Problem 561 - Advanced Calculus

Question

Minimize the distance from a point \(p \in R^{n}\) to the hyperplane \(<\mathrm{x}, \mathrm{a}>+\mathrm{b}=0\) where \(\mathrm{a} \in \mathrm{R}^{n}\) and \(\mathrm{b} \in \mathrm{R}\). (Assume \(\left.\mathrm{a} \neq 0 .\right)\)

Step-by-Step Solution

Answer

The minimum distance \(D(p)\) between the point \(p\) and the hyperplane \( + b = 0\) is given by \(\frac{| - b|}{|a|}\), and the point \(x_0\) on the hyperplane closest to the point \(p\) satisfies \( + b = 0\).

Step 1: Write down the distance formula between point and hyperplane

The formula for the distance between a point \(p\) in \(R^n\) and a hyperplane \( + b = 0\) is given by: \(D(p) = \frac{| + b|}{|a|}\).

Step 2: Find the position vector for the point on the hyperplane closest to \(p\)

To minimize the distance, we want to find a point \(x_0\) on the hyperplane that is closest to \(p\). The vector connecting \(p\) and \(x_0\) is orthogonal to the normal vector of the hyperplane, which is given by the vector \(a\). Let \(x_0\) be the point on the hyperplane closest to \(p\) with the position vector \(x_0\). Then, the vector connecting \(p\) and \(x_0\) is given by \((p - x_0)\). The fact that the vector connecting \(p\) and \(x_0\) is orthogonal to the normal vector of the hyperplane can be written as: \(<(p - x_0), a> = 0\).

Step 3: Solve for \(x_0\)

Using the equation \( + b = 0\) and substituting \(x_0\) for \(x\), we have: \( + b = 0\). Now, using the equation from step 2 and substituting the derived equation for \(x_0\), we have: \(<(p - x_0), a> = - = 0\). Adding \(b\) to both sides, we have \( - + b = b\). Since \( + b = 0\), we can substitute \(-\) for \(b\), and we have: \( + b = b\). Therefore, the point \(x_0\) on the hyperplane closest to \(p\) satisfies: \( + b = 0\).

Step 4: Calculate the minimum distance \(D(p)\) from the point \(p\) to the hyperplane

Now we can use the expression for \(x_0\) to calculate the minimum distance between \(p\) and the hyperplane. We know that: \(D(p) = \frac{| + b|}{|a|}\). Since we have \( + b = 0\), we can substitute \(0\) for \(-b\) and have: \(D(p) = \frac{| - b|}{|a|}\). This expression gives us the minimum distance between the point \(p\) and the hyperplane \( + b = 0\). To summarize, the minimum distance \(D(p)\) is given by \(\frac{| - b|}{|a|}\), and the point \(x_0\) on the hyperplane closest to the point \(p\) satisfies \( + b = 0\).

Key Concepts

Distance Formula

Understanding the distance formula is crucial for measuring the shortest path between a point and a hyperplane. Think of it as a ruler that stretches exactly from one object to another, not following any curves or taking detours. In mathematical terms, the distance from a point p to a hyperplane can be measured using the formula:
\[ D(p) = \frac{|\langle p, a \rangle + b|}{|a|} \]
Here's the breakdown: p is your starting point, and a denotes the hyperplane's normal vector—think of this as an arrow shooting perpendicularly from the flat surface. The symbol b is essentially a scalar that adjusts the plane in space. The absolute value signs make sure we get a positive distance (since distance can't be negative), and dividing by the magnitude (or length) of a scales the value correctly. It's like measuring in the right units to get the distance right.

Normal Vector

The term 'normal vector' might sound a bit odd, but all it refers to is a vector that stands at a right angle (perpendicular) to a surface—in this case, our hyperplane. Imagine a wall: the normal vector would be like a stick pushed straight out from it. Mathematically, this vector is critical because it defines the orientation of the hyperplane in space. In the distance formula, the normal vector's role is to point out directionally how far you would have to travel from the hyperplane to hit the point p if you could move in a straight, orthogonal line.

Orthogonal Vectors

Vectors are like arrows flying through space, and when two of them are orthogonal, it means they're doing their own thing—neither one affects the other's direction; they cross at a neat 90 degree angle. This idea is key when we want to minimize the distance from a point to a hyperplane, as the shortest line connecting the point and the plane will always be orthogonal to the plane, aligning perfectly with our normal vector. It's a bit like dropping a plumb line from your point straight down to the hyperplane below.

Minimize Distance

When we talk about minimizing distance, we're trying to be as efficient as possible. It's like finding the quickest route to the store—you wouldn't want to walk in zigzags if you can help it! In the world of geometry, this means finding the shortest line from our point to the hyperplane. Here's a nifty fact: that shortest line is always going to be the one that hits the hyperplane straight on, which again brings us back to the perpendicular or normal vector we talked about. So when we say 'minimize distance', we're just trying to find that most direct path.

Vector Algebra

Vector algebra is the branch of mathematics where we play around with vectors—those arrows in space. It includes adding, subtracting vectors and scaling them (stretching or squishing), much like how you might stack blocks or stretch silly putty. In our problem, we use vector algebra to subtract the point x_0 from p, to get the vector that represents our shortest path. Just remember, vectors have both direction and magnitude, so when we manipulate them, we're juggling both how strong they are (their length) and which way they're pointing.