根據(jù)：|a| =√a² = 1 ； |b| =√b² = “3” ； a• b = |a| * |b| cosθ |3a - b| = √(3a - b)² = √(9a² - 6a• b + b²) = √(9|a|² - 6|a|* |b| cosθ + |b|²) = ...b的模等于5|b| = 5過程類似，略答案：|3a - b| = √(9|a|² - 6|a|* |b| cosθ+ |b|²) = √(9 - 6*1* 5 cos120°+ 25) = 7