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Question: \(a,b,c\) are three vectors, such that \(a + b + c = 0\), \(|a| = 1,|b| = 2,|c| = 3\), then \(a.b + ...

a,b,ca,b,c are three vectors, such that a+b+c=0a + b + c = 0, ∣a∣=1,∣b∣=2,∣c∣=3|a| = 1,|b| = 2,|c| = 3, then a.b+b.c+c.aa.b + b.c + c.a is equal to

A

0

B

– 7

C

7

D

1

Answer

– 7

Explanation

Solution

a+b+c=0\mathbf{a} + \mathbf{b} + \mathbf{c} = 0⇒ (a+b+c).(a+b+c)(\mathbf{a} + \mathbf{b} + \mathbf{c}).(\mathbf{a} + \mathbf{b} + \mathbf{c}) = 0

⇒ ∣a∣2+∣b∣2+∣c∣2+2(a.b+b.c+c.a)=0|\mathbf{a}|^{2} + |\mathbf{b}|^{2} + |\mathbf{c}|^{2} + 2(\mathbf{a}.\mathbf{b} + \mathbf{b}.\mathbf{c} + \mathbf{c}.\mathbf{a}) = 0

⇒ a.b+b.c+c.a=−1−4−92=−7.\mathbf{a}.\mathbf{b} + \mathbf{b}.\mathbf{c} + \mathbf{c}.\mathbf{a} = \frac{- 1 - 4 - 9}{2} = - 7.