Given the direction ratios of two lines: \((a, b, c)\) and \((b - c, c - a, a - b)\). We need to find the angle between these two lines.
The cosine of the angle \(\theta\) between two lines with direction ratios \((p_1, q_1, r_1)\) and \((p_2, q_2, r_2)\) is given by:
\(\cos(\theta) = \frac{p_1 p_2 + q_1 q_2 + r_1 r_2}{\sqrt{p_1^2 + q_1^2 + r_1^2} \sqrt{p_2^2 + q_2^2 + r_2^2}}\)
Substitute the given direction ratios:
\(\cos(\theta) = \frac{a(b-c) + b(c-a) + c(a-b)}{\sqrt{a^2 + b^2 + c^2} \sqrt{(b-c)^2 + (c-a)^2 + (a-b)^2}}\)
Simplifying the numerator:
\(a(b-c) + b(c-a) + c(a-b) = ab - ac + bc - ab + ca - cb\)
All terms cancel out, so the numerator becomes 0:
\(ab - ac + bc - ab + ca - cb = 0\)
Thus, the cosine of the angle is:
\(\cos(\theta) = \frac{0}{\sqrt{a^2 + b^2 + c^2} \sqrt{(b-c)^2 + (c-a)^2 + (a-b)^2}} = 0\)
When \(\cos(\theta) = 0\), the angle \(\theta\) is \(90^\circ\).
Thus, the correct answer is: 90°.