Y RAJYALAKSHMI
Last Activity: 10 Years ago
|a – b|2 + |b – c|2 + |c – a|2 = 9
=> 2(|a|2 + |b|2 + |c|2) – 2(a.b + b.c + c.a) = 9
=> 2 * 3 – 2(a.b + b.c + c.a) = 9 (Since a, b, c are unit vectors)
=> (cos (a, b) + cos (b, c) + cos(c, a) = – 3/2 [Since a, b, c are unit vectors & a.b = |a||b| cos(a, b) where cos (a, b) , cos (b, c) , cos(c, a) are cosine of angles between these vecotrs ]
This gives us cos (a, b) = cos (b, c) = cos(c, a) = –1/2
=> a.b = b.c = c.a = –1/2
|2a + 5b + 5c|2 = 4|a|2 + 25|b|2 + 25|c|2 + 20 a. b + 50 b. c + 20 c.a
= 4 + 25 +25 + 20 * (–1/2) + 50 * (–1/2) + 20 * (–1/2) = 9
Therefore, |2a + 5b + 5c| = 3