文章目录
-
- 向量之间的位置关系
- 混合积的定义与性质
-
- 性质
- 用坐标计算混合积
- 三向量共面的条件
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- 从线性代数的角度理解
- 参考
- 06向量及其坐标表示、向量的方向角与方向余弦、向量组共线与共面的条件、向量的加法与数乘运算、向量组的线性组合、二维向量的基向量分解、三维向量的基向量分解、用坐标做向量的数乘
- 07向量的点积、数量积、两向量垂直的条件、投影与投影向量、向量的正交分解、几个不等式、用坐标计算数量积
- 08向量的叉积、向量积、用坐标行列式计算向量积、二重外积
- 09向量的混合积、向量之间的位置关系、用坐标行列式计算混合积、三向量共面的条件
- 10空间直线方程、参数方程、向量式方程、点向式方程、两点式方程、一般方程、空间直线的一般方程化为点向式方程
- 11空间平面方程、参数方程、向量式方程、行列式方程、三点式方程、点法式方程、一般方程
向量之间的位置关系
垂直 ⇔ a ⋅ b = 0 \Leftrightarrow \boldsymbol{a} \cdot \boldsymbol{b}=0 \quad ⇔a⋅b=0
平行 ⇔ a × b = 0 \Leftrightarrow \boldsymbol{a} \times \boldsymbol{b}=\mathbf{0} \quad ⇔a×b=0
共面 ⇔ \Leftrightarrow ⇔ a , b , c \boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c} a,b,c 线性相关
投影、面积与体积的计算
Π a ( b ) = b ⋅ a ∘ \Pi_{\boldsymbol{a}}(\boldsymbol{b})=\boldsymbol{b} \cdot \boldsymbol{a}^{\circ} Πa(b)=b⋅a∘
平行四边形的面积 S = ∣ a × b ∣ \begin{array}{c} \text { 平行四边形的面积 } \\ S=|\boldsymbol{a} \times \boldsymbol{b}| \end{array} 平行四边形的面积 S=∣a×b∣
问题：如何求平行六面体的体积？
以 a , b , c a, b, c a,b,c 为棱作平行六面体, 则底面积: A = ∣ a × b ∣ A=|\boldsymbol{a} \times \boldsymbol{b}| A=∣a×b∣ ，高: h = ∣ c ∣ ∣ cos α ∣ h=|\boldsymbol{c}||\cos \alpha| h=∣c∣∣cosα∣
故平行六面体体积为
V 六面体 = A h = ∣ a × b ∣ ∣ c ∥ cos α ∣ = ∣ ( a × b ) ⋅ c ∣ \begin{aligned} V_{\text {六面体 }} &=A h=|\boldsymbol{a} \times \boldsymbol{b}| \mid \boldsymbol{c} \| \cos \alpha \mid \\ &=|(\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}| \end{aligned} V六面体 =Ah=∣a×b∣∣c∥cosα∣=∣(a×b)⋅c∣
( 注 ) \Large\color{violet}{(注) } (注) 当 a , b , c \boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c} a,b,c 为右手系时, cos α > 0 \quad \cos \alpha>0 cosα>0, 则 ( a × b ) ⋅ c > 0 (\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c} >0 (a×b)⋅c>0 .
混合积的定义与性质
定 义 1 \large\color{magenta}{\boxed{\color{brown}{定义1} }} 定义1 \quad 向量 a , b , c \boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c} a,b,c 的混合积（记作 ( a b c ) (\boldsymbol{a} \boldsymbol{b} \boldsymbol{c}) (abc) ) 是一个数量：
( a b c ) = ( a × b ) ⋅ c (\boldsymbol{a} \boldsymbol{b} \boldsymbol{c})=(\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c} (abc)=(a×b)⋅c
混合积的几何意义
V 平行六面体 = ∣ ( a b c ) ∣ . V_{\text {平行六面体 }}=|(\boldsymbol{a} \boldsymbol{b} \boldsymbol{c})| . V平行六面体 =∣(abc)∣.
( 注 1 ) \Large\color{violet}{(注1) } (注1) ∣ ( a b c ) ∣ ≤ ∣ a ∣ ∣ b ∣ ∣ c ∣ |(\boldsymbol{a} \boldsymbol{b} \boldsymbol{c})| \leq|\boldsymbol{a}||\boldsymbol{b}||\boldsymbol{c}| ∣(abc)∣≤∣a∣∣b∣∣c∣.
( 注 2 ) \Large\color{violet}{(注2) } (注2) 以 a , b , c a, b, c a,b,c 为棱的四面体的体积
V 四面体 = 1 6 ∣ ( a b c ) ∣ = 1 6 ∣ ( a × b ) ⋅ c ∣ V_{\text {四面体 }}=\frac{1}{6}|(\boldsymbol{a} \boldsymbol{b} \boldsymbol{c})|=\frac{1}{6}|(\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}| V四面体 =61∣(abc)∣=61∣(a×b)⋅c∣
性质
(1) 轮换对称性 ( a b c ) = ( b c a ) = ( c a b ) . \quad(\boldsymbol{a} \boldsymbol{b} \boldsymbol{c})=(\boldsymbol{b} \boldsymbol{c} \boldsymbol{a})=(\boldsymbol{c} \boldsymbol{a} \boldsymbol{b}) . (abc)=(bca)=(cab).
(2) 反交换律 ( a b c ) = − ( b a c ) = − ( c b a ) . \quad(\boldsymbol{a} \boldsymbol{b} \boldsymbol{c})=-(\boldsymbol{b} \boldsymbol{a} \boldsymbol{c})=-(\boldsymbol{c} \boldsymbol{b} \boldsymbol{a}) . (abc)=−(bac)=−(cba).
(3) ( a × b ) ⋅ c = a ⋅ ( b × c ) (\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}=\boldsymbol{a} \cdot(\boldsymbol{b} \times \boldsymbol{c}) (a×b)⋅c=a⋅(b×c)
(4) 当 a , b , c \boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c} a,b,c 中有两个向量相同时，它们的混合积为 0. 0 . 0.
例 1 \Large\color{violet}{例1} 例1 证明拉格朗日恒等式：
( a × b ) ⋅ ( c × d ) = ∣ a ⋅ c a ⋅ d b ⋅ c b ⋅ d ∣ . \quad(\boldsymbol{a} \times \boldsymbol{b}) \cdot(\boldsymbol{c} \times \boldsymbol{d})=\left|\begin{array}{ll}\boldsymbol{a} \cdot \boldsymbol{c} & \boldsymbol{a} \cdot \boldsymbol{d} \\ \boldsymbol{b} \cdot \boldsymbol{c} & \boldsymbol{b} \cdot \boldsymbol{d}\end{array}\right| . (a×b)⋅(c×d)=∣∣∣∣a⋅cb⋅ca⋅db⋅d∣∣∣∣.
【证】 记 h = c × d \boldsymbol{h}=\boldsymbol{c} \times \boldsymbol{d} h=c×d, 则
( a × b ) ⋅ ( c × d ) = ( a × b ) ⋅ h = a ⋅ ( b × h ) (\boldsymbol{a} \times \boldsymbol{b}) \cdot(\boldsymbol{c} \times \boldsymbol{d})=(\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{h}=\boldsymbol{a} \cdot(\boldsymbol{b} \times \boldsymbol{h}) (a×b)⋅(c×d)=(a×b)⋅h=a⋅(b×h)
又 b × h = b × ( c × d ) = ( b ⋅ d ) c − ( b ⋅ c ) d \quad \boldsymbol{b} \times \boldsymbol{h}=\boldsymbol{b} \times(\boldsymbol{c} \times \boldsymbol{d})=(\boldsymbol{b} \cdot \boldsymbol{d}) \boldsymbol{c}-(\boldsymbol{b} \cdot \boldsymbol{c}) \boldsymbol{d} b×h=b×(c×d)=(b⋅d)c−(b⋅c)d, 故
( a × b ) ⋅ ( c × d ) = ( a ⋅ c ) ( b ⋅ d ) − ( a ⋅ d ) ( b ⋅ c ) = ∣ a ⋅ c a ⋅ d b ⋅ c b ⋅ d ∣ (\boldsymbol{a} \times \boldsymbol{b}) \cdot(\boldsymbol{c} \times \boldsymbol{d})=(\boldsymbol{a} \cdot \boldsymbol{c})(\boldsymbol{b} \cdot \boldsymbol{d})-(\boldsymbol{a} \cdot \boldsymbol{d})(\boldsymbol{b} \cdot \boldsymbol{c})=\left|\begin{array}{ll}\boldsymbol{a} \cdot \boldsymbol{c} & \boldsymbol{a} \cdot \boldsymbol{d} \\ \boldsymbol{b} \cdot \boldsymbol{c} & \boldsymbol{b} \cdot \boldsymbol{d}\end{array}\right| (a×b)⋅(c×d)=(a⋅c)(b⋅d)−(a⋅d)(b⋅c)=∣∣∣∣a⋅cb⋅ca⋅db⋅d∣∣∣∣
用坐标计算混合积
定 理 1 \large\color{magenta}{\boxed{\color{brown}{定理1} }} 定理1 设 a = ( a 1 , a 2 , a 3 ) , b = ( b 1 , b 2 , b 3 ) , c = ( c 1 , c 2 , c 3 ) \boldsymbol{a}=\left(a_{1}, a_{2}, a_{3}\right), \boldsymbol{b}=\left(b_{1}, b_{2}, b_{3}\right), \quad \boldsymbol{c}=\left(c_{1}, c_{2}, c_{3}\right) a=(a1,a2,a3),b=(b1,b2,b3),c=(c1,c2,c3), 有
( a × b ) ⋅ c = ( a 2 b 3 − a 3 b 2 ) c 1 − ( a 1 b 3 − a 3 b 1 ) c 2 + ( a 1 b 2 − a 2 b 1 ) c 3 (\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}=\left(a_{2} b_{3}-a_{3} b_{2}\right) c_{1}-\left(a_{1} b_{3}-a_{3} b_{1}\right) c_{2}+\left(a_{1} b_{2}-a_{2} b_{1}\right) c_{3} (a×b)⋅c=(a2b3−a3b2)c1−(a1b3−a3b1)c2+(a1b2−a2b1)c3
【证】 a × b = ( a 2 b 3 − a 3 b 2 ) i − ( a 1 b 3 − a 3 b 1 ) j + ( a 1 b 2 − a 2 b 1 ) k \boldsymbol{a} \times \boldsymbol{b}=\left(a_{2} b_{3}-a_{3} b_{2}\right) \boldsymbol{i}-\left(a_{1} b_{3}-a_{3} b_{1}\right) \boldsymbol{j}+\left(a_{1} b_{2}-a_{2} b_{1}\right) \boldsymbol{k} a×b=(a2b3−a3b2)i−(a1b3−a3b1)j+(a1b2−a2b1)k
故 ( a × b ) ⋅ c = ( a × b ) ⋅ ( c 1 i + c 2 j + c 3 k ) \quad(\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}=(\boldsymbol{a} \times \boldsymbol{b}) \cdot\left(c_{1} \boldsymbol{i}+c_{2} \boldsymbol{j}+c_{3} k\right) (a×b)⋅c=(a×b)⋅(c1i+c2j+c3k)
= ( a 2 b 3 − a 3 b 2 ) c 1 − ( a 1 b 3 − a 3 b 1 ) c 2 + ( a 1 b 2 − a 2 b 1 ) c 3 =\left(a_{2} b_{3}-a_{3} b_{2}\right) c_{1}-\left(a_{1} b_{3}-a_{3} b_{1}\right) c_{2}+\left(a_{1} b_{2}-a_{2} b_{1}\right) c_{3} =(a2b3−a3b2)c1−(a1b3−a3b1)c2+(a1b2−a2b1)c3
利用混合积的行列式计算公式，很容易推出混合积的性质
( a × b ) ⋅ c = ( a 2 b 3 − a 3 b 2 ) c 1 − ( a 1 b 3 − a 3 b 1 ) c 2 + ( a 1 b 2 − a 2 b 1 ) c 3 = ∣ a 2 a 3 b 2 b 3 ∣ c 1 − ∣ a 1 a 3 b 1 b 3 ∣ c 2 + ∣ a 1 a 2 b 1 b 2 ∣ c 3 = ∣ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ∣ \begin{aligned} (\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c} &=\left(a_{2} b_{3}-a_{3} b_{2}\right) c_{1}-\left(a_{1} b_{3}-a_{3} b_{1}\right) c_{2}+\left(a_{1} b_{2}-a_{2} b_{1}\right) c_{3} \\ &=\left|\begin{array}{ll} a_{2} & a_{3} \\ b_{2} & b_{3} \end{array}\right| c_{1}-\left|\begin{array}{ll} a_{1} & a_{3} \\ b_{1} & b_{3} \end{array}\right| c_{2}+\left|\begin{array}{ll} a_{1} & a_{2} \\ b_{1} & b_{2} \end{array}\right| c_{3} \\ &=\left|\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right| \end{aligned} (a×b)⋅c=(a2b3−a3b2)c1−(a1b3−a3b1)c2+(a1b2−a2b1)c3=∣∣∣∣a2b2a3b3∣∣∣∣c1−∣∣∣∣a1b1a3b3∣∣∣∣c2+∣∣∣∣a1b1a2b2∣∣∣∣c3=∣∣∣∣∣∣a1b1c1a2b2c2a3b3c3∣∣∣∣∣∣
( a × b ) ⋅ c = ∣ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ∣ (\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}=\left|\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3}\end{array}\right| (a×b)⋅c=∣∣∣∣∣∣a1b1c1a2b2c2a3b3c3∣∣∣∣∣∣
性质 \quad 对于向量 a , b , c , d \boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}, \boldsymbol{d} a,b,c,d 及实数 λ \lambda λ, 有
(1) ( a b c + d ) = ( a b c ) + ( a b d ) (\boldsymbol{a} \boldsymbol{b} \boldsymbol{c}+\boldsymbol{d})=(\boldsymbol{a} \boldsymbol{b} \boldsymbol{c})+(\boldsymbol{a} \boldsymbol{b} \boldsymbol{d}) (abc+d)=(abc)+(abd).
(2) ( λ a b c ) = λ ( a b c ) (\lambda \boldsymbol{a} \boldsymbol{b} \boldsymbol{c})=\lambda(\boldsymbol{a} \boldsymbol{b} \boldsymbol{c}) (λabc)=λ(abc)
例 2 \Large\color{violet}{例2} 例2 化简混合积 ( a + b b + c c + a ) (a+b b+c c+a) (a+bb+cc+a)
( a + b b + c c + a ) = ( a + b b + c c ) + ( a + b b + c a ) = ( a + b b c ) + ( a + b c c ) + ( a + b b a ) + ( a + b c a ) = ( a b c ) + ( b b c ) + 0 + ( a b a ) + ( b b a ) + ( a c a ) + ( b c a ) = ( a b c ) + 0 + 0 + 0 + 0 + 0 + ( b c a ) = 2 ( a b c ) . \begin{aligned} (a+&b b+c c+a)=(a+b b+c c)+(a+b b+c a) \\ &=(a+b b c)+(a+b c c)+(a+b b a)+(a+b c a) \\ &=(a b c)+(b b c)+0+(a b a)+(b b a)+(a c a)+(b c a) \\ &=(a b c)+0+0+0+0+0+(b c a) \\ &=2(a b c) . \end{aligned} (a+bb+cc+a)=(a+bb+cc)+(a+bb+ca)=(a+bbc)+(a+bcc)+(a+bba)+(a+bca)=(abc)+(bbc)+0+(aba)+(bba)+(aca)+(bca)=(abc)+0+0+0+0+0+(bca)=2(abc).
三向量共面的条件
由混合积的几何意义 V = ∣ ( a b c ) ∣ V=|(\boldsymbol{a} \boldsymbol{b} \boldsymbol{c})| V=∣(abc)∣ 知,三向量 a , b , c \boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c} a,b,c 共面的充分必要条件是它们的混合积为 0, 即
( a b c ) = ∣ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ∣ = 0 (\boldsymbol{a} \boldsymbol{b} \boldsymbol{c})=\left|\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right|=0 (abc)=∣∣∣∣∣∣a1b1c1a2b2c2a3b3c3∣∣∣∣∣∣=0
因 a × b a \times b a×b 同时垂直向量 a , b \boldsymbol{a}, \boldsymbol{b} a,b, 则 a × b \boldsymbol{a} \times \boldsymbol{b} a×b 垂直 a , b \boldsymbol{a}, \boldsymbol{b} a,b 所在的平面 π \pi π.所以，三向量 a , b , c \boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c} a,b,c 共面的充要条件是 a × b \boldsymbol{a} \times \boldsymbol{b} a×b 垂直 c \boldsymbol{c} c, 即
( a × b ) ⋅ c = ( a b c ) = ∣ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ∣ = 0 (\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}=(\boldsymbol{a} \boldsymbol{b} \boldsymbol{c})=\left|\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right|=0 (a×b)⋅c=(abc)=∣∣∣∣∣∣a1b1c1a2b2c2a3b3c3∣∣∣∣∣∣=0
从线性代数的角度理解
三向量 a , b , c a, b, c a,b,c 共面的充要条件是 a , b , c a, b, c a,b,c 线性相关，即存在不全为0的实数 k 1 , k 2 , k 3 k_{1}, k_{2}, k_{3} k1,k2,k3, 使得 k 1 a + k 2 b + k 3 c = 0 k_{1} a+k_{2} b+k_{3} c=0 k1a+k2b+k3c=0, 即该关于 k 1 , k 2 , k 3 k_{1}, k_{2}, k_{3} k1,k2,k3 的方程组有非零解，其充要条件是系数行列式为0, 即
∣ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ∣ = 0. \left|\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right|=0 . ∣∣∣∣∣∣a1b1c1a2b2c2a3b3c3∣∣∣∣∣∣=0.
例 3 \Large\color{violet}{例3} 例3 设空间直角坐标系中四点 A , B , C , D A, B, C, D A,B,C,D 的坐标分别为
A = ( 1 , 1 , 2 ) , B = ( − 1 , − 2 , 0 ) , C = ( 1 , − 1 , 1 ) , D = ( 3 , 0 , 2 ) A=(1,1,2), B=(-1,-2,0), C=(1,-1,1), D=(3,0,2) A=(1,1,2),B=(−1,−2,0),C=(1,−1,1),D=(3,0,2)
试证点A, B , C B, C B,C, D D D 共面.
【证】
A B → = ( − 2 , − 3 , − 2 ) , A C → = ( 0 , − 2 , − 1 ) , A D → = ( 2 , − 1 , 0 ) \overrightarrow{A B}=(-2,-3,-2), \quad \overrightarrow{A C}=(0,-2,-1), \quad \overrightarrow{A D}=(2,-1,0) AB
=(−2,−3,−2),AC
=(0,−2,−1),AD
=(2,−1,0)
则有 ( A B → × A C → ) ⋅ A D → = ∣ − 2 − 3 − 2 0 − 2 − 1 2 − 1 0 ∣ = ∣ 0 − 4 − 2 0 − 2 − 1 2 − 1 0 ∣ = 0 \quad(\overrightarrow{A B} \times \overrightarrow{A C}) \cdot \overrightarrow{A D}=\left|\begin{array}{ccc}-2 & -3 & -2 \\ 0 & -2 & -1 \\ 2 & -1 & 0\end{array}\right|=\left|\begin{array}{ccc}0 & -4 & -2 \\ 0 & -2 & -1 \\ 2 & -1 & 0\end{array}\right|=0 (AB
×AC
)⋅AD
=∣∣∣∣∣∣−202−3−2−1−2−10∣∣∣∣∣∣=∣∣∣∣∣∣002−4−2−1−2−10∣∣∣∣∣∣=0,
所以 A B → , A C → , A D → \overrightarrow{A B}, \overrightarrow{A C}, \overrightarrow{A D} AB
,AC
,AD
共面, 即点 A , B , C , D A, B, C, D A,B,C,D 共面.
参考
空间解析几何_国防科技大学
北京大学数学系几何与代数教研室代数小组，《高等代数》（第四版）
高等代数，林亚南，高等教育出版社
高等代数学习辅导，林亚南，林鹭，杜妮，陈清华，高等教育出版社
高等代数 电子科技大学
高等代数_安阳师范学院
《高等代数》（第五版）