已知|a|=|b|=1,且|a+b|=根号3|a-b|,求|3a-2b|
|a|=|b|=1
→ a²= b²=1
|a+b|=√3|a-b|
→ (a+b)²=3(a-b)²
→ a²+b²+2ab= 3a²+3b²-6ab
→ ab=(2a²+2b²)/8=(2×1+2×1)/8=1/2
|3a-2b|=√[(3a-2b) ²]
=(√9a²+4b²-12ab)
=√(9×1+4×1-12×1/2)
=√7