作业帮设L为过原点的直线,OX轴到直线L的角为30度,A是以直线L为反射轴的反射变换,求变换A的矩阵
来源：学生作业帮 编辑：作业帮 分类：英语作业 时间：2024/05/18 02:37:39
设L为过原点的直线,OX轴到直线L的角为30度,A是以直线L为反射轴的反射变换,求变换A的矩阵
Reflection:Let V be a Euclidean space and v∈V be a unit vector.Define Aa:V→V
a|→a-2(a,v)v
It is called a reflection with respect to the subspace L(v)⊥.
According to your question:∠lOx=30.
So the unit vector v=(√3/2 ,1/2).
Now we take any vector a such that
A(a)=a-2(a,v)v.
The responding matrix is clearly soluted.
For example,suppose the dimmension is 2.
we let a=(x,y)
A(a)=(√3/2 ,1/2)-(√3x+y)(x,y)
=.
=A*(x,y)
where A is the matrix.
you can compute easily.
If the dimension is 3,you can suppose a=(x,y,z)
and comput
A(a)=A*(x,y,z)
Here A is a 3*3 matrix.
a|→a-2(a,v)v
It is called a reflection with respect to the subspace L(v)⊥.
According to your question:∠lOx=30.
So the unit vector v=(√3/2 ,1/2).
Now we take any vector a such that
A(a)=a-2(a,v)v.
The responding matrix is clearly soluted.
For example,suppose the dimmension is 2.
we let a=(x,y)
A(a)=(√3/2 ,1/2)-(√3x+y)(x,y)
=.
=A*(x,y)
where A is the matrix.
you can compute easily.
If the dimension is 3,you can suppose a=(x,y,z)
and comput
A(a)=A*(x,y,z)
Here A is a 3*3 matrix.
设L为过原点的直线,OX轴到直线L的角为30度,A是以直线L为反射轴的反射变换,求变换A的矩阵
已知点A(2,0)到直线l的距离为根号3,且直线l过原点,求直线l的方程
已知直线l过点A(0,根号10) ,且原点O到直线l的距离为根号5,求直线l的方程.
若原点在直线L上的射影为(a,b),求直线L的方程.
一道关于矩阵的题!若矩阵〔a 0 〕把直线L：2x+y-7=0变换为直线L':9x+y-91=0,试求a,-1 b
设直线L过点（0,0）,倾斜角为a,如果L绕坐标原点按逆时针方向旋转四十五度,得到L1,求L1的方程
直线l过点A（2,1）,且点B（1,3）到直线l的距离为1,求直线l的方程.
已知直线l过点A(-2,3),且点B(1-1)到该直线l的距离为3,求直线l的方程
直线l在y轴上的截距为10,且原点到直线的距离为8,求直线l的方程
直线l在x轴上的截距为10且原点到直线的距离为6,求直线l的方程
直线l在X轴上的截距为10,且原点到直线l的距离为6,求l方程!
直线l经过点A(5,10),切原点到它的距离为5,求直线方程