设有两个向量 a ⃗ , b ⃗ , 两向量夹角为 θ \vec a,\vec b,两向量夹角为\theta a,b,两向量夹角为θ,称数
∣ a ⃗ ∣ ∣ b ⃗ ∣ cos θ |\vec a||\vec b|\cos \theta ∣a∣∣b∣cosθ
叫做向量 a ⃗ , b ⃗ \vec a,\vec b a,b的数量积,记作 a ⃗ ⋅ b ⃗ \vec a\cdot\vec b a⋅b,即
a ⃗ ⋅ b ⃗ = ∣ a ⃗ ∣ ∣ b ⃗ ∣ cos θ \vec a\cdot\vec b=|\vec a||\vec b|\cos\theta a⋅b=∣a∣∣b∣cosθ
说明:
F ⃗ 沿直线位移 s ⃗ \vec F沿直线位移\vec s F沿直线位移s所做的功为: W = F ⃗ ⋅ s ⃗ W = \vec F\cdot\vec s W=F⋅s
a ⃗ ⋅ 0 ⃗ = 0 \vec a \cdot \vec0=0 a⋅0=0
投影的数量积表示
a ⃗ ⋅ b ⃗ = ∣ a ⃗ ∣ ⋅ ∣ b ⃗ ∣ ⋅ cos θ = ∣ a ⃗ ∣ ⋅ P i j a ⃗ b ⃗ ∴ P i j a ⃗ b ⃗ = a ⃗ ⋅ b ⃗ ∣ a ⃗ ∣ = e ⃗ a ⃗ ⋅ ⃗ b \vec a \cdot\vec b=|\vec a|\cdot|\vec b|\cdot\cos\theta\\ =|\vec a|\cdot Pij_{\vec a}\vec b \\ \therefore Pij_{\vec a}\vec b = \frac{\vec a\cdot\vec b}{|\vec a|}=\vec e_{\vec a}\vec\cdot b a⋅b=∣a∣⋅∣b∣⋅cosθ=∣a∣⋅Pijab∴Pijab=∣a∣a⋅b=ea⋅b
两个重要的推论:
注:
设 a ⃗ = ( x 1 , y 1 , z 1 ) , b ⃗ = ( x 2 , y 2 , z 2 ) \vec a=(x_1,y_1,z_1),\vec b=(x_2,y_2,z_2) a=(x1,y1,z1),b=(x2,y2,z2),则
例1 已知三点 M ( 1 , 1 , 1 ) , A ( 2 , 2 , 1 ) , B ( 2 , 1 , 2 ) ,求 ∠ A M B M(1,1,1),A(2,2,1),B(2,1,2),求\angle{AMB} M(1,1,1),A(2,2,1),B(2,1,2),求∠AMB
解: M A ⃗ = ( 1 , 1 , 0 ) , M B ⃗ = ( 1 , 0 , 1 ) ∠ A M B = M A ⃗ ⋅ M B ⃗ ∣ M A ⃗ ∣ ∣ M B ⃗ ∣ = 1 2 ⋅ 2 = 1 2 ∴ ∠ A M B = π 3 解:\vec{MA}=(1,1,0),\vec{MB}=(1,0,1)\\ \angle{AMB}=\frac{\vec{MA}\cdot\vec{MB}}{|\vec{MA}||\vec{MB}|}\\ =\frac{1}{\sqrt2\cdot\sqrt2}=\frac{1}{2}\\ \therefore \angle{AMB}=\frac{\pi}{3} 解:MA=(1,1,0),MB=(1,0,1)∠AMB=∣MA∣∣MB∣MA⋅MB=2⋅21=21∴∠AMB=3π
例2 已知 a ⃗ = ( 2 , 2 , 1 ) , b ⃗ = ( 3 , 4 , − 2 ) ,求 P i j a ⃗ b ⃗ \vec a=(2,2,1),\vec b=(3,4,-2),求Pij_{\vec a}\vec b a=(2,2,1),b=(3,4,−2),求Pijab
解: ∣ a ⃗ ∣ = 3 , e ⃗ a ⃗ = ( 2 3 , 2 3 , 1 3 ) ∴ P i j a ⃗ b ⃗ = e ⃗ a ⃗ ⋅ b ⃗ = 2 + 8 3 − 2 3 = 4 解:|\vec a|=3,\vec e_{\vec a}=(\frac{2}{3},\frac{2}{3},\frac{1}{3})\\ \therefore Pij_{\vec a}\vec b=\vec e_{\vec a}\cdot \vec b=2+\frac{8}{3}-\frac{2}{3}=4 解:∣a∣=3,ea=(32,32,31)∴Pijab=ea⋅b=2+38−32=4
设向量 c ⃗ \vec c c有两个向量 a ⃗ , b ⃗ ,夹角为 θ \vec a,\vec b,夹角为\theta a,b,夹角为θ按下列方式定出:
∣ c ⃗ ∣ = ∣ a ⃗ ∣ ∣ b ⃗ ∣ sin θ |\vec c| = |\vec a||\vec b|\sin\theta ∣c∣=∣a∣∣b∣sinθ; c ⃗ \vec c c的方向垂直于 a ⃗ 于 b ⃗ \vec a于\vec b a于b所决定的平面,方向按右手规则从 a ⃗ \vec a a转向 b ⃗ \vec b b确定。向量 c ⃗ \vec c c叫做向量 a ⃗ 与 b ⃗ \vec a与\vec b a与b的向量积,记作 a ⃗ × b ⃗ \vec a\times \vec b a×b.
注:右手规则,如下图2.1-1所示:
四指握的方向为 a ⃗ 到 b ⃗ \vec a到\vec b a到b的转向,大拇指指的方向即为向量积的方向。
注:
a ⃗ × a ⃗ = 0 ⃗ \vec a\times \vec a=\vec0 a×a=0
两非零向量 a ⃗ , b ⃗ \vec a,\vec b a,b,则 a ⃗ ∥ b ⃗ ⇔ a ⃗ × b ⃗ = 0 ⃗ \vec a\parallel \vec b\Leftrightarrow \vec a\times\vec b=\vec0 a∥b⇔a×b=0
如下图2.3-1所示:
以 a ⃗ , b ⃗ \vec a,\vec b a,b为邻边做平行四边形,则 ∣ a ⃗ × b ⃗ ∣ = ∣ a ⃗ ∣ ⋅ ∣ b ⃗ ∣ sin θ = S 平行四边形 |\vec a\times\vec b|=|\vec a|\cdot|\vec b|\sin\theta=S_{平行四边形} ∣a×b∣=∣a∣⋅∣b∣sinθ=S平行四边形,即 ∣ a ⃗ × b ⃗ ∣ |\vec a\times\vec b| ∣a×b∣表示以 a ⃗ , b ⃗ \vec a,\vec b a,b为邻边的平行四边形的面积。
设 a ⃗ = ( x 1 , y 1 , z 1 ) , b ⃗ = ( x 2 , y 2 , z 2 ) \vec a=(x_1,y_1,z_1),\vec b=(x_2,y_2,z_2) a=(x1,y1,z1),b=(x2,y2,z2),则
a ⃗ × b ⃗ = ( x 1 i ⃗ + y 1 j ⃗ + z 1 k ⃗ ) × ( x 2 i ⃗ + y 2 j ⃗ + z 2 k ⃗ ) = ( y 1 z 2 − z 1 y 2 ) i ⃗ + ( x 1 z 2 − z 1 x 2 ) j ⃗ + ( x 1 y 2 − y 1 x 2 ) k ⃗ \vec a\times\vec b=(x_1\vec i+y_1\vec j+z_1\vec k)\times(x_2\vec i+y_2\vec j+z_2\vec k)\\ =(y_1z_2-z_1y_2)\vec i+(x_1z_2-z_1x_2)\vec j+(x_1y_2-y_1x_2)\vec k a×b=(x1i+y1j+z1k)×(x2i+y2j+z2k)=(y1z2−z1y2)i+(x1z2−z1x2)j+(x1y2−y1x2)k
用三阶行列式表示为:
a ⃗ × b ⃗ = ∣ i ⃗ j ⃗ k ⃗ x 1 y 1 z 1 x 2 y 2 z 2 ∣ = i ⃗ ⋅ ∣ y 1 z 1 y 2 z 2 ∣ − j ⃗ ∣ x 1 z 1 x 2 z 2 ∣ + k ⃗ ⋅ ∣ x 1 y 1 x 2 y 2 ∣ = ( y 1 z 2 − z 1 y 2 ) ⋅ i ⃗ − ( x 1 z 2 − z 1 x 2 ) ⋅ j ⃗ + ( x 1 y 2 − y 1 x 2 ) ⋅ k \vec a\times\vec b= \begin{vmatrix} \vec i&\vec j&\vec k\\ x_1&y_1&z_1\\ x_2&y_2&z_2 \end{vmatrix}\\ =\vec i\cdot \begin{vmatrix} y_1&z_1\\ y_2&z_2\\ \end{vmatrix} - \vec j \begin{vmatrix} x_1&z_1\\ x_2&z_2 \end{vmatrix}+\vec k\cdot \begin{vmatrix} x_1&y_1\\ x_2&y_2 \end{vmatrix}\\ =(y_1z_2-z_1y_2)\cdot\vec i-(x_1z_2-z_1x_2)\cdot\vec j+(x_1y_2-y_1x_2)\cdot k a×b= ix1x2jy1y2kz1z2 =i⋅ y1y2z1z2 −j x1x2z1z2 +k⋅ x1x2y1y2 =(y1z2−z1y2)⋅i−(x1z2−z1x2)⋅j+(x1y2−y1x2)⋅k
例4 已知 c ⃗ ⊥ a ⃗ , 且 c ⃗ ⊥ b ⃗ . ∣ c ⃗ ∣ = 3 ,求 c ⃗ \vec c \perp \vec a,且\vec c \perp \vec b. |\vec c|=3,求\vec c c⊥a,且c⊥b.∣c∣=3,求c.其中, a ⃗ = ( 2 , 1 , − 1 ) , b ⃗ = ( 1 , − 1 , 2 ) \vec a=(2,1,-1),\vec b=(1,-1,2) a=(2,1,−1),b=(1,−1,2)
解: a ⃗ × b ⃗ = ∣ i ⃗ j ⃗ k ⃗ 2 1 − 1 1 − 1 2 ∣ = i ⃗ − 5 j ⃗ + ( − 3 ) k ⃗ e a ⃗ × b ⃗ = ( 1 35 , − 5 35 , − 3 35 ) ∴ c ⃗ = ± 3 e a ⃗ × b ⃗ = ( ± 3 35 , ± 15 35 , ± 9 35 ) 解:\vec a\times\vec b= \begin{vmatrix} \vec i&\vec j&\vec k\\ 2&1&-1\\ 1&-1&2\\ \end{vmatrix}\\ =\vec i-5\vec j+(-3)\vec k\\ e_{\vec a\times\vec b}=(\frac{1}{\sqrt{35}},\frac{-5}{\sqrt{35}},\frac{-3}{\sqrt{35}})\\ \therefore \vec c=\pm3e_{\vec a\times \vec b}=(\pm\frac{3}{\sqrt{35}},\pm\frac{15}{\sqrt{35}},\pm\frac{9}{\sqrt{35}}) 解:a×b= i21j1−1k−12 =i−5j+(−3)kea×b=(351,35−5,35−3)∴c=±3ea×b=(±353,±3515,±359)
例5 已知 A ( 1 , 2 , 3 ) , B ( 3 , 4 , 5 ) , C ( 2 , 4 , 7 ) , 求 S △ A B C A(1,2,3),B(3,4,5),C(2,4,7),求S_{\triangle{ABC}} A(1,2,3),B(3,4,5),C(2,4,7),求S△ABC
S △ A B C = 1 2 ∣ A B ⃗ × A C ⃗ ∣ = 1 2 ∥ i ⃗ j ⃗ k ⃗ 2 2 2 1 2 4 ∥ = 1 2 4 2 + ( − 6 ) 2 + 2 2 = 14 S_{\triangle{ABC}}=\frac{1}{2}|\vec{AB}\times\vec{AC}|\\ =\frac{1}{2} \begin{Vmatrix} \vec i&\vec j&\vec k\\ 2&2&2\\ 1&2&4\\ \end{Vmatrix}\\ =\frac{1}{2}\sqrt{4^2+(-6)^2+2^2}=\sqrt{14} S△ABC=21∣AB×AC∣=21 i21j22k24 =2142+(−6)2+22=14
设已知三个向量 a ⃗ , b ⃗ , c ⃗ \vec a,\vec b,\vec c a,b,c,先做向量 a ⃗ 和 b ⃗ 的向量积 a ⃗ × b ⃗ \vec a和\vec b的向量积\vec a\times\vec b a和b的向量积a×b,把得到的向量与 c ⃗ 做数量积 ( a ⃗ × b ⃗ ) ⋅ c ⃗ \vec c做数量积(\vec a\times\vec b)\cdot\vec c c做数量积(a×b)⋅c,这样得到的数量叫做三向量的混合积,记作 [ a ⃗ b ⃗ c ⃗ ] [\vec a\vec b\vec c] [abc]
设 a ⃗ = ( x 1 , y 1 , z 1 ) , b ⃗ = ( x 2 , y 2 , z 2 ) , c ⃗ = ( x 3 , y 3 , z 3 ) \vec a=(x_1,y_1,z_1),\vec b=(x_2,y_2,z_2),\vec c=(x_3,y_3,z_3) a=(x1,y1,z1),b=(x2,y2,z2),c=(x3,y3,z3),则
[ a ⃗ b ⃗ c ⃗ ] = ( a ⃗ × b ⃗ ) ⋅ c ⃗ = ∣ x 1 y 1 z 1 x 2 y 2 z 2 x 3 y 3 z 3 ∣ [\vec a\vec b\vec c]=(\vec a\times\vec b)\cdot\vec c=\\ \begin{vmatrix} x_1&y_1&z_1\\ x_2&y_2&z_2\\ x_3&y_3&z_3\\ \end{vmatrix} [abc]=(a×b)⋅c= x1x2x3y1y2y3z1z2z3
如上图所示向量积的混合积绝对值表示以 a ⃗ , b ⃗ , c ⃗ \vec a,\vec b,\vec c a,b,c为棱的平行六面体的体积。
a ⃗ , b ⃗ , c ⃗ 共面 ⇔ [ a ⃗ b ⃗ c ⃗ ] = 0 \vec a,\vec b,\vec c共面\Leftrightarrow [\vec a\vec b\vec c]=0 a,b,c共面⇔[abc]=0
例i6 已知 ( a ⃗ × b ⃗ ) ⋅ c ⃗ = 2 , 则 [ ( a ⃗ + b ⃗ ) × ( b ⃗ + c ⃗ ) ] ⋅ ( c ⃗ + a ⃗ ) = ? (\vec a\times\vec b)\cdot\vec c=2,则[(\vec a+\vec b)\times(\vec b+\vec c)]\cdot(\vec c+\vec a)=? (a×b)⋅c=2,则[(a+b)×(b+c)]⋅(c+a)=?
解: [ ( a ⃗ + b ⃗ ) × ( b ⃗ + c ⃗ ) ] ⋅ ( c ⃗ + a ⃗ ) = ( a ⃗ × b ⃗ + a ⃗ × c ⃗ + b ⃗ × b ⃗ + b ⃗ × c ⃗ ) ⋅ ( c ⃗ + a ⃗ ) = [ a ⃗ b ⃗ c ⃗ ] + [ a ⃗ b ⃗ a ⃗ ] + [ a ⃗ c ⃗ c ⃗ ] + [ a ⃗ c ⃗ a ⃗ ] + [ b ⃗ c ⃗ c ⃗ ] + [ b ⃗ c ⃗ a ⃗ ] = 2 [ a ⃗ b ⃗ c ⃗ ] = 4 解:\\ [(\vec a+\vec b)\times(\vec b+\vec c)]\cdot(\vec c+\vec a)=\\ (\vec a\times\vec b+\vec a\times\vec c+\vec b\times\vec b+\vec b\times\vec c)\cdot(\vec c+\vec a)\\ =[\vec a\vec b\vec c]+[\vec a\vec b\vec a]+[\vec a\vec c\vec c]+[\vec a\vec c\vec a]+[\vec b\vec c\vec c]+[\vec b\vec c\vec a]=2[\vec a\vec b\vec c]=4 解:[(a+b)×(b+c)]⋅(c+a)=(a×b+a×c+b×b+b×c)⋅(c+a)=[abc]+[aba]+[acc]+[aca]+[bcc]+[bca]=2[abc]=4
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参考:
[1]同济大学数学系.高等数学 第七版 下册[M].北京:高等教育出版社,2014.7.p14-22.
[2]同济七版《高等数学》全程教学视频[CP/OL].2020-04-16.p52.