1.3(1) 空间向量及其运算的坐标表示
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【基础过关系列】 高二数学上学期同步知识点剖析精品讲义(人教A版2019)
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选择性必修第一册同步巩固,难度2颗星!
基础知识
空间直角坐标系
1 空间直角坐标系
在空间选定一点\(O\)和一个单位正交基底 \(\{\vec{i}, \vec{j}, \vec{k}\}\),以点\(O\)为原点,分别以 \(\vec{i}, \vec{j}, \vec{k}\)的方向为正方向、以它们的长为单位长度建立三条数轴:\(x\)轴、\(y\)轴、\(z\)轴,它们都叫做坐标轴,这时我们就建立了一个空间直角坐标系\(Oxyz\),\(O\)叫做原点, \(\vec{i}, \vec{j}, \vec{k}\)都叫做坐标向量,通过每两条坐标轴的平面叫做坐标平面,分别称为\(Oxy\)平面,\(Oxz\)平面,\(Oyz\)平面,它们把空间分成八个部分.
在空间直角坐标系中,让右手拇指指向\(x\)轴的正方向,食指指向\(y\)轴的正方向,如果中指指向\(z\)轴的正方向,则称这个坐标系为右手直角坐标系.
2 空间直角坐标系中的坐标
在空间直角坐标系\(O-x y z\)中,对空间任一点\(A\),存在唯一的有序实数组\((x ,y ,z)\),使 \(\overrightarrow{O A}=x \vec{i}+y \vec{j}+z \vec{k}\) , 有序实数组\((x ,y ,z)\)叫作向量\(A\)在空间直角坐标系中的坐标,记作 \(A(x ,y ,z)\),\(x\)叫横坐标,\(y\)叫纵坐标,\(z\)叫竖坐标.
![](https://img2023.cnblogs.com/blog/1943255/202305/1943255-20230518103350025-597788813.png )
【例1】在空间直角坐标系中标出下列各点
\(A(0,2,0)\),\(B(0,2,4)\),\(C(-1,0,5)\),\(D(1,-3,4)\)
解析 略
【例2】求点\(A(1,-2,3)\)关于\(Oxy\)平面,\(Oxz\)平面,\(Oyz\)平面的对称点.
解析 \(A_1 (1,-2,-3)\),\(A_2 (1,2,3)\),\(A_3 (-1,-2,3)\).
空间向量的直角坐标运算律
① 若 \(\vec{a}=\left(a_{1}, a_{2}, a_{3}\right)\),\(\vec{b}=\left(b_{1}, b_{2}, b_{3}\right)\)
则 \(\vec{a}+\vec{b}=\left(a_{1}+b_{1}, a_{2}+b_{2}, a_{3}+b_{3}\right)\), \(\vec{a}-\vec{b}=\left(a_{1}-b_{1}, a_{2}-b_{2}, a_{3}-b_{3}\right)\),
\(\lambda \vec{a}=\left(\lambda a_{1}, \lambda a_{2}, \lambda a_{3}\right) \quad (λ∈R)\),
\(\vec{a} \cdot \vec{b}=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\),
\(\vec{a} \| \vec{b} \Rightarrow a_{1}=\lambda b_{1}, a_{2}=\lambda b_{2}, a_{3}=\lambda b_{3}(\lambda \in R)\),
\(\vec{a} \perp \vec{b} \Rightarrow \vec{a} \cdot \vec{b}=0 \Rightarrow a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}=0\).
② 若\(A(x_1 ,y_1 ,z_1 )\) , \(B(x_2 ,y_2 ,z_2 )\) ,则 \(\overrightarrow{A B}=\left(x_{2}-x_{1}, y_{2}-y_{1}, z_{2}-z_{1}\right)\).
③ 模长公式
若 \(\vec{a}=\left(a_{1}, a_{2}, a_{3}\right)\),则 \(|\vec{a}|=\sqrt{\vec{a} \cdot \vec{a}}=\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}\).
④ 夹角公式
\(\cos \langle\vec{a}, \vec{b}\rangle=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}=\dfrac{a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}}{\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}} \sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}}\),
\(∆ABC\)中 , \(\overrightarrow{A B} \cdot \overrightarrow{A C}>0 \Rightarrow A\)为锐角, \(\overrightarrow{A B} \cdot \overrightarrow{A C}<0 \Rightarrow\)为钝角.
⑤ 两点间的距离公式
若\(A(x_1 ,y_1 ,z_1)\),\(B(x_2 ,y_2 ,z_2)\),
则 \(|\overrightarrow{A B}|=\sqrt{\overrightarrow{A B}^{2}}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}\)
或 \(d_{A B}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}\).
【例1】若 \(\vec{a}=\left(a_{1}, a_{2}, a_{3}\right)\),\(\vec{b}=\left(b_{1}, b_{2}, b_{3}\right)\),证明 \(\vec{a} \cdot \vec{b}=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\).
证明 设 \(\vec{i}, \vec{j}, \vec{k}\)为空间的一个单位正交基底,
则 \(\vec{a}=a_{1} \vec{i}+a_{2} \vec{j}+a_{3} \vec{k}\), \(\vec{b}=b_{1} \vec{i}+b_{2} \vec{j}+b_{3} \vec{k}\),
所以 \(\vec{a} \cdot \vec{b}=\left(a_{1} \vec{i}+a_{2} \vec{j}+a_{3} \vec{k}\right) \cdot\left(b_{1} \vec{i}+b_{2} \vec{j}+b_{3} \vec{k}\right)\)
利用向量数量积的分配律以及\(\vec{i} \cdot \vec{i}=\vec{j} \cdot \vec{j}=\vec{k} \cdot \vec{k}=1\), \(\vec{i} \cdot \vec{j}=\vec{j} \cdot \vec{k}=\vec{k} \cdot \vec{i}=0\)
得 \(\vec{a} \cdot \vec{b}=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\).
【例2】 已知 \(\vec{a}=(1,0,1)\), \(\vec{b}=(x,-1,2)\),且 \(\vec{a} \cdot \vec{b}=3\),则向量 \(\vec{a}\)与 \(\vec{b}\)的夹角为\(\underline{\quad \quad}\).
解析 \(∵\vec{a}=(1,0,1)\), \(\vec{b}=(x,-1,2)\),且 \(\vec{a} \cdot \vec{b}=3\),
\(\therefore \vec{a} \cdot \vec{b}=x+2=3\),解得\(x=1\),
\(\therefore \cos \langle\vec{a}, \vec{b}\rangle=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}=\dfrac{3}{\sqrt{2} \cdot \sqrt{6}}=\dfrac{\sqrt{3}}{2}\),
\(∴\)向量 \(\vec{a}\)与 \(\vec{b}\)的夹角为 \(\dfrac{\pi}{6}\).
基本方法
【题型1】空间坐标系
【典题1】 在长方体\(OABC-D' A' B' C'\)中,\(OA=3\),\(OC=4\),\(OD'=2\),以 \(\left\{\dfrac{1}{3} \overrightarrow{O A}, \dfrac{1}{4} \overrightarrow{O C}, \dfrac{1}{2} \overrightarrow{O D^{\prime}}\right\}\)为单位正交基底,建立如图所示的空间直角坐标系\(Oxyz\),
(1)写出\(D',B'\)四点的坐标;(2)写出向量 \(\overrightarrow{A^{\prime} B^{\prime}}, \overrightarrow{A^{\prime} C^{\prime}}\)的坐标.
解析 (1)点\(D'\)在\(z\)轴上,且\(OD'=2\),所以 \(\overrightarrow{O D^{\prime}}=0 \vec{j}+0\vec{b}+2\vec{k}\),
所以点\(D'\)的坐标是\((0,0,2)\);
点\(B'\)在\(x\)轴、\(y\)轴、\(z\)轴上的射影分别为\(A,C,D'\),它们在坐标轴上的坐标分别为\(3,4,2\),
所以点\(B'\)的坐标是\((3,4,2)\);
(2) \(\overrightarrow{A^{\prime} B^{\prime}}=\overrightarrow{O C}=0 \vec{i}+4 \vec{j}+0 \vec{k}=(0,4,0)\); \(\overrightarrow{A^{\prime} C^{\prime}}=\overrightarrow{A^{\prime} D^{\prime}}+\overrightarrow{D^{\prime} C^{\prime}}=-3 \vec{i}+4 \vec{j}+0 \vec{k}\).
巩固练习
1 (多选)如图,在长方体\(ABCD-A_1 B_1 C_1 D_1\)中,\(AB=5\),\(AD=4\),\(AA_1=3\),以直线\(DA\),\(DC\),\(DD_1\)分别为\(x\)轴、\(y\)轴、\(z\)轴,建立空间直角坐标系,则 ( )
A.点\(B_1\)的坐标为\((4,5,3)\)
B.点\(C_1\)关于点\(B\)对称的点为\((5,8, -3)\)
C.点\(A\)关于直线\(BD_1\)对称的点为\((0,5,3)\)
D.点\(C\)关于平面\(ABB_1 A_1\)对称的点为\((8,5,0)\)
2 在空间直角坐标系\(Oxyz\)中,
(1)哪个坐标平面与\(x\)轴垂直?哪个坐标平面与\(y\)轴垂直?哪个坐标平面与\(z\)轴垂直?
(2)写出点\(P(2,3,4)\)在三个坐标平面内的射影的坐标;
(3)写出点\(P(1,3,5)\)关于原点成中心对称的点的坐标.
参考答案
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答案 \(ACD\)
解析 由图形及其已知可得:点\(B_1\)的坐标为\((4,5,3)\),点\(C_1 (0,5,3)\)关于点\(B\)对称的点为\((8,5,-3)\),
点\(A\)关于直线\(BD_1\)对称的点为\(C_1 (0,5,3)\),
点\(C(0,5,0)\)关于平面\(ABB_1 A_1\)对称的点为\((8,5,0)\).
因此\(ACD\)正确.
故选:\(ACD\). -
答案 (1)平面\(Oyz\),平面\(Oxz\),平面\(Oxy\);
(2)在平面\(Oyz\)上的射影坐标是\((0,3,4)\),在平面\(Oxy\)上的射影坐标是\((2,3,0)\),在平面\(Oxz\)上的射影坐标是\((2,0,4)\);
(3)\((-1,-3,-5)\).
【题型2】空间向量坐标运算
【典题1】已知 \(\vec{a}=(x, 4,1)\), \(\vec{b}=(-2, y,-1)\), \(\vec{c}=(3,-2, z)\), \(\vec{a} / / \vec{b}\), \(\vec{b} \perp \vec{c}\),求:
(1) \(\vec{a}, \vec{b}, \vec{c}\); \(\qquad \qquad\) (2) \(\vec{a}+\vec{c}\)与 \(\vec{b}+\vec{c}\)所成角的余弦值.
解析(1) \(\because \vec{a} / / \vec{b}\), \(\therefore \dfrac{x}{-2}=\dfrac{4}{y}=\dfrac{1}{-1}\),解得\(x=2,y=-4\),
故 \(\vec{a}=(2,4,1)\), \(\vec{b}=(-2,-4,-1)\),
又因为 \(\vec{b} \perp \vec{c}\),所以 \(\vec{b} \cdot \vec{c}=0\),即\(-6+8-z=0\),解得\(z=2\),
故 \(\vec{c}=(3,-2,2)\);
(2)由(1)可得 \(\vec{a}+\vec{c}=(5,2,3)\), \(\vec{b}+\vec{c}=(1,-6,1)\),
设向量\(\vec{a}+\vec{c}\)与 \(\vec{b}+\vec{c}\)所成的角为\(θ\),
则 \(\cos \theta=\dfrac{5-12+3}{\sqrt{38} \cdot \sqrt{38}}=-\dfrac{2}{19}\).
巩固练习
1 已知 \(\vec{a}=(4,2,-4)\), \(\vec{b}=(6,-3,3)\),则 \(|\vec{a}-\vec{b}|=\)\(\underline{\quad \quad}\).
2 已知向量 \(\vec{a}=\left(1,-2, \dfrac{m}{2}-2\right)\), \(\vec{b}=\left(m, 3,-\dfrac{m}{2}-2\right)\),若 \((\vec{a}+\vec{b}) \perp(\vec{a}-\vec{b})\),则\(m\)的值为\(\underline{\quad \quad}\).
3 已知空间向量 \(\vec{m}=(3,1,3)\), \(\vec{n}=(-1, \lambda,-1)\),且 \(\vec{m} / / \vec{n}\),则实数\(λ=\)\(\underline{\quad \quad}\) .
4 若 \(\vec{a}=(1, \lambda, 2)\), \(\vec{b}=(2,-1,1)\), \(\vec{a}\)与 \(\vec{b}\)的夹角为\(60^{\circ}\),则\(λ\)的值为\(\underline{\quad \quad}\).
5 若向量 \(\vec{a}=(1, \lambda, 1)\), \(\vec{b}=(2,-1,-2)\),且 \(\vec{a}\)与 \(\vec{b}\)的夹角余弦为 \(\dfrac{\sqrt{2}}{6}\),则\(λ\)等于\(\underline{\quad \quad}\).
参考答案
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答案 \(\sqrt{78}\)
解析 \(∵\vec{a}=(4,2,-4)\), \(\vec{b}=(6,-3,3)\), \(\therefore \vec{a}-\vec{b}=(-2,5,-7)\),
\(\therefore|\vec{a}-\vec{b}|=\sqrt{(-2)^{2}+5^{2}+(-7)^{2}}=\sqrt{78}\). -
答案 \(-2\)
解析 根据题意, \(\vec{a}+\vec{b}=(1+m, 1,-4)\), \(\vec{a}-\vec{b}=(1-m,-5, m)\),
所以由 \((\vec{a}+\vec{b}) \perp(\vec{a}-\vec{b})\),则有 \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=(1+m)(1-m)-5-4 m=0\)
所以\(m=-2\). -
答案 \(-\dfrac{1}{3}\)
解析 \(\because \vec{m} / / \vec{n}\),\(∴\)可设 \(k \vec{m}=\vec{n}\),
\(\therefore\left\{\begin{array}{l} -1=3 k \\ \lambda=k \\ -1=3 k \end{array}\right.\),解得 \(\lambda=k=-\dfrac{1}{3}\). -
答案 \(-17\)或\(1\)
解析 \(∵\vec{a}=(1, \lambda, 2)\), \(\vec{b}=(2,-1,1)\),
\(\therefore|\vec{a}|=\sqrt{5+\lambda^{2}},|\vec{b}|=\sqrt{6}\) ,\(\vec{a} \cdot \vec{b}=4-\lambda\)
又 \(\vec{a}\)与 \(\vec{b}\)的夹角为\(60^{\circ}\),
\(\therefore \cos 60^{\circ}=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\dfrac{4-\lambda}{\sqrt{5+\lambda^{2}} \cdot \sqrt{6}}=\dfrac{1}{2}\),解得:\(λ=-17\)或\(1\). -
答案 \(-\sqrt{2}\)
解析 \(∵\)向量\(\vec{a}=(1, \lambda, 1)\), \(\vec{b}=(2,-1,-2)\), \(\vec{a}\)与 \(\vec{b}\)的夹角余弦为 \(\dfrac{\sqrt{2}}{6}\),
\(\therefore \cos <\vec{a}, \vec{b}>=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}=\dfrac{-\lambda}{\sqrt{2+\lambda^{2}} \cdot \sqrt{9}}=\dfrac{\sqrt{2}}{6}\),
解得 \(\lambda=-\sqrt{2}\).
【题型3】建立空间坐标系处理几何问题
【典题1】 在棱长为\(1\)的正方体\(ABCD-A' B' C' D'\)中,\(M\)为\(BC'\)的中点,\(E_1,F_1\)分别在棱\(A' B'\),\(C' D'\)上, \(B^{\prime} E_{1}=\dfrac{1}{4} A^{\prime} B^{\prime}\), \(D^{\prime} F_{1}=\dfrac{1}{4} C^{\prime} D^{\prime}\),
(1)求\(AM\)的长; \(\qquad \qquad\) (2)求\(BE_1\)与\(DF_1\)所成角的余弦值.
解析 如图建立空间直角坐标系\(Dxyz\),
则\(A(1,0,0)\), \(M\left(\dfrac{1}{2}, 1, \dfrac{1}{2}\right)\),
所以 \(A M=\sqrt{\left(\dfrac{1}{2}-1\right)^{2}+(1-0)^{2}+\left(\dfrac{1}{2}-0\right)^{2}}=\dfrac{\sqrt{6}}{2} \text {, }\),
(2)由已知,得\(B(1,1,0)\), \(E_{1}\left(1, \dfrac{3}{4}, 1\right)\),\(D(0,0,0)\), \(F_{1}\left(0, \dfrac{1}{4}, 1\right)\),
所以 \(\overrightarrow{B E_{1}}=\left(1, \dfrac{3}{4}, 1\right)-(1,1,0)=\left(0,-\dfrac{1}{4}, 1\right)\), \(\overrightarrow{D F_{1}}=\left(0, \dfrac{1}{4}, 1\right)-(0,0,0)=\left(0, \dfrac{1}{4}, 1\right)\),
\(\left|\overrightarrow{B E_{1}}\right|=\dfrac{\sqrt{17}}{4}\), \(\left|\overrightarrow{D F_{1}}\right|=\dfrac{\sqrt{17}}{4}\),
所以 \(\overrightarrow{B E_{1}} \cdot \overrightarrow{D F_{1}}=0 \times 0+\left(-\dfrac{1}{4} \times \dfrac{1}{4}\right)+1 \times 1=\dfrac{15}{16}\),
所以 \(\cos \left\langle\overrightarrow{B E_{1}}, \overrightarrow{D F_{1}}\right\rangle=\dfrac{\overrightarrow{B E_{1}} \cdot \overrightarrow{D F_{1}}}{\left|\overrightarrow{B E_{1}}\right|\left|\overrightarrow{D F_{1}}\right|}=\dfrac{\dfrac{15}{16}}{\dfrac{\sqrt{17}}{4} \times \dfrac{\sqrt{17}}{4}}=\dfrac{15}{17}\),
所以\(BE_1\)与\(DF_1\)所成角的余弦值是 \(\dfrac{15}{17}\).
巩固练习
1 如图,正方体\(OABC-D' A' B' C'\)的棱长为\(3\),点\(N,M\)分别在\(AC\),\(BC'\)上,\(AN=2CN\),\(BM=2MC'\),求\(MN\)的长.
2 如图,在正方体\(ABCD-A' B' C' D'\)中,\(M\)是\(AB\)的中点,求\(DB'\)与\(CM\)所成角的余弦值.
参考答案
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答案 \(\sqrt{5}\)
解析 建立如图的空间直角坐标系,
则\(A(3,0,0)\),\(C(0,3,0)\),
又由\(AN=2CN\),所以\(N(1,2,0)\),
因为\(B(3,3,0)\),\(C'(0,3,3)\),\(BM=2MC'\),所以\(M(1,3,2)\),
所以 \(\overrightarrow{N M}=(1,3,2)-(1,2,0)=(0,1,2)\),即 \(|\overrightarrow{N M}|=\sqrt{0^{2}+1^{2}+2^{2}}=\sqrt{5}\),
即\(MN=\sqrt{5}\). -
答案 \(\dfrac{\sqrt{15}}{15}\)
解析 建立如图的空间直角坐标系,不妨设边长为\(1\),
则\(D(0,0,0)\),\(B' (1,1,1)\),\(C(0,1,0)\), \(M\left(1, \dfrac{1}{2}, 0\right)\),
所以 \(\overrightarrow{D B^{\prime}}=(1,1,1)\), \(\overrightarrow{C M}=\left(1,-\dfrac{1}{2}, 0\right)\),
所以 \(\cos <\overrightarrow{D B^{\prime}}, \overrightarrow{C M}>=\dfrac{\overrightarrow{D B^{\prime}} \cdot \overrightarrow{C M}}{\left|\overrightarrow{D B^{\prime}}\right||\overrightarrow{C M}|}=\dfrac{1-\dfrac{1}{2}+0}{\sqrt{3} \cdot \sqrt{\dfrac{5}{4}}}=\dfrac{\sqrt{15}}{15}\).
所以\(DB_1\)与\(CM\)所成角的余弦值为 \(\dfrac{\sqrt{15}}{15}\).
分层练习
【A组---基础题】
1 若向量 \(\vec{a}=(1,-1,2)\), \(\vec{b}=(2,1,-3)\),则 \(|2 \vec{a}+\vec{b}|=\)( )
A. \(\sqrt{7}\) \(\qquad \qquad \qquad\qquad\) B. \(2\sqrt{2}\) \(\qquad \qquad \qquad\qquad\) C.\(3\) \(\qquad \qquad \qquad\qquad\) D. \(3\sqrt{2}\)
2 已知 \(\vec{a}=(2,-1,3)\), $\vec{b}=(-4, x+1, y-2) $,若 \(\vec{a} \| \vec{b}\),则\(x+y=\)( )
A.\(-6\) \(\qquad\qquad \qquad \qquad\) B.\(-5\) \(\qquad\qquad \qquad \qquad\) C.\(-4\) \(\qquad \qquad \qquad\qquad\)D.\(-3\)
3 已知向量 \(\vec{a}=(1,1,1)\), \(\vec{b}=(1,-2,2)\),且 \(k \vec{a}\)与 \(\vec{a}+\vec{b}\)互相垂直,则\(k\)的值为( )
A.\(2\) \(\qquad\qquad \qquad \qquad\) B.\(0\) \(\qquad\qquad \qquad \qquad\) C.\(-2\) \(\qquad\qquad \qquad \qquad\) D.\(1\)
4 若点\(A(-2,2,1)\)关于\(y\)轴的对称点为\(A'\),则向量 \(\overrightarrow{AA^{\prime}}\)的坐标为\(\underline{\quad \quad}\) .
5如图,以长方体\(ABCD-A_1 B_1 C_1 D_1\)的顶点\(D\)为坐标原点,过\(D\)的三条棱所在的直线为坐标轴,建立空间直角坐标系,若 \(\overrightarrow{D B_{1}}\)的坐标为 \((2,3,4)\),则 \(\overrightarrow{A C_{1}}\)的坐标为\(\underline{\quad \quad}\).
6 若\(A(2,-4,-1)\),\(B(-1,5,1)\),\(C(3,-4,1)\),则 \(\overrightarrow{C A} \cdot \overrightarrow{C B}=\)\(\underline{\quad \quad}\) .
7 已知 \(\vec{a}=(1,0,1)\), \(\vec{b}=(x,-1,2)\),且 \(\vec{a} \cdot \vec{b}=3\),则向量 \(\vec{a}\)与 \(\vec{b}\)的夹角为\(\underline{\quad \quad}\) .
8 在正方体\(ABCD-A_1 B_1 C_1 D_1\)中,\(E,F\)分别是\(BB_1\),\(D_1 B_1\)的中点,求证\(EF⊥DA_1\).
9如图,直三棱柱\(ABC-A_1 B_1 C_1\)底面\(△ABC\)中,\(CA=CB=1\),\(∠BCA=90^∘\),棱\(AA_1=2\),\(M\)是\(A_1 B_1\)的中点.
(1)求 \(\cos <{\overrightarrow{B A_{1}}}, \overrightarrow{C B_{1}}>\)的值;\(\qquad \qquad\) (2)求证:\(A_1 B⊥C_1 M\).
参考答案
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答案 \(D\)
解析 \(∵\)向量 \(\vec{a}=(1,-1,2)\), \(\vec{b}=(2,1,-3)\), \(\therefore 2 \vec{a}+\vec{b}=(4,-1,1)\),
\(\therefore|2 \vec{a}+\vec{b}|=\sqrt{16+1+1}=3 \sqrt{2}\).
故选:\(D\). -
答案 \(D\)
解析 \(∵\vec{a} \| \vec{b}\),\(∴\)存在实数\(k\)使得 \(\overrightarrow{k a}=\vec{b}\),
\(\left\{\begin{array}{l} -4=2 k \\ x+1=-k \\ y-2=3 k \end{array}\right.\),解得\(k=-2\),\(x=1\),\(y=-4\).
则\(x+y=-3\).
故选:\(D\). -
答案 \(B\)
解析 \(∵\)向量 \(\vec{a}=(1,1,1)\), \(\vec{b}=(1,-2,2)\),
\(\therefore k \vec{a}=(k, k, k)\), \(\vec{a}+\vec{b}=(2,-1,3)\),
\(∵ k \vec{a}\)与 \(\vec{a}+\vec{b}\)互相垂直,
\(\therefore(k \vec{a}) \cdot(\vec{a}+\vec{b})=2 k-k+3 k=0\),解得\(k=0\).
故选:\(B\). -
答案 \((4,0,-2)\)
解析 因为点\(A(-2,2,1)\),所以点\(A\)关于\(y\)轴的对称点为\(A' (2,2,-1)\),
故 \(\overrightarrow{AA^{\prime}}=(4,0,-2)\). -
答案 \((-2,3,4)\)
解析 \(\overrightarrow{D B_{1}}=(2,3,4)\),\(∴B_1 (2,3,4)\).
\(∴A(2,0,0)\),\(C_1 (0,3,4)\), \(\therefore \overrightarrow{A C_{1}}=(0,3,4)-(2,0,0)=(-2,3,4)\). -
答案 \(4\)
解析 由已知,\(\overrightarrow{C A} =(2-3,-4-(-4),-1-1)=(-1,0,-2)\)
\(\overrightarrow{C B}=(-1-3,5-(-4),1-1)=(-4,9,0)\),
\(\therefore \overrightarrow{C A} \cdot \overrightarrow{C B}=4+0+0=4\). -
答案 \(\dfrac{\pi}{6}\)
解析 \(∵\vec{a}=(1,0,1)\), \(\vec{b}=(x,-1,2)\),且 \(\vec{a} \cdot \vec{b}=3\),
\(\therefore \vec{a} \cdot \vec{b}=x+2=3\),解得\(x=1\),
\(\therefore \cos \langle\vec{a}, \vec{b}\rangle=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}=\dfrac{3}{\sqrt{2} \cdot \sqrt{6}}=\dfrac{\sqrt{3}}{2}\),
\(∴\)向量 \(\vec{a}\)与 \(\vec{b}\)的夹角为 \(\dfrac{\pi}{6}\). -
证明 不妨设正方体的棱长为\(1\),建立如图的空间直角坐标系\(Oxyz\),
则 \(E\left(1,1, \dfrac{1}{2}\right)\), \(F\left(\dfrac{1}{2}, \dfrac{1}{2}, 1\right)\),所以 \(\overrightarrow{E F}=\left(-\dfrac{1}{2},-\dfrac{1}{2}, \dfrac{1}{2}\right)\),
又\(A_1 (1,0,1)\),\(D(0,0,0)\),所以 \(\overrightarrow{\mathrm{DA}_{1}}=(1,0,1)\),
所以 \(\overrightarrow{E F} \cdot \overrightarrow{D A_{1}}=\left(-\dfrac{1}{2},-\dfrac{1}{2}, \dfrac{1}{2}\right) \cdot(1,0,1)=0\),
所以 \(\overrightarrow{E F} \perp \overrightarrow{D A_{1}}\),即\(EF⊥DA_1\). -
答案 (1) \(\dfrac{\sqrt{30}}{10}\)(2)略
解析 (1)以\(C\)为原点,\(CA,CB,CC_1\)所在直线分别为\(x,y,z\)轴,建立空间直角坐标系,
\(A_1 (1,0,2)\),\(B(0,1,0)\),\(C(0,0,0)\),\(B_1 (0,1,2)\),
\(\overrightarrow{B A_{1}}=(1,-1,2)\),\(\overrightarrow{C B_{1}}=(0,1,2)\),
\(\therefore \cos <{\overrightarrow{B A_{1}}} , \overrightarrow{C B_{1}}>=\dfrac{\overrightarrow{B A_{1}} \cdot \overrightarrow{C B_{1}}}{\left|\overrightarrow{B A_{1}}\right| \cdot\left|\overrightarrow{C B_{1}}\right|}=\dfrac{3}{\sqrt{6} \cdot \sqrt{5}}=\dfrac{\sqrt{30}}{10}\).
证明 :(2)\(A_1 (1,0,2)\),\(B(0,1,0)\),\(C_1 (0,0,2)\), \(M\left(\dfrac{1}{2}, \dfrac{1}{2}, 2\right)\),
\(\overrightarrow{A_{1} B}=(-1,1,-2)\), \(\overrightarrow{C_{1} M}=\left(\dfrac{1}{2}, \dfrac{1}{2}, 0\right)\) ,
\(\therefore \overrightarrow{A_{1} B} \cdot \overrightarrow{C_{1} M}=0\),
\(∴A_1 B⊥C_1 M\).
【B组---提高题】
1 如图,\(BC=4\)原点\(O\)是\(BC\)的中点,点 \(A\left(\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}, 0\right)\),点\(D\)在平面\(yOz\)上,且\(∠BDC=90^{\circ}\),\(∠DCB=30^{\circ}\),则\(AD\)的长度为 \(\underline{\quad \quad}\).
2 已知向量 \(\vec{a}=(1,2,3)\), \(\vec{b}=\left(x, x^{2}+y-2, y\right)\),并且 \(\vec{a} 、 \vec{b}\)共线且方向相同,则\(x+y=\)\(\underline{\quad \quad}\).
3 已知点\(A(1,-2,11)\)、\(B(4,2,3)\),\(C(6,-1,4)\),则\(△ABC\)中角\(C\)的大小是\(\underline{\quad \quad}\).
4 \(△ABC\)的三个顶点分别是\(A(1,-1,2)\),\(B(5,-6,2)\),\(C(1,3,-1)\),则\(AC\)边上的高\(BD\)长为\(\underline{\quad \quad}\).
参考答案
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答案 \(\sqrt{6}\)
解析 \(∵\)点\(D\)在平面\(yoz\)上,\(∴\)点\(D\)的横坐标为\(0\),
又\(∵BC=4\),原点\(O\)是\(BC\)的中点,\(∴∠BDC=90^{\circ}\),\(∠DCB=30^{\circ}\),
\(∴\)竖坐标为 \(z=4 \sin 30^{\circ} \sin 60^{\circ}=\sqrt{3}\),纵坐标为 \(y=-\left(2-4 \sin 30^{\circ} \cos 60^{\circ}\right)=-1\),
\(\therefore D(0,-1, \sqrt{3})\),
\(\therefore|A D|=\sqrt{\left(\dfrac{\sqrt{3}}{2}\right)^{2}+\left(\dfrac{1}{2}+1\right)^{2}+(\sqrt{3})^{2}}=\sqrt{6}\). -
答案 \(4\)
解析 向量 \(\vec{a}=(1,2,3)\), \(\vec{b}=\left(x, x^{2}+y-2, y\right)\),并且同向,
\(\therefore\left\{\begin{array}{l} x^{2}+y-2=2 x \\ y=3 x \end{array}\right.\) ,解得 \(\left\{\begin{array}{l} x=1 \\ y=3 \end{array}\right.\)或 \(\left\{\begin{array}{l} x=-2 \\ y=-6 \end{array}\right.\);
当\(x=-2,y=-6\)时,反向,应舍去;
\(∴x,y\)的值为 \(\left\{\begin{array}{l} x=1 \\ y=3 \end{array}\right.\).
\(∴x+y=4\). -
答案 \(90^{\circ}\)
解析 \(∵A(1,-2,11)\)、\(B(4,2,3)\),\(C(6,-1,4)\),
\(\therefore|\overrightarrow{A C}|=\sqrt{(1-6)^{2}+(-2+1)^{2}+(11-4)^{2}}=\sqrt{75}\)
\(|\overrightarrow{B C}|=\sqrt{(4-6)^{2}+(2+1)^{2}+(3-4)^{2}}=\sqrt{14}\)
又 \(\because \overrightarrow{C A}=(-5,-1,7)\),\(\overrightarrow{C B}=(-2,3,-1)\)
\(\therefore \overrightarrow{C A} \cdot \overrightarrow{C B}=(-5) \times(-2)+(-1) \times 3+7 \times(-1)=0\)
可得 \(\cos \angle A C B=\dfrac{\overrightarrow{C A} \cdot \overrightarrow{C B}}{|\overrightarrow{C A}||\times| \overrightarrow{C B} \mid}=0\)
\(∵∠ACB∈(0^{\circ},180^{\circ})\) \(∴∠ACB=90^{\circ}\)
故答案为\(90^{\circ}\) -
答案 \(5\)
解析 设 \(\overrightarrow{A D}=\lambda \overrightarrow{A C}\),
则 \(\overrightarrow{O D}=\overrightarrow{O A}+\lambda \overrightarrow{A C}=(1,-1,2)+\lambda(0,4,-3)=(1,-1+4 \lambda, 2-3 \lambda)\),
\(\therefore \overrightarrow{B D}=\overrightarrow{O D}-\overrightarrow{O B}=(-4,5+4 \lambda,-3 \lambda)\),
\(\because \overrightarrow{B D} \perp \overrightarrow{A C}\),
\(\therefore \overrightarrow{B D} \cdot \overrightarrow{A C}=0+4(5+4 \lambda)+9 \lambda=0\),解得 \(\lambda=-\dfrac{4}{5}\).
\(\therefore \overrightarrow{B D}=\left(-4, \dfrac{9}{5}, \dfrac{12}{5}\right)\),
\(\therefore|\overrightarrow{B D}|=\sqrt{4^{2}+\left(\dfrac{9}{5}\right)^{2}+\left(\dfrac{12}{5}\right)^{2}}=5\).
【C组---拓展题】
1 如图三棱柱\(ABC-A_1 B_1 C_1\)中,侧面\(BB_1 C_1 C\)是边长为\(2\)菱形,\(∠CBB_1=60^∘\),\(BC_1\)交\(B_1C\) 于点\(O\),\(AO⊥\)侧面\(BB_1 C_1 C\),且\(△AB_1 C\)为等腰直角三角形,如图建立空间直角坐标系\(O-xyz\),则点\(A_1\)的坐标为( )
A. \((-1, \sqrt{3}, 1)\) \(\qquad \qquad\) B. \((-\sqrt{3}, 1,1)\) \(\qquad \qquad\) C. \((-1,2, \sqrt{3})\) \(\qquad \qquad\) D. \((-2,1, \sqrt{3})\)
2 设向量 \(\vec{u}=(a, b, 0)\), \(\vec{v}=(c, d, 1)\),其中\(a^2+b^2=c^2+d^2=1\),则下列判断错误的是( )
A.向量 \(\vec{v}\)与\(z\)轴正方向的夹角为定值(与\(c,d\)之值无关)
B. \(\vec{u} \cdot \vec{v}\)的最大值为 \(\sqrt{2}\)
C. \(\vec{u}\)与 \(\vec{v}\)的夹角的最大值为 \(\dfrac{3 \pi}{4}\)
D.\(ad-bc\)的最大值为\(1\).
3 已知长方体\(ABCD-A_1 B_1 C_1 D_1\)中,\(AB=4\),\(BC=3\),\(AA_1=2\),空间中存在一动点\(P\)满足 \(\left|\overrightarrow{B_{1} P}\right|=1\),记 \(I_{1}=\overrightarrow{A B} \cdot \overrightarrow{A P}\), \(I_{2}=\overrightarrow{A D} \cdot \overrightarrow{A P}\), \(I_{3}=\overrightarrow{A C_{1}} \cdot \overrightarrow{A P}\),则( )
A.存在点\(P\),使得\(I_1=I_2\) \(\qquad \qquad\) B.存在点\(P\),使得\(I_1=I_3\)
C.对任意的点\(P\),有\(I_1>I_2\) \(\qquad \qquad\) D.对任意的点\(P\),有\(I_2>I_3\)
参考答案
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答案 \(B\)
解析 三棱柱\(ABC-A_1 B_1 C_1\)中,侧面\(BB_1 C_1 C\)是边长为\(2\)菱形,\(∠CBB_1=60°\),
\(BC_1\)交\(B_1 C\)于点\(O\),\(AO⊥\)侧面\(BB_1 C_1 C\),且\(△AB_1 C\)为等腰直角三角形,
如图建立空间直角坐标系\(O-xyz\),
过\(A_1\)作\(A_1 E⊥\)平面\(BCC_1 B_1\),垂足是\(E\),连结\(B_1 E\),\(C_1 E\),
则\(B_1 E∥OC_1\),\(C_1 E∥OB_1\),\(A_1 E∥AO\),
\(∴\)点\(A_1\)的坐标为 \((-\sqrt{3}, 1,1)\).
故选:\(B\).
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答案 \(B\)
解析 由向量 \(\vec{u}=(a, b, 0)\), \(\vec{v}=(c, d, 1)\),其中\(a^2+b^2=c^2+d^2=1\),知:
在\(A\)中,设\(z\)轴正方向的方向向量 \(\vec{z}=(0,0, t)\),
向量 \(\vec{v}\)与\(z\)轴正方向的夹角的余弦值:
\(\cos \alpha=\dfrac{\bar{z} \cdot \vec{v}}{|\vec{z}| \cdot|\vec{v}|}=\dfrac{t}{t \cdot \sqrt{c^{2}+d^{2}+1}}=\dfrac{\sqrt{2}}{2}\), \(\therefore \alpha=45^{\circ}\),
\(∴\)向量 \(\vec{v}\)与\(z\)轴正方向的夹角为定值\(45°\)(与\(c,d\)之值无关),故\(A\)正确;
在\(B\)中, \(\vec{u} \cdot \vec{v}=a c+b d \leq \dfrac{a^{2}+c^{2}}{2}+\dfrac{b^{2}+d^{2}}{2}=\dfrac{a^{2}+b^{2}+c^{2}+d^{2}}{2}=1\),
且仅当\(a=c\),\(b=d\)时取等号,因此 \(\vec{u} \cdot \vec{v}\)的最大值为\(1\),故\(B\)错误;
在\(C\)中,由\(B\)可得: \(|\vec{u} \cdot \vec{v}| \leq 1\), \(\therefore-1 \leq \vec{u} \cdot \vec{v} \leq 1\),
\(\therefore \cos <\vec{u}, \vec{v}>=\dfrac{\vec{u} \cdot \vec{v}}{|\vec{u}| \cdot|\vec{v}|}=\dfrac{a c+b d}{\sqrt{a^{2}+b^{2}} \cdot \sqrt{c^{2}+d^{2}+1}}\)\(\geq-\dfrac{1}{1 \times \sqrt{2}}=-\dfrac{\sqrt{2}}{2}\),
\(∴ \vec{u}\)与 \(\vec{v}\)的夹角的最大值为 \(\dfrac{3 \pi}{4}\),故\(C\)正确;
在\(D\)中, \(a d-b c \leq \dfrac{a^{2}+d^{2}}{2}+\dfrac{b^{2}+c^{2}}{2}=\dfrac{a^{2}+b^{2}+c^{2}+d^{2}}{2}=1\),
\(∴ad-bc\)的最大值为\(1\).故\(D\)正确.
故选:\(B\). -
答案 \(C\)
解析 如图所示
建立如图所示的空间直角坐标系,以\(B_1 A_1\)为\(x\)轴,\(B_1 C_1\)为\(y\)轴,\(B_1 B\)为\(z\)轴,\(B_1\)为坐标原点,由题意则\(B(0,0,2)\),\(A(4,0,2)\),\(D(4,3,2)\),\(C_1 (0,3,0)\),设\(P(x,y,z)\),
所以 \(\overrightarrow{A B}=(-4,0,0)\), \(\overrightarrow{A P}=(x-4, y, z-2)\), \(\overrightarrow{A D}=(0,3,0)\),\(\overrightarrow{A C}_{1}=(-4,3,-2)\)
因为满足 \(\left|\overrightarrow{B_{1} P}\right|=1\),
所以\(x^2+y^2+z^2=1\),\(x∈[-1,1]\),\(y∈[-1,1]\),\(z∈[-1,1]\),
\(\therefore I_{1}=\overrightarrow{A B} \cdot \overrightarrow{A P}=-4(x-4)\),
\(\therefore I_{2}=\overrightarrow{A D} \cdot \overrightarrow{A P}=3 y\)
\(\therefore I_{3}=\overrightarrow{A C_{1}} \cdot \overrightarrow{A P}=-4(x-4)+3 y-2(z-2)\),
\(I_1-I_2=-4(x-4)-3y=16-4x-3y>0\)恒成立,故\(C\)正确,\(A\)不正确;
\(I_1-I_3=-3y+2(z-2)=-4-3y+2z<0\)恒成立,所以\(B\)不正确,
\(I_2-I_3=4(x-4)+2(z-2)=-12+4x+2z<0\)恒成立,所以\(D\)不正确;
故选:\(C\).