每日一题 · 2026-10-31

已知向量 \(\boldsymbol{a}=(1,0)\),\(\boldsymbol{b}=(0,1)\),\(\boldsymbol{c}=(\cos\theta,\sin\theta)\),\(\theta\in[0,2\pi)\).若 \(\boldsymbol{a}\cdot\boldsymbol{b}=[\boldsymbol{a}\cdot\boldsymbol{c}]+[\boldsymbol{b}\cdot\boldsymbol{c}]\)（其中 \([x]\) 表示不超过 \(x\) 的最大整数）,则 \(|\boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c}|\) 的取值范围为\(\underline{\qquad\qquad}\)

答案： \((\sqrt{5},1+\sqrt{2}]\)

解析：

\(\boldsymbol{a}\cdot\boldsymbol{b}=0\),\(\boldsymbol{a}\cdot\boldsymbol{c}=\cos\theta\),\(\boldsymbol{b}\cdot\boldsymbol{c}=\sin\theta\),故 \([\cos\theta]+[\sin\theta]=0\).
符合条件的 \(\theta\in(0,\dfrac{\pi}{2})\),此时 \(\boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c}=(1+\cos\theta,1+\sin\theta)\),

\[|\boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c}|^2=(1+\cos\theta)^2+(1+\sin\theta)^2=3+2(\cos\theta+\sin\theta)=3+2\sqrt{2}\sin\left(\theta+\dfrac{\pi}{4}\right), \]

\(\sin\left(\theta+\dfrac{\pi}{4}\right)\in\left(\dfrac{\sqrt{2}}{2},1\right]\),故 \(3+2\sqrt{2}\sin\left(\theta+\dfrac{\pi}{4}\right)\in(5,3+2\sqrt{2}]\),
因此 \(|\boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c}|\in(\sqrt{5},1+\sqrt{2}]\).

来源：福建省福州市高三3月质量检测