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, then the value of a.b + b.c + c.a isReview Later13-3/2None of the Above
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.
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