If a , ~b , c are perpendicular to b + c , c + a and a + b respectively and | a + b |=2,| b + c |=6,| c + a |=4 , then | a + b + c |=

Step-by-step solution

Given vectors a , b , c with perpendicularity conditions: a ( b + c ) , b ( c + a ) , c ( a + b ) , and magnitudes | a + b | = 2 , | b + c | = 6 , | c + a | = 4 . The perpendicularity conditions yield three dot product equations: a b + a c = 0 b c + b a = 0 c a + c b = 0 Adding these gives 2( a b + b c + c a ) = 0 , hence: a b + b c + c a = 0 Squaring the given magnitudes: | a + b |^2 = | a |^2 + | b |^2 + 2 a b = 4 | b + c |^2 = | b |^2 + | c |^2 + 2 b c = 36 | c + a |^2 = | c |^2 + | a |^2 + 2 c a = 16 Summing th