人工智能教程 - 数学基础课程1.2 - 数学分析（二）-2

行列式和叉积

点积应用

3)向量A沿着某单位向量u的分量(components of $\overrightarrow{A}$Aalong a direction $\widehat{u}$u)

component of A along $\widehat{u}$u = $|\overrightarrow{A}|$∣A∣cos($\Theta$Θ)

=$|\overrightarrow{A}||\widehat{u}|.cos\Theta$∣A∣∣u∣.cosΘ
=$\overrightarrow{A}.\widehat{u}$A.u

4)物理

F沿切向方向的分量使摆来回摆动

• component of F along the target direction $\widehat{T}$T is what causes the pendulum to swing
• component of $\widehat{N}$N

Ex:Area?

Easier:area of triangle

we could find $cos\Theta$cosΘ,the solve $sin^2s\Theta+cos^2\Theta=1$sin2sΘ+cos2Θ=1

more esaier:行列式(determinants)

$\overrightarrow{A}'=\overrightarrow{A}$A′=A rotated $90^o$90o

$\Theta '=\frac{\pi}{2}-\Theta$Θ′=2π​−Θ

$cos(\Theta ')=sin(\Theta)$cos(Θ′)=sin(Θ)

$|\overrightarrow{A}|.|\overrightarrow{B}|.sin\Theta$∣A∣.∣B∣.sinΘ
$=|\overrightarrow{A}'|.|\overrightarrow{B}|.cos(\Theta')$=∣A′∣.∣B∣.cos(Θ′)
$=\overrightarrow{A}'.\overrightarrow{B}$=A′.B

$\overrightarrow{A}=<a_1,a_2>$A=<a1​,a2​>
$\overrightarrow{A}'=<-a_2,a_1>$A′=<−a2​,a1​>

$\overrightarrow{A}'.\overrightarrow{B}=<-a_2,a_1>.<b_1,b_2>$A′.B=<−a2​,a1​>.<b1​,b2​>
$=a_1.b_2-a_2.b_1$=a1​.b2​−a2​.b1​
$=det(\overrightarrow{A},\overrightarrow{B})$=det(A,B)
$=\begin{vmatrix} a_1 & a_2\\ b_1& b_2 \end{vmatrix}$=∣∣∣∣​a1​b1​​a2​b2​​∣∣∣∣​

So,determinant of $\overrightarrow{A}$A and $\overrightarrow{B}$B = area

空间中的行列式(Determinant in space)

3 vectors $\overrightarrow{A},\overrightarrow{B},\overrightarrow{C}$A,B,C:
$det(\overrightarrow{A},\overrightarrow{B},\overrightarrow{C})=\begin{vmatrix} a_1 & a_2&a_3\\ b_1& b_2&b_3\\ c_1& c_2&c_3 \end{vmatrix}$det(A,B,C)=∣∣∣∣∣∣​a1​b1​c1​​a2​b2​c2​​a3​b3​c3​​∣∣∣∣∣∣​
$=a_1\begin{vmatrix} b_2&b_3\\ c_2&c_3 \end{vmatrix}-a_2\begin{vmatrix} b_1&b_3\\ c_1 & c_3 \end{vmatrix}+a_3\begin{vmatrix} b_1&b_2\\ c_1 & c_2 \end{vmatrix}$=a1​∣∣∣∣​b2​c2​​b3​c3​​∣∣∣∣​−a2​∣∣∣∣​b1​c1​​b3​c3​​∣∣∣∣​+a3​∣∣∣∣​b1​c1​​b2​c2​​∣∣∣∣​

note:竖线里(insides verticals bars)

定理(Theorem)

几何学上(Geometrically)，$det(\overrightarrow{A},\overrightarrow{B},\overrightarrow{C})=\pm$det(A,B,C)=± 正负平行六面体的面积(volume of the parallelepiped)
)

叉积(cross product)

of 2 vectors in 3-space

def:$\overrightarrow{A}$AX$\overrightarrow{B}$B= is a vector,not dot product

$\overrightarrow{A}$AX$\overrightarrow{B}=\begin{vmatrix} \widehat{i}& \widehat{j}&\widehat{k}\\ a_1 & a_2&a_3\\ b_1& b_2&b_3 \end{vmatrix}$B=∣∣∣∣∣∣​ia1​b1​​j​a2​b2​​ka3​b3​​∣∣∣∣∣∣​

$=\begin{vmatrix} a_2&a_3\\ b_2&b_3 \end{vmatrix}\widehat{i}-\begin{vmatrix} a_1&a_3\\ b_1&b_3 \end{vmatrix}\widehat{j}+\begin{vmatrix} a_1&a_2\\ b_1&b_2 \end{vmatrix}\widehat{k}$=∣∣∣∣​a2​b2​​a3​b3​​∣∣∣∣​i−∣∣∣∣​a1​b1​​a3​b3​​∣∣∣∣​j​+∣∣∣∣​a1​b1​​a2​b2​​∣∣∣∣​k

Theorem:

• $|\overrightarrow{A}$∣AX$\overrightarrow{B}|$B∣=area of parallelogram

叉积的模长(length)

• dir($\overrightarrow{A}$AX$\overrightarrow{B})\perp$B)⊥to plane of a parallelogram（垂直于平行四边形所在的平面）

)

使用右手定则判断垂直的方向(with right-hand rule)

• right hand points parallel to vector A (//$\overrightarrow{A}$A)
• Fingers point //$\overrightarrow{B}$B
• thumb points //$\overrightarrow{A}$AX$\overrightarrow{B}$B

Another look at volume
)
Volume = area(base) X height
=$|\overrightarrow{B}$∣BX$\overrightarrow{C}|$C∣.($|\overrightarrow{A}$∣A.$\widehat{n}$n)

$det(\overrightarrow{A},\overrightarrow{B},\overrightarrow{C})=\overrightarrow{A}.(\overrightarrow{B}$det(A,B,C)=A.(B X $\overrightarrow{C})$C)

$=a_1\begin{vmatrix} b_2&b_3\\ c_2&c_3 \end{vmatrix}-a_2\begin{vmatrix} b_1&b_3\\ c_1 & c_3 \end{vmatrix}+a_3\begin{vmatrix} b_1&b_2\\ c_1 & c_2 \end{vmatrix}$=a1​∣∣∣∣​b2​c2​​b3​c3​​∣∣∣∣​−a2​∣∣∣∣​b1​c1​​b3​c3​​∣∣∣∣​+a3​∣∣∣∣​b1​c1​​b2​c2​​∣∣∣∣​