1.

Express cross and dot product of two vectors in Cartesian coordinate.

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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