matlab 向量旋转

A = rand(2, 10)% 2*10

modA = hypot(A(1,:), A(2,:))

direction = pi/4% "已知方向"，这里假设方位角是π/4

D = [cos(direction)sin(direction)]

B = [modA*D(1)modA*D(2)]% 2*10

旋转矩阵为T，

T=[cos(a) sin(a)

-sin(a) cos(a)]

被旋转向量为A，

则TA即可。

注：a 的单位为弧度。

引入角度alpha和半径R，假设曲线z = f(R)是绕z轴旋转: clc,clear alpha=0:pi/50:2*pi%角度[0,2*pi] R=0:0.1:10%半径 x = R.*cos(alpha)y = R.*sin(alpha)[X,Y] = meshgrid(x,y)z = f(sqrt(X.^2+Y.^2))%z = f(R)，R^2 = x^2 + y^2 me.

---