31立体几何解法第四招:移花接木-平移法求求异面直线夹角
立体几何解法第四招:移花接木-平移法求求异面直线夹角
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异面直线所成角的大小,是由空间一点分别引它们的平行线所成的锐角(或直角)来定义的。因此,通常求异面直线所成的角要通过平移直线,形成角,把异面问题转化为共面问题,然后在某个三角形或平行四边形中求出角来得到异面直线所成角的大小。在这一方法中,平移直线是求异面直线所成角的关键。
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1.平移法求异面直线夹角的步骤:
选点→平移→定角→计算(异面直线所成角的范围为
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)
2.
两种常见平移方法:
三角形:
利用中点构造三角形中位线,实现直线的平移,将异面直线转化为相交直线
平行四边形:
利用平行线的传递性,构造平行四边形,实现直线的平移,将异面直线转化为相交直线
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(2018·天津文)如图,在四面体
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中,
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是等边三角形,平面
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平面
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,点
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为棱
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的中点,
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,
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,
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.
(Ⅰ)求证:
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;
(Ⅱ)求异面直线
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与
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所成角的余弦值;
(Ⅲ)求直线
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与平面ABD所成角的正弦值.
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【答案】见解析
【解析】(Ⅰ)证明:由平面
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平面
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,平面
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平面
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,
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,可得
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平面
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,故
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.
(Ⅱ)取棱
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的中点
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,连接
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,
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.又因为
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为棱
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的中点,故
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.
所以
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(或其补角)为异面直线
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与
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所成的角.
在
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中,
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,故
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.因为
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平面
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,故
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.
在
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中,
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,故
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.
在等腰三角形
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中,
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,可得
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.
所以,异面直线
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与
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所成角的余弦值为
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.
(Ⅲ)连接
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.因为
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为等边三角形,
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为边
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的中点,故
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,
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.又因为平面
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平面
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,而
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平面
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,故
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平面
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.所以,
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为直线
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与平面
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所成的角.
在
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中,
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.
在
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中,
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.
所以,直线
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与平面
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所成角的正弦值为
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.
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1.(2020届河南南阳一中开学考)在长方体
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中,
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,
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,
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,点
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为长方形
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对角线的交点,
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为棱
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的中点,则异面直线
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与
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所成的角为( )
A.
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B.
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C.
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D.
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2.如图,正方体
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中,直线
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与
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所成角大小为__________.
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3.(2020届全国大联考高三联考)已知三棱锥
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,
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,
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是边长为
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的正三角形,
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,
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分别是
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、
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的中点,
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为棱
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上一动点(点
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除外),
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,若异面直线
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与
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所成的角为
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,且
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,则
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__________.