Two line whose equations are `(x-3)/(2)=(y-2)/(3)=(z-1)/(3) and (x-2)/(3)=(y-3)/(2)=(z-2)/(3)` find the angle between them

Options

A. `(phi)/(3)`
B. `(phi)/(2)`
C. `(phi)/(6)`
D. `cos^(-1)((2)/(sqrt(186)))`

Answer

To find the angle between the two lines given by their equations, we can follow these steps:

### Step 1: Identify Direction Ratios
The equations of the lines are given in symmetric form. We can extract the direction ratios from each line.

1. For the first line:
\[ \frac{x-3}{2} = \frac{y-2}{3} = \frac{z-1}{3} \]
The direction ratios \( (a_1, b_1, c_1) \) are \( (2, 3, 3) \).

2. For the second line:
\[ \frac{x-2}{3} = \frac{y-3}{2} = \frac{z-2}{3} \]
The direction ratios \( (a_2, b_2, c_2) \) are \( (3, 2, 3) \).

### Step 2: Use the Formula for the Angle Between Two Lines
The formula for the cosine of the angle \( \theta \) between two lines with direction ratios \( (a_1, b_1, c_1) \) and \( (a_2, b_2, c_2) \) is:
\[ \cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}} \]

### Step 3: Calculate the Numerator
Substituting the values into the formula:
\[ \cos \theta = \frac{(2)(3) + (3)(2) + (3)(3)}{\sqrt{2^2 + 3^2 + 3^2} \cdot \sqrt{3^2 + 2^2 + 3^2}} \]
Calculating the numerator:
\[ = 6 + 6 + 9 = 21 \]

### Step 4: Calculate the Denominator
Now, we calculate the denominators:
1. For the first line:
\[ \sqrt{2^2 + 3^2 + 3^2} = \sqrt{4 + 9 + 9} = \sqrt{22} \]

2. For the second line:
\[ \sqrt{3^2 + 2^2 + 3^2} = \sqrt{9 + 4 + 9} = \sqrt{22} \]

Thus, the denominator becomes:
\[ \sqrt{22} \cdot \sqrt{22} = 22 \]

### Step 5: Final Calculation
Putting it all together:
\[ \cos \theta = \frac{21}{22} \]

### Step 6: Find the Angle
To find the angle \( \theta \):
\[ \theta = \cos^{-1}\left(\frac{21}{22}\right) \]

### Summary of the Solution
The angle between the two lines is given by:
\[ \theta = \cos^{-1}\left(\frac{21}{22}\right) \]