If ab + bc + ca = 12 and a2 + b2 + c2 = 40, then possible value of \(\frac{1}{2}\) (a + b +c) [(a - b)2 + (b - c)2 + (c - a)2] is:
Answer & explanation
Correct answer: option 2
ab + bc + ca = 12
a2 + b2 + c2 = 40
⇒ (a + b + c)2 = [a2 + b2 + c2 + 2(ab + bc + ca)]
= 40 + 2 × 12 = 40 + 24 = 64
⇒ (a + b +c) = 8
Now,
⇒ \(\frac{1}{2}\) (a + b +c) [(a - b)2 + (b - c)2 + (c - a)2]
= \(\frac{1}{2}\) (a + b +c) [2(a2 + b2 + c2) - 2(ab + bc + ca)]
= \(\frac{1}{2}\) × 8 [(2 × 40) - (2 × 12)]
= 4 (80 - 24) = 224