The angle between the lines `(x+4)/(1) = (y-3)/(2) = (z+2)/(3) and (x)/(3) = (y)/(-2) = (z)/(1)` is

Options

A. `sin^(-1) ((1)/(7))`
B. `cos^(-1) ((2)/(7))`
C. `cos^(-1) ((1)/(7))`
D. none of these

Answer

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

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

For the first line:
\[ \frac{x + 4}{1} = \frac{y - 3}{2} = \frac{z + 2}{3} \]
The direction ratios (DR) for the first line \(L_1\) are:
\[ (1, 2, 3) \]

For the second line:
\[ \frac{x}{3} = \frac{y}{-2} = \frac{z}{1} \]
The direction ratios (DR) for the second line \(L_2\) are:
\[ (3, -2, 1) \]

### Step 2: Represent Direction Ratios as Vectors
We can represent the direction ratios as vectors:
\[ \vec{a} = \langle 1, 2, 3 \rangle \]
\[ \vec{b} = \langle 3, -2, 1 \rangle \]

### Step 3: Calculate the Dot Product
The dot product of two vectors \(\vec{a}\) and \(\vec{b}\) is given by:
\[ \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \]
Calculating the dot product:
\[ \vec{a} \cdot \vec{b} = (1)(3) + (2)(-2) + (3)(1) = 3 - 4 + 3 = 2 \]

### Step 4: Calculate the Magnitudes of the Vectors
The magnitude of vector \(\vec{a}\) is:
\[ |\vec{a}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]
The magnitude of vector \(\vec{b}\) is:
\[ |\vec{b}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \]

### Step 5: Use the Dot Product to Find Cosine of the Angle
Using the dot product formula:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
Substituting the values we found:
\[ 2 = (\sqrt{14})(\sqrt{14}) \cos \theta \]
\[ 2 = 14 \cos \theta \]
\[ \cos \theta = \frac{2}{14} = \frac{1}{7} \]

### Step 6: Find the Angle
To find the angle \(\theta\), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{1}{7}\right) \]

### Final Answer
The angle between the two lines is:
\[ \theta = \cos^{-1}\left(\frac{1}{7}\right) \]

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