平面向量的实际背景及基本概念 · 历年高考数学真题与解析

本页汇总 高考真题检索 的「平面向量的实际背景及基本概念」高考数学真题共 5 道,覆盖 2008–2015 年,最常出题型为 单选题;含完整答案与解析。

5questions
Past questions included
2008–2015
Years covered
区分题为主
Overall difficulty
Single-choice
Most common question type
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Core competencies应用理解

Past Question List

2015 National Gaokao Fill in the blank Differentiating Question 13 2015_新课标 II 卷 (2015·理)

13.(5分)设向量 $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ 不平行,向量 $\lambda \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$ 与 $\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{~b}}$ 平行,则实数 $\lambda=-\frac{1}{2}$
$\_\_\_\_$ $\frac{1}{2}$ .

2014 National Gaokao Single choice Differentiating Question 5 2014_退役省自主命题 (2014·文)

5.设 $\vec{a}, \vec{b}, \vec{c}$ 是非零向量,已知命题 $\mathrm{P}:$ 若 $\vec{a} \cdot \vec{b}=0, \vec{b} \cdot \vec{c}=0$ ,则 $\vec{a} \cdot \vec{c}=0$ ;命题 $\mathrm{q}: ~$ 若 $\vec{a} / / \vec{b}, \vec{b} / / \vec{c}$ ,则 $\vec{a} / / \vec{c}$ ,则下列命题中真命题是

A. $p \vee q$
B. $p \wedge q$
C. $(\neg p) \wedge(\neg q)$
D. $p \vee(\neg q)$
2013 National Gaokao Single choice Differentiating Question 4 2013_退役省自主命题 (2013·文)

(3)已知点 $A(1,3), B(4,-1)$,则与向量 $\overrightarrow{A B}$ 同方向的单位向量为

A. $\left(\frac{3}{5},-\frac{4}{5}\right)$
B. $\left(\frac{4}{5},-\frac{3}{5}\right)$
C. $\left(-\frac{3}{5}, \frac{4}{5}\right)$
D. $\left(-\frac{4}{5}, \frac{3}{5}\right)$
2008 National Gaokao Single choice Differentiating Question 8 2008_老新课标卷 (2008·理)

8.平面向量 $\boldsymbol{a}, \boldsymbol{b}$ 共线的充要条件是

A. $\boldsymbol{a}, \boldsymbol{b}$ 方向相同
B. $\boldsymbol{a}, \boldsymbol{b}$ 两向量中至少有一个为零向量
C. $\exists \lambda \in \mathbf{R}, \quad \boldsymbol{b}=\lambda \boldsymbol{a}$
D. 存在不全为零的实数 $\lambda_{1}, \lambda_{2}, \lambda_{1} \boldsymbol{a}+\lambda_{2} \boldsymbol{b}=\mathbf{0}$
2008 National Gaokao Single choice Differentiating Question 9 2008_老新课标卷 (2008·文)

9、平面向量 $\vec{a}, \vec{b}$ 共线的充要条件是

A. $\vec{a}, \vec{b}$ 方向相同
B. $\vec{a}, \vec{b}$ 两向量中至少有一个为零向量
C. $\exists \lambda \in R, \quad \vec{b}=\lambda \vec{a}$
D. 存在不全为零的实数 $\lambda_{1}, \lambda_{2}, \lambda_{1} \vec{a}+\lambda_{2} \vec{b}=\overrightarrow{0}$

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