approach to solving the problem may need to be adjusted.

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