To find the minimum length of the perpendicular segment \( PN \) drawn from a point \( P \) on the circle \( x^2 + y^2 = 1 \) to the line defined by \( x = y = z \), we can follow these steps:
### Step 1: Define the point \( P \) on the circle
The equation of the circle is given by \( x^2 + y^2 = 1 \). Any point \( P \) on this circle can be represented in parametric form as:
\[
P = (\cos \theta, \sin \theta, 0)
\]
where \( \theta \) is the angle parameter.
### Step 2: Define the point \( N \) on the line
The line \( x = y = z \) can be represented by points of the form \( N = (t, t, t) \) for some scalar \( t \).
### Step 3: Find the vector \( PN \)
The vector \( PN \) from point \( P \) to point \( N \) can be expressed as:
\[
PN = N - P = (t - \cos \theta, t - \sin \theta, t - 0) = (t - \cos \theta, t - \sin \theta, t)
\]
### Step 4: Set the condition for perpendicularity
For the vector \( PN \) to be perpendicular to the line \( x = y = z \), the dot product of \( PN \) with the direction vector of the line \( (1, 1, 1) \) must be zero:
\[
(t - \cos \theta) + (t - \sin \theta) + t = 0
\]
This simplifies to:
\[
3t - (\cos \theta + \sin \theta) = 0
\]
Thus, we can solve for \( t \):
\[
t = \frac{\cos \theta + \sin \theta}{3}
\]
### Step 5: Substitute \( t \) back into \( PN \)
Substituting \( t \) back into the expression for \( PN \):
\[
PN = \left(\frac{\cos \theta + \sin \theta}{3} - \cos \theta, \frac{\cos \theta + \sin \theta}{3} - \sin \theta, \frac{\cos \theta + \sin \theta}{3}\right)
\]
This simplifies to:
\[
PN = \left(\frac{-2\cos \theta + \sin \theta}{3}, \frac{\cos \theta - 2\sin \theta}{3}, \frac{\cos \theta + \sin \theta}{3}\right)
\]
### Step 6: Find the length of \( PN \)
The length of \( PN \) is given by the magnitude:
\[
|PN| = \sqrt{\left(\frac{-2\cos \theta + \sin \theta}{3}\right)^2 + \left(\frac{\cos \theta - 2\sin \theta}{3}\right)^2 + \left(\frac{\cos \theta + \sin \theta}{3}\right)^2}
\]
Factoring out \( \frac{1}{9} \):
\[
|PN| = \frac{1}{3} \sqrt{(-2\cos \theta + \sin \theta)^2 + (\cos \theta - 2\sin \theta)^2 + (\cos \theta + \sin \theta)^2}
\]
### Step 7: Simplify the expression
Calculating the squares:
1. \((-2\cos \theta + \sin \theta)^2 = 4\cos^2 \theta - 4\cos \theta \sin \theta + \sin^2 \theta\)
2. \((\cos \theta - 2\sin \theta)^2 = \cos^2 \theta - 4\cos \theta \sin \theta + 4\sin^2 \theta\)
3. \((\cos \theta + \sin \theta)^2 = \cos^2 \theta + 2\cos \theta \sin \theta + \sin^2 \theta\)
Combining these gives:
\[
= 4\cos^2 \theta + \sin^2 \theta + \cos^2 \theta + 4\sin^2 \theta + \cos^2 \theta + 2\cos \theta \sin \theta
\]
\[
= 6\cos^2 \theta + 6\sin^2 \theta - 2\cos \theta \sin \theta
\]
Using \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[
= 6 - 2\cos \theta \sin \theta
\]
### Step 8: Find the minimum value
To minimize \( |PN| \), we need to maximize \( \cos \theta \sin \theta \), which occurs at \( \theta = \frac{\pi}{4} \) where \( \cos \theta \sin \theta = \frac{1}{2} \). Thus:
\[
|PN| = \frac{1}{3} \sqrt{6 - 1} = \frac{1}{3} \sqrt{5}
\]
### Final Answer
The minimum length of \( PN \) is:
\[
\frac{\sqrt{5}}{3}
\]
