If the direction ratios of two lines are `lt a, b, c gt and lt b-c, c-a, a-b gt`, find the angle between them.

Answer

To find the angle between the two lines with direction ratios \( \langle a, b, c \rangle \) and \( \langle b-c, c-a, a-b \rangle \), we will use the formula for the cosine of the angle between two vectors.

### Step-by-Step Solution:

1. **Identify the Direction Ratios:**
Let \( \mathbf{A} = \langle a, b, c \rangle \) and \( \mathbf{B} = \langle b-c, c-a, a-b \rangle \).

2. **Calculate the Dot Product:**
The dot product \( \mathbf{A} \cdot \mathbf{B} \) is given by:
\[ \mathbf{A} \cdot \mathbf{B} = a(b-c) + b(c-a) + c(a-b) \]
Expanding this, we get:
\[ = ab - ac + bc - ba + ca - cb \]
Rearranging the terms:
\[ = ab - ba + bc - cb - ac + ca \]
Notice that \( ab - ba = 0 \), \( bc - cb = 0 \), and \( -ac + ca = 0 \). Therefore:
\[ \mathbf{A} \cdot \mathbf{B} = 0 \]

3. **Calculate the Magnitudes:**
The magnitude of \( \mathbf{A} \) is:
\[ |\mathbf{A}| = \sqrt{a^2 + b^2 + c^2} \]
The magnitude of \( \mathbf{B} \) is:
\[ |\mathbf{B}| = \sqrt{(b-c)^2 + (c-a)^2 + (a-b)^2} \]
Expanding \( |\mathbf{B}| \):
\[ = \sqrt{(b^2 - 2bc + c^2) + (c^2 - 2ca + a^2) + (a^2 - 2ab + b^2)} \]
Combining like terms:
\[ = \sqrt{2a^2 + 2b^2 + 2c^2 - 2(ab + ac + bc)} = \sqrt{2(a^2 + b^2 + c^2 - ab - ac - bc)} \]

4. **Use the Cosine Formula:**
The cosine of the angle \( \theta \) between the two vectors is given by:
\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} \]
Since \( \mathbf{A} \cdot \mathbf{B} = 0 \):
\[ \cos \theta = \frac{0}{|\mathbf{A}| |\mathbf{B}|} = 0 \]

5. **Determine the Angle:**
If \( \cos \theta = 0 \), then:
\[ \theta = 90^\circ \text{ or } \frac{\pi}{2} \text{ radians} \]

### Final Answer:
The angle between the two lines is \( 90^\circ \) or \( \frac{\pi}{2} \) radians.