用matlab完成问题:x为3行2列的矩阵,由命令x=randi (100,3,2)随机生成。 (1)以x每行数据为圆心,在xy直角坐标系中画出3个半径为 1的圆,x和y 轴的范围均为[-10,110]; (2)在圆内按x的行索引分别标注编号1、2、3; (3)由编号1的圆圈向编号⒉圆圈画一个黑色箭头,箭头两端要落在圆圈上, 箭头所在直线应经过圆心,箭头大小r=1、角度o= 15° (见下图)﹔ (4)同(3),由编号2(和3)圆圈向编号3(和1)圆圈画箭头
时间: 2023-02-27 10:35:02 浏览: 77
Matlab代码如下:
1. 随机生成矩阵x:
```
x = randi(100,3,2);
```
2. 画出3个半径为1的圆:
```
t = linspace(0,2*pi,100);
for i = 1:3
x_c = x(i,1);
y_c = x(i,2);
x_circle = x_c + cos(t);
y_circle = y_c + sin(t);
plot(x_circle, y_circle);
hold on;
end
axis([-10 110 -10 110]);
```
3. 在圆内按x的行索引分别标注编号1、2、3:
```
for i = 1:3
text(x(i,1), x(i,2), num2str(i), 'HorizontalAlignment', 'center', 'VerticalAlignment', 'middle');
end
```
4. 画出箭头:
```
for i = 1:3
for j = i+1:3
x_c1 = x(i,1);
y_c1 = x(i,2);
x_c2 = x(j,1);
y_c2 = x(j,2);
x_arrow = [x_c1 x_c2];
y_arrow = [y_c1 y_c2];
plot(x_arrow, y_arrow, 'k', 'LineWidth', 2);
hold on;
arrow_angle = atan2(y_c2 - y_c1, x_c2 - x_c1);
x_head = x_c2 - 1 * cos(arrow_angle - pi / 6);
y_head = y_c2 - 1 * sin(arrow_angle - pi / 6);
plot([x_c2 x_head], [y_c2 y_head], 'k', 'LineWidth', 2);
x_head = x_c2 - 1 * cos(arrow_angle + pi / 6);
y_head = y_c2 - 1 * sin(arrow_angle + pi / 6);
plot([x_c2 x_head], [y_c2 y_head], 'k', 'LineWidth', 2);
end
end
```