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Problem 561

# 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 $$\left.\mathrm{a} \neq 0 .\right)$$

Expert verified
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 $$+ b = 0$$.
See the step by step solution

## 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 $$+ b = 0$$.

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