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If a, b, c are unit vectors such that a + b + c = 0, then the value of a.b + b.c + c.a isReview Later13-3/2None of the Above

Question

If a, b, c are unit vectors such that a+b+c=0 a + b + c = 0 , then the value of a.b+b.c+c.a a.b + b.c + c.a is

Review Later

  1. 13
  2. -3/2
  3. None of the Above
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Solution

Given that a, b, c are unit vectors and a + b + c = 0.

We know that the square of the magnitude of a vector is equal to the dot product of the vector with itself.

So, we square both sides of the equation:

(a + b + c) . (a + b + c) = 0 . 0

Expanding the left side using the distributive property of dot product gives:

a.a + a.b + a.c + b.a + b.b + b.c + c.a + c.b + c.c = 0

Since a, b, and c are unit vectors, a.a = b.b = c.c = 1. Also, the dot product is commutative, meaning a.b = b.a and b.c = c.b and c.a = a.c.

So, the equation simplifies to:

1 + a.b + a.c + a.b + 1 + b.c + a.c + b.c + 1 = 0

Grouping like terms gives:

3 + 2(a.b + b.c + c.a) = 0

Subtract 3 from both sides:

2(a.b + b.c + c.a) = -3

Finally, divide both sides by 2 to solve for a.b + b.c + c.a:

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

So, the value of a.b + b.c + c.a is -3/2.

This problem has been solved

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