早教吧作业答案频道 -->其他-->
MATLAB已知3点求夹角>>x1=1;y1=1;x2=0;y2=0;x3=0;y3=3;theta=acosd(dot([x1-x2,y1-y2],[x3-x2,y3-y2])/(norm([x1-x2,y1-y2])*norm([x1-x2,y3-y2])))theta=47.8696>>x1=1;y1=1;x2=0;y2=0;x3=0;y3=3;theta=acos(dot([x1-x2,y1-y2],[x3-x2,y3-y2])/(norm([x1-x2,
题目详情
MATLAB 已知3点求夹角
>> x1=1;y1=1;x2=0;y2=0;x3=0;y3=3;
theta=acosd(dot([x1-x2,y1-y2],[x3-x2,y3-y2])/(norm([x1-x2,y1-y2])*norm([x1-x2,y3-y2])))
theta =
47.8696
>>
x1=1;y1=1;x2=0;y2=0;x3=0;y3=3;
theta=acos(dot([x1-x2,y1-y2],[x3-x2,y3-y2])/(norm([x1-x2,y1-y2])*norm([x1-x2,y3-y2])))
theta =
0.8355
>> pi/4
ans =
0.7854
为什么上面的3个点明显夹角是45度,但是算出来确实 47.8696呢
问题已解决来人送分
>> x1=1;y1=1;x2=0;y2=0;x3=0;y3=3;
theta=acosd(dot([x1-x2,y1-y2],[x3-x2,y3-y2])/(norm([x1-x2,y1-y2])*norm([x1-x2,y3-y2])))
theta =
47.8696
>>
x1=1;y1=1;x2=0;y2=0;x3=0;y3=3;
theta=acos(dot([x1-x2,y1-y2],[x3-x2,y3-y2])/(norm([x1-x2,y1-y2])*norm([x1-x2,y3-y2])))
theta =
0.8355
>> pi/4
ans =
0.7854
为什么上面的3个点明显夹角是45度,但是算出来确实 47.8696呢
问题已解决来人送分
▼优质解答
答案和解析
你给出的3个点的夹角就是47.8度,你应该该y3=2
看了MATLAB已知3点求夹角>>...的网友还看了以下:
MATLAB已知3点求夹角>>x1=1;y1=1;x2=0;y2=0;x3=0;y3=3;theta 2020-03-30 …
已知一次函数y=-2x+b,过点A(-1,y1)和点B(3,y2),则y1与y2的大小关系是?(A 2020-06-03 …
如图,已知抛物线y1=−3x2+3,直线y2=3x+3,当x任取一值时,x对应的函数值分别为y1, 2020-07-16 …
已知y1是x的正比例函数,y2是x的反比例函数,并且当自变量x=1时,y1-y2=-3当自变量x= 2020-07-25 …
已知函数y=-x²+x-1\5,当自变量x取m时,对应的函数值大于0,当自变量分别取m+2、m+1 2020-08-01 …
如图,已知抛物线y1=-x2+1,直线y2=-x+1,当x任取一值时,x对应的函数值分别为y1,y 2020-08-01 …
直线两点式y-y1/y2-y1=x-x1/x2-x1这个设法的字母顺序有固定顺序吗?能设为y-y2 2020-08-01 …
(x1+x2)(x1-x2)/2+(y1-y2)(y1+y2)=0,是怎么到[(y1-y2)/(x1 2020-10-31 …
如何求解带参数的四元三次方程组symsy1y2y3an(n是参数)9*(n-4)*y2*y3+9*y 2020-10-31 …
(2013•衡水二模)如图,已知抛物线y1=-x2+1,直线y2=-x+1,当x任取一值时,x对应的 2020-11-12 …