To find the equation of the plane that passes through the point (1, 1, 1) and has direction ratios of a normal vector that are the prime factors of 2001, we will follow these steps:
### Step 1: Find the prime factors of 2001
First, we need to factor 2001 into its prime components.
- 2001 can be divided by 3 (since the sum of the digits 2 + 0 + 0 + 1 = 3 is divisible by 3).
- Dividing 2001 by 3 gives us 667.
- Next, we factor 667. It can be divided by 23 (since 667 = 23 * 29).
- Thus, the prime factorization of 2001 is:
\[
2001 = 3 \times 23 \times 29
\]
### Step 2: Identify the direction ratios
According to the problem, we need to arrange the prime factors in increasing order:
- The prime factors are 3, 23, and 29.
- Therefore, we assign:
- \( a = 3 \)
- \( b = 23 \)
- \( c = 29 \)
### Step 3: Write the normal vector
The direction ratios of the normal vector to the plane are given as \( b, c, a \). Thus, the normal vector \( \vec{n} \) can be expressed as:
\[
\vec{n} = (b, c, a) = (23, 29, 3)
\]
### Step 4: Use the point-normal form of the plane equation
The equation of a plane in point-normal form is given by:
\[
\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}
\]
where \( \vec{r} = (x, y, z) \) is the position vector, \( \vec{n} \) is the normal vector, and \( \vec{a} \) is the position vector of the point through which the plane passes.
Here, \( \vec{a} = (1, 1, 1) \) and \( \vec{n} = (23, 29, 3) \).
### Step 5: Calculate \( \vec{a} \cdot \vec{n} \)
Now we calculate the dot product:
\[
\vec{a} \cdot \vec{n} = (1, 1, 1) \cdot (23, 29, 3) = 1 \times 23 + 1 \times 29 + 1 \times 3 = 23 + 29 + 3 = 55
\]
### Step 6: Write the equation of the plane
Now we can write the equation of the plane:
\[
(x, y, z) \cdot (23, 29, 3) = 55
\]
This expands to:
\[
23x + 29y + 3z = 55
\]
### Final Answer
The equation of the plane is:
\[
23x + 29y + 3z = 55
\]
