找到一个方法,跟各位分享下,如果有更好的办法,希望各位大侠指教,下面是我的解决办法:
0 Y; Y+ o5 c" \cos(sym(pi/2)),这样运行后结果就是零了,之前的矩阵结果是:T40 =0 K2 D' }9 o, \' n2 a
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[ (2^(1/2)*3^(1/2)*sin(s1))/3, (3^(1/2)*sin(s1)^2)/3 - cos(s1)^2, cos(s1)*sin(s1) + (3^(1/2)*cos(s1)*sin(s1))/3, a1*cos(s1) - a3*((3^(1/2)*sin(s1)^2)/3 - cos(s1)^2) + d4*(cos(s1)*sin(s1) + (3^(1/2)*cos(s1)*sin(s1))/3) + (2^(1/2)*3^(1/2)*d3*sin(s1))/3]2 s- p8 D, l' K" k$ F
[ -(2^(1/2)*3^(1/2)*cos(s1))/3, - cos(s1)*sin(s1) - (3^(1/2)*cos(s1)*sin(s1))/3, sin(s1)^2 - (3^(1/2)*cos(s1)^2)/3, d4*(sin(s1)^2 - (3^(1/2)*cos(s1)^2)/3) + a1*sin(s1) + a3*(cos(s1)*sin(s1) + (3^(1/2)*cos(s1)*sin(s1))/3) - (2^(1/2)*3^(1/2)*d3*cos(s1))/3]: [1 z8 r. M& p9 u9 |1 @
[ 3^(1/2)/3, -(2^(1/2)*3^(1/2)*sin(s1))/3, -(2^(1/2)*3^(1/2)*cos(s1))/3, (3^(1/2)*d3)/3 - (2^(1/2)*3^(1/2)*d4*cos(s1))/3 + (2^(1/2)*3^(1/2)*a3*sin(s1))/3]8 o$ d; i0 ~2 }* W3 z7 Y
[ 0, 0, 0, 1] |