高三数学一轮复习试题：平面向量的概念
<p><strong>导读：</strong>高考，比的不是智商高低，比的是谁的耐心好，经过一轮、二轮、三轮复习的摧残还能有几个小伙伴说自己屹立不倒的?今天数学网小编末宝就给大家带来了高考数学一轮复习的同步练习，快来看看吧。</p><p class="MsoNormal">1.<span>设</span><b><i>a</i></b><span>是非零向量，</span><i>λ</i><span>是非零实数，下列结论中正确的是</span>(<span> </span>)</p><p class="MsoNormal">A.<b><i>a</i></b><span>与</span><i>λ</i><b><i>a</i></b><span>的方向相反</span> B.<b><i>a</i></b><span>与</span><i>λ</i><sup>2</sup><b><i>a</i></b><span>的方向相同</span> </p><p class="MsoNormal">C.|<span>－</span><i>λ</i><b><i>a</i></b>|≥|<b><i>a</i></b>| D.|<span>－</span><i>λ</i><b><i>a</i></b>|≥|<i>λ</i>|·<b><i>a</i></b> </p><p class="MsoNormal"><b><i><br /></i></b> </p><p class="MsoNormal">5.<span>已知点</span><i>O</i><span>，</span><i>A</i><span>，</span><i>B</i><span>不在同一条直线上，点</span><i>P</i><span>为该平面上一点，且</span>2→(OP)<span>＝</span>2→(OA)<span>＋</span>→(BA)<span>，则</span>(<span> </span>)</p><p class="MsoNormal">A.<span>点</span><i>P</i><span>在线段</span><i>AB</i><span>上</span> </p><p class="MsoNormal">B.<span>点</span><i>P</i><span>在线段</span><i>AB</i><span>的反向延长线上</span> </p><p class="MsoNormal">C.<span>点</span><i>P</i><span>在线段</span><i>AB</i><span>的延长线上</span> </p><p class="MsoNormal">D.<span>点</span><i>P</i><span>不在直线</span><i>AB</i><span>上</span> </p><p class="MsoNormal"><span>解析 因为</span>2→(OP)<span>＝</span>2→(OA)<span>＋</span>→(BA)<span>，所以</span>2→(AP)<span>＝</span>→(BA)<span>，所以点</span><i>P</i><span>在线段</span><i>AB</i><span>的反向延长线上，故选</span>B.</p><p class="MsoNormal"><span>答案 </span>B</p><p class="MsoNormal"> </p><p class="MsoNormal"><br /></p><p class="MsoNormal">7.<span>若点</span><i>O</i><span>是</span>△<i>ABC</i><span>所在平面内的一点，且满足</span>|→(OB)<span>－</span>→(OC)<span>＝</span>|→(OB)<span>＋</span>→(OC)<span>－</span>2→(OA)|<span>，则</span>△<i>ABC</i><span>的形状为</span>________.</p><p class="MsoNormal"> </p><p class="MsoNormal">8.<span>向量</span><b><i>e</i></b><sub>1</sub><span>，</span><b><i>e</i></b><sub>2</sub><span>不共线，</span>→(AB)<span>＝</span>3(<b><i>e</i></b><sub>1</sub><span>＋</span><b><i>e</i></b><sub>2</sub>)<span>，</span>→(CB)<span>＝</span><b><i>e</i></b><sub>2</sub><span>－</span><b><i>e</i></b><sub>1</sub><span>，</span>→(CD)<span>＝</span>2<b><i>e</i></b><sub>1</sub><span>＋</span><b><i>e</i></b><sub>2</sub><span>，给出下列结论：</span>①<i>A</i><span>，</span><i>B</i><span>，</span><i>C</i><span>共线；</span>②<i>A</i><span>，</span><i>B</i><span>，</span><i>D</i><span>共线；</span>③<i>B</i><span>，</span><i>C</i><span>，</span><i>D</i><span>共线；</span>④<i>A</i><span>，</span><i>C</i><span>，</span><i>D</i><span>共线，其中所有正确结论的序号为</span>________.</p><p class="MsoNormal"><span>解析 由</span>→(AC)<span>＝</span>→(AB)<span>－</span>→(CB)<span>＝</span>4<b><i>e</i></b><sub>1</sub><span>＋</span>2<b><i>e</i></b><sub>2</sub><span>＝</span>2→(CD)<span>，且</span>→(AB)<span>与</span>→(CB)<span>不共线，可得</span><i>A</i><span>，</span><i>C</i><span>，</span><i>D</i><span>共线，且</span><i>B</i><span>不在此直线上</span>.</p><p class="MsoNormal"><span>答案 </span>④</p><p class="MsoNormal">9.<span>已知</span>△<i>ABC</i><span>和点</span><i>M</i><span>满足</span>→(MA)<span>＋</span>→(MB)<span>＋</span>→(MC)<span>＝</span>0<span>，若存在实数</span><i>m</i><span>使得</span>→(AB)<span>＋</span>→(AC)<span>＝</span><i>m</i>→(AM)<span>成立，则</span><i>m</i><span>＝</span>________.</p><p class="MsoNormal"><span>解析 由已知条件得</span>→(MB)<span>＋</span>→(MC)<span>＝－</span>→(MA)<span>，如图，延长</span><i>AM</i><span>交</span><i>BC</i><span>于</span><i>D</i><span>点，则</span><i>D</i><span>为</span><i>BC</i><span>的中点</span>.<span>延长</span><i>B</i> <i>M</i><span>交</span><i>AC</i><span>于</span><i>E</i><span>点，延长</span><i>CM</i><span>交</span><i>AB</i><span>于</span><i>F</i><span>点，同理可证</span><i>E</i><span>、</span><i>F</i><span>分别为</span><i>AC</i><span>、</span><i>AB</i><span>的中点，即</span><i>M</i><span>为</span>△<i>ABC</i><span>的重心，</span>∴→(AM)<span>＝</span>3(2)→(AD)<span>＝</span>3(1)(→(AB)<span>＋</span>→(AC))<span>，即</span>→(AB)<span>＋</span>→(AC)<span>＝</span>3→(AM)<span>，则</span><i>m</i><span>＝</span>3.</p><p></p><p class="MsoNormal">12.<span>已知</span><i>O</i><span>，</span><i>A</i><span>，</span><i>B</i><span>是不共线的三点，且</span>→(OP)<span>＝</span><i>m</i>→(OA)<span>＋</span><i>n</i>→(OB)(<i>m</i><span>，</span><i>n</i>∈<b>R</b>).</p><p class="MsoNormal">(1)<span>若</span><i>m</i><span>＋</span><i>n</i><span>＝</span>1<span>，求证：</span><i>A</i><span>，</span><i>P</i><span>，</span><i>B</i><span>三点共线；</span> </p><p class="MsoNormal">(2)<span>若</span><i>A</i><span>，</span><i>P</i><span>，</span><i>B</i><span>三点共线，求证：</span><i>m</i><span>＋</span><i>n</i><span>＝</span>1.</p><p class="MsoNormal"><span>证明 </span>(1)<span>若</span><i>m</i><span>＋</span><i>n</i><span>＝</span>1<span>，</span> </p><p class="MsoNormal"><span>则</span>→(OP)<span>＝</span><i>m</i>→(OA)<span>＋</span>(1<span>－</span><i>m</i>)→(OB)<span>＝</span>→(OB)<span>＋</span><i>m</i>(→(OA)<span>－</span>→(OB))<span>，</span> </p><p class="MsoNormal">∴→(OP)<span>－</span>→(OB)<span>＝</span><i>m</i>(→(OA)<span>－</span>→(OB))<span>，</span> </p><p class="MsoNormal"><span>即</span>→(BP)<span>＝</span><i>m</i>→(BA)<span>，</span>∴→(BP)<span>与</span>→(BA)<span>共线</span>.</p><p class="MsoNormal"><span>又</span>∵→(BP)<span>与</span>→(BA)<span>有公共点</span><i>B</i><span>，则</span><i>A</i><span>、</span><i>P</i><span>、</span><i>B</i><span>三点共线，</span> </p><p class="MsoNormal"> </p><p>更多数学复习资讯，尽在数学网。</p><p class="p"><b>末宝带你游数学：</b> </p><p 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