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Question: In the triangle ABC, \(\overset{\rightarrow}{AB} = \mathbf{a},\overset{\rightarrow}{AC} = \mathbf{c}...

In the triangle ABC, AB→=a,AC→=c,BC→=b\overset{\rightarrow}{AB} = \mathbf{a},\overset{\rightarrow}{AC} = \mathbf{c},\overset{\rightarrow}{BC} = \mathbf{b}, then

A

a+b+c=0\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}

B

a+b−c=0\mathbf{a} + \mathbf{b} - \mathbf{c} = \mathbf{0}

C

a−b+c=0\mathbf{a} - \mathbf{b} + \mathbf{c} = \mathbf{0}

D

−a+b+c=0- \mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}

Answer

a+b−c=0\mathbf{a} + \mathbf{b} - \mathbf{c} = \mathbf{0}

Explanation

Solution

AB→+BC→+CA→=0⇒a+b−c=0.\overset{\rightarrow}{AB} + \overset{\rightarrow}{BC} + \overset{\rightarrow}{CA} = 0 \Rightarrow \mathbf{a} + \mathbf{b} - \mathbf{c} = 0.