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Question

Suppose the vectors a,b,c on a plane satisfy the condition that |a|=|b|=|c|=|a+b|=1;c is perpendicular to a and b.c>0, then

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Solution

¯¯¯a+¯¯c2=¯¯¯a2+¯¯c2+2¯¯¯a.¯¯c
¯¯¯aisperpendicularto¯¯c.
¯¯¯a¯¯c=0(i)
¯¯¯a+¯¯c2=¯¯¯a2+¯¯¯a2
¯¯¯a+¯¯b2=¯¯¯a2+¯¯b2+2¯¯¯a.¯¯b
¯¯¯a=¯¯b=¯¯¯a+¯¯b=1
1=1+1+2¯¯¯a.¯¯b
¯¯¯a.¯¯b=12
2¯¯¯a.¯¯b=1
¯¯¯a.¯¯b=12
¯¯¯a.¯¯bcosθ=12
cosθ=12
θ=ππ3=2π3
Anglebetween¯¯¯aand¯¯b=2π3
¯¯¯a.¯¯c=0from(i)
¯¯¯a.¯¯ccosα=0
cosα=0impliesα=π2
Anglebetween¯¯¯aand¯¯c=π2

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