1. Given vectors $\boldsymbol{a}, \boldsymbol{b}$ satisfy $|\boldsymbol{a}-\boldsymbol{b}|=3,|\boldsymbol{a}+2 \boldsymbol{b}|=6, \boldsymbol{a}^{2}+\boldsymbol{a} \cdot \boldsymbol{b}-2 \boldsymbol{b}^{2}=-9$, then $|\boldsymbol{b}|=$
Answer: $\sqrt{7}$
Analysis: From the condition, we know that $a^{2}+a \cdot b-2 b^{2}=(a-b) \cdot(a+2 b)=-9$, so $|3 b|=|(a+2 b)-(a-b)|$ $=\sqrt{[(a+2 b)-(a-b)]^{2}}=\sqrt{(a+2 b)^{2}+(a-b)^{2}-2(a+2 b) \cdot(a-b)}=\sqrt{63}=3 \sqrt{7}$, thus $|b|=\sqrt{7}$.