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Question: A, B, C, D are any four points, then \[\overset{\rightarrow}{AB}.\overset{\rightarrow}{CD} + \overs...

A, B, C, D are any four points, then

AB→.CD→+BC→.AD→+CA→.BD→=\overset{\rightarrow}{AB}.\overset{\rightarrow}{CD} + \overset{\rightarrow}{BC}.\overset{\rightarrow}{AD} + \overset{\rightarrow}{CA}.\overset{\rightarrow}{BD} =

A

2AB→.BC→.CD→2\overset{\rightarrow}{AB}.\overset{\rightarrow}{BC}.\overset{\rightarrow}{CD}

B

AB→+BC→+CD→\overset{\rightarrow}{AB} + \overset{\rightarrow}{BC} + \overset{\rightarrow}{CD}

C

535\sqrt{3}

D

0

Answer

0

Explanation

Solution

AD→=AB→+BC→+CD→=a+b+c\overset{\rightarrow}{AD} = \overset{\rightarrow}{AB} + \overset{\rightarrow}{BC} + \overset{\rightarrow}{CD} = \mathbf{a} + \mathbf{b} + \mathbf{c}

AC→=AB→+BC→=a+b\overset{\rightarrow}{AC} = \overset{\rightarrow}{AB} + \overset{\rightarrow}{BC} = \mathbf{a} + \mathbf{b} or CA→=−(a+b)\overset{\rightarrow}{CA} = - (\mathbf{a} + \mathbf{b})

BD→=BC→+CD→=b+c\overset{\rightarrow}{BD} = \overset{\rightarrow}{BC} + \overset{\rightarrow}{CD} = \mathbf{b} + \mathbf{c}

Therefore, AB→.CD→+BC→.AD→+CA→.BD→\overset{\rightarrow}{AB}.\overset{\rightarrow}{CD} + \overset{\rightarrow}{BC}.\overset{\rightarrow}{AD} + \overset{\rightarrow}{CA}.\overset{\rightarrow}{BD}

=a.c+b.(a+b+c)+(−a−b).(b+c)= \mathbf{a}.\mathbf{c} + \mathbf{b}.(\mathbf{a} + \mathbf{b} + \mathbf{c}) + ( - \mathbf{a} - \mathbf{b}).(\mathbf{b} + \mathbf{c})

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