Question

If a,b and c are three unit vectors such that

a.b+b.c-c.a=3/2 then

Answer 1

a+b+C=0(a+b+C)² =0|a|² + |b|² + |c|² +2(a.b +b.c + c.a)=0As it is given in the question that a, b, C are the unit vectors. Therefore, |a|=|b|=|c|=1.So,      1+1+1+2(a.b + b.c + c.a)=0       2(a.b + b.c + c.a)= -3       a.b + b.c + c.a = -3/2

Hope This Helps :)

Answer 2

Answer:

Then $a.b+b.c+c.a = \frac{-3}{2}$

Step-by-step explanation:

• We are given that a,b,c are the unit vectors hence the magnitude of a = b = c=1
• So when we sum these and square it we get
  $(a+b+c)^2 = 0\\a^2+b^2+c^2+2(a.b+b.c+c.a) = 0$
• Now as we know that
  $a.a =b.b=c.c=1$
• So the equation becomes
  $1+1+1+2(a.b+b.c+c.a) = 0\\a.b+b.c+c.a = \frac{-3}{2}$
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