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Express cross and dot product of two vectors in Cartesian coordinate. |
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Answer» Solution :Let `vecA and vecB` be the two vectors. `vecA=vecA_(x)hati+vecA_(s)HATJ+vecA_(z)hatk, vecB=vecB_(x)hati+vecB_(y)hatj+vecB_(z)hatk` Cross product of `vecA and vecB`. `VECAXXVECB=(vecA_(x)hati+vecA_(y)hatj+vecA_(z)hatk)XX(vecB_(x)hati+vecB_(y)hatj+vecB_(z)hatk)` `=vecA_(x)vecB_(x)hatixxhati+vecA_(x)vecB_(y)hatixxhatj+vecA_(x)vecB_(z)hatixxhatk` `+vecA_(y)vecB_(y)hatjxxhati+vecA_(y)vecB_(z)hatjxxhatj+vecA_(z)B_(z)atkxxhatk` `vecAxxvecB=vecA_(x)vecB_(y)(hatk)+A_(x)B_(z)(-hatj)+vecA_(y)vecB_(x)(-hatk)+vecA_(y)vecB_(y)(0)` `+vecA_(y)vecB_(z)(hati)+vecA_(z)vecB_(x)(hatj)+vecA_(z)vecB_(z)(0)` `(vecAxxvecB)=(vecA_(y)vecB_(z)-vecA_(z)vecB_(y))hati+(vecA_(z)vecB_(x)-vecA_(x)vecB_(z))hatj+(vecA_(x)vecB_(y)-vecA_(y)vecB_(x))hatk` `[because hatixx hati=hatjxxhatj=hatkxxhatk=0` `hatixxhatj=k, hatixxhatk=-hatj, hatjxxhati=-k,hatjxxhatk=hati, hatk xxhati=+hatjxxhatj=-hatk]` It can be in DETERMINANT form as, `vecA xx vecB=|(hati,hatj,hatjk),(A_(x),A_(y),A_(z)),(B_(x),B_(y),B_(z))|` `=hati(A_(y)B_(z)-B_(y)A_(z))-hatj(A_(x)B_(z)-B_(y)A_(z))+hatk(A_(x)B_(y)-B_(x)A_(y))` DOT product of `vecA and vecB`. `vecA=vecA_(x)hati+vecA_(y)hatj+vecA_(z)hatk, vecB=vecB_(x)hati+vecB_(x)hatj_vecB_(z)hatk` `vecA.vecB=(vecA_(x)hati+vecA_(y)hatj+vecA_(z)hatk).(vecB_(x)hati+vecB_(y)hatj+vecB_(z)hatk)` `(vecA_(x)vecB_(x)(hati.hatj)+vecA_(x)vecB_(y)(hati.hatj)` `+vecA_(x)vecB_(z)(hati.hatk)+vecA_(y)vecB_(x)(hatj.hatk)+vecA_(y)vecB_(y) (hati.hatj)` `+vecA_(y)vecB_(z)(hatj.hatk)+vecA_(z)vecB_(x)(hatk.hati)+vecA_(z)vecB_(y)(hatk.hatj)` `+A_(z)B_(z)(hatk.hatk)` `vecA.vecB=vec(A_(x))vec(B_(x))(1)+vec(A_(x))vec(B_(y))(0)+vecA_(x)vec(B_(z))(0)+vec(A_(y))vec(B_(z))(0)+vec(A_(y))vec(B_(y))(1)` `+vec(A_(y))vec(B_(z))(0)+vec(A_(z))vec(B_(x))(0)+vec(A_(z))v_(y)(0)+A_(z)B_(z)(1)` `vecA.vecB=vecA_(x)vecB_(x)+vec(A_(y))vec(B_(y))+vecA_(z)vecB_(z)` `[because hati.hati=hatj.hatj=hatk.hatk=1` `hati.hatj=hati.hatk=0, hatj.hati=hatj.hatk=0, hatk.hati=hatk.hatj=0]` |
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