(1)证明：\s\up14(→(→)＝\s\up14(→(→)＋\s\up14(→(→)，\s\up14(→(→)＝\s\up14(→(→)＋\s\up14(→(→).

因为BB1⊥平面ABC，

所以\s\up14(→(→)·\s\up14(→(→)＝0，\s\up14(→(→)·\s\up14(→(→)＝0.

又△ABC为正三角形，

所以〈\s\up14(→(→)，\s\up14(→(→)〉＝π－〈\s\up14(→(→)，\s\up14(→(→)〉＝π－＝.

因为\s\up14(→(→)·\s\up14(→(→)＝(\s\up14(→(→)＋\s\up14(→(→))·(\s\up14(→(→)＋\s\up14(→(→))＝\s\up14(→(→)·\s\up14(→(→)＋\s\up14(→(→)·\s\up14(→(→)＋\s\up14(→(→)2＋\s\up14(→(→)·\s\up14(→(→)＝|\s\up14(→(→)|·|\s\up14(→(→)|·cos〈\s\up14(→(→)·\s\up14(→(→)〉＋\s\up14(→(→)2＝－1＋1＝0，

所以AB1⊥BC1.

(2)解：结合(1)知\s\up14(→(→)·\s\up14(→(→)＝|\s\up14(→(→)|·|\s\up14(→(→)|·cos〈\s\up14(→(→)，\s\up14(→(→)〉＋\s\up14(→(→)2＝

\s\up14(→(→)2－1.

又|\s\up14(→(→)|＝ \s\up14(→(（\o(AB,\s\up14(→)＝ \s\up14(→(2＋\o(BB1,\s\up14(→)＝|\s\up14(→(→)|，

所以cos〈\s\up14(→(→)，\s\up14(→(→)〉＝\s\up14(→(BB1,\s\up14(→)＝.

所以|\s\up14(→(→)|＝2，即侧棱长为2.

B级　能力提升

1．已知空间向量a，b，c，两两夹角为60°，其模都为1，则|a－b＋2c|＝(　　)

A. B．5 C．6 D.

解析：因为|a|＝|b|＝|c|＝1，〈a，b〉＝〈b，c〉＝〈c，a〉＝60°，

所以a·b＝b·c＝a·c＝，a2＝b2＝c2＝1.

所以|a－b＋2c|＝