If the direction ratios of two lines are a, b, c and (b-c), (c-a), (a…

The correct answer is Option (3) → $\frac{\pi}{2}$

The direction ratios of the two lines are:

Line 1 : $(a,\; b,\; c)$

Line 2 : $(b-c,\; c-a,\; a-b)$

Angle $\theta$ between two lines satisfies:

$\cos\theta=\frac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^2+b^2+c^2}\,\sqrt{(b-c)^2+(c-a)^2+(a-b)^2}}$

Compute the numerator:

$a(b-c)+b(c-a)+c(a-b)$

$=ab-ac+bc-ba+ca-bc$

All terms cancel:

$=0$

Therefore:

$\cos\theta = 0$

$\theta = 90^\circ$

Angle between the lines = $90^\circ$