《2025年新课程同步导学高中数学选择性必修第二册湘教版》

例1 已知$\boldsymbol{a} = (2, 3, -4)$, $\boldsymbol{b} = (-4, -3, -2)$, $\boldsymbol{b} = \frac{1}{2}\boldsymbol{x} - 2\boldsymbol{a}$, 则$\boldsymbol{x} =$ (B)

A.$(0, 3, -6)$
B.$(0, 6, -20)$
C.$(0, 6, -6)$
D.$(6, 6, -6)$

答案： B

例2 已知$\boldsymbol{a} = (2, -1, -2)$, $\boldsymbol{b} = (0, -1, 4)$, 求$\boldsymbol{a} + \boldsymbol{b}$, $\boldsymbol{a} - \boldsymbol{b}$, $2\boldsymbol{a} - 3\boldsymbol{b}$.$\boldsymbol{a} + \boldsymbol{b} = (2 + 0, -1 + (-1), -2 + 4) = (2, -2, 2)$；
$\boldsymbol{a} - \boldsymbol{b} = (2 - 0, -1 - (-1), -2 - 4) = (2, 0, -6)$；
$2\boldsymbol{a} - 3\boldsymbol{b} = 2(2, -1, -2) - 3(0, -1, 4) = (4, -2, -4) - (0, -3, 12) = (4 - 0, -2 - (-3), -4 - 12) = (4, 1, -16)$。

答案： $\boldsymbol{a} + \boldsymbol{b} = (2 + 0, -1 + (-1), -2 + 4) = (2, -2, 2)$；
$\boldsymbol{a} - \boldsymbol{b} = (2 - 0, -1 - (-1), -2 - 4) = (2, 0, -6)$；
$2\boldsymbol{a} - 3\boldsymbol{b} = 2(2, -1, -2) - 3(0, -1, 4) = (4, -2, -4) - (0, -3, 12) = (4 - 0, -2 - (-3), -4 - 12) = (4, 1, -16)$。