回答:
矢量是
说明:
垂直于2个向量的向量用行列式(叉积)计算
哪里
在这里,我们有
因此,
通过做2点产品进行验证
所以,
<0,8,5>和<-1,-1,2>的叉积是多少?
<21,-5,8> We know that vecA xx vecB = ||vecA|| * ||vecB|| * sin(theta) hatn, where hatn is a unit vector given by the right hand rule. So for of the unit vectors hati, hatj and hatk in the direction of x, y and z respectively, we can arrive at the following results. color(white)( (color(black){hati xx hati = vec0}, color(black){qquad hati xx hatj = hatk}, color(black){qquad hati xx hatk = -hatj}), (color(black){hatj xx hati = -hatk}, color(black){qquad hatj xx hatj = vec0}, color(black){qquad hatj xx hatk = hati}), (color(black){hatk xx hati = hatj}, color(black){qquad hatk xx hatj = -hati}, color(black){qquad hatk xx hatk
[0,8,5]和[1,2,-4]的叉积是多少?
[0,8,5] xx [1,2,-4] = [-42,5,-8] vecA和vecB的叉积由vecA给出xx vecB = || vecA || * || vecB || * sin(theta)hatn,其中theta是vecA和vecB之间的正角度,而hatn是一个单位向量,其方向由右手规则给出。对于单位向量hati,hatj和hatk分别在x,y和z的方向上,颜色(白色)((颜色(黑色){hati xx hati = vec0},颜色(黑色){qquad hati xx hatj = hatk} ,颜色(黑色){qquad hati xx hatk = -hatj}),(颜色(黑色){hatj xx hati = -hatk},颜色(黑色){qquad hatj xx hatj = vec0},颜色(黑色){qquad hatj xx hatk = hati}),(color(black){hatk xx hati = hatj},color(black){qquad hatk xx hatj = -hati},color(black){qquad hatk xx hatk = vec0}))另外,交叉积是分布式的,这意味着vecA xx(vecB + vecC)= vecA xx vecB + vecA xx vecC。对于这个问题,[0,8,5] xx [1,2,-4] =(8hatj + 5hatk)xx(hati + 2hat
[-1,0,1]和[0,1,2]的叉积是多少?
叉积数= < - 1,2,-1>用行列式|计算叉积(veci,vecj,veck),(d,e,f),(g,h,i)|其中<d,e,f>和<g,h,i>是2个向量这里,我们有veca = < - 1,0,1>和vecb = <0,1,2>因此,| (veci,vecj,veck),( - 1,0,1),(0,1,2)| = VECI | (0,1),(1,2)| -vecj | (-1,1),(0,2)| + veck | (-1,0),(0,1)| = veci(-1)-vecj(-2)+ veck(-1)= < - 1,2,-1> = vecc通过做2点产品验证<-1,2,-1>。< - 1, 0,1> = 1 + 0-1 = 0 <-1,2,-1>。<0,1,2> = 0 + 2-2 = 0因此,vecc垂直于veca和vecb