已知 a =(1, 3 ) ， b =(-1,0) ，则 a 与 b 夹角的余弦值为＿＿＿＿＿＿。 - 1 2 数量积与夹角夹角公式坐标运算数量积山东卷 2020

− 2 1 数量积： a <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"/> ⋅ b <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"/> = 1 × ( − 1 ) + 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/> × 0 = − 1 。两个模： ∣ a <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"/> ∣ = 1 + 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/> = 2 ， ∣ b <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"/> ∣ = 1 。由夹角公式 cos θ = ∣ a <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"/> ∣∣ b <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"/> ∣ a <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"/> ⋅ b <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"/> = 2 × 1 − 1 = − 2 1 。

高考数学考点题库/平面向量/数量积与夹角

2020 年山东卷·第13题

填空题★★★☆☆ 中档改编题通法

已知

a = (1, 3)

，

b = (- 1, 0)

，则

a

与

b

夹角的余弦值为＿＿＿＿＿＿。

答案

- \frac{1}{2}

分步解析

1数量积： $a \cdot b = 1 \times (- 1) + 3 \times 0 = - 1$ 。
2两个模： $∣ a ∣ = 1 + 3 = 2$ ， $∣ b ∣ = 1$ 。
3由夹角公式 $cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣} = \frac{- 1}{2 \times 1} = - \frac{1}{2}$ 。

#夹角公式#坐标运算#数量积

2020 年真题山东卷

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