If li,mi,ni where i=1,m=2,n=3 denote the direction cosines of 3 mutually perpendicular vectors in space ,prove that AA^T=I wher A is a matrix and A^T is its transpose and I is a indentity matrix

A=l1m1n1l2m2n2l3m3n3 AT=l1l2l3m1m2m3n1n2n3Now AAT=l1m1n1l2m2n2l3m3n3×l1l2l3m1m2m3n1n2n3 AAT=l12+m12+n12l1l2+m1m2+n1n2l1l3+m1m3+n1n3l2l1+m2m1+n2n1l22+m22+n22l2l3+m2m3+n2n3l3l1+m3m1+n3n1l3l2+m3m2+n3n2l32+m32+n32Now since( l1,m1,n1),(l2,m2,n2) and (l3,m3,n3) denotes the direction cosines of 3 mutually perpendicular linesHence l1l2+m1m2+n1n2=0,l1l3+m1m3+n1n3=0,l2l3+m2m3+n2n3=0Also since( l1,m1,n1),(l2,m2,n2) and (l3,m3,n3) denotes the direction cosines of 3 mutually perpendicular linesHence we always have l12+m12+n12=1,l22+m22+n22=1,l32+m32+n32=1Now putting the values , we get AAT=100010001 which is an identity matrix AAT=I    proved

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