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State and explain the charateristics of vector product of two vectors. |
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Answer» Solution :(1) `vecaxxvecb=vecbxxveca` The vector product of two vector is not commutative ut `vecaxxvecb=-vecbxxveca` is opposite to each other. However `|vecaxxvecb|=|vecbxxveca|` (2) Scalar product act behave like reflection (taking image in mirror) `xrarr-x,yrarr-yandzrarr-z`. In reflection occurrence all components change sign mean positive vector becomes negative. So, `vecaxxvecbrarr(-veca)xx(-vecb)=vecaxxvecb` Hence, in reflection sign is not change in resultant. (3) Vector product obeys distributive law : `vecaxx(vecb+vecc)=vecaxxvecb+vecaxxvecc` (4) For two non-zero vectors `vecaxxveca=vec0` where `vec0` is vector of zero modulus Here `vecaxxveca=(a)(a)SIN0^(@)hatn` `=vec0` (`because` Angle between `veca and veca` is `0^(@)`) Hence, condition of parallel or anti parallel of two non-zero vectors is that its vector product should be zero. (5) If two non-zero vector is perpendicular, then `vecaxxvecb=ab SIN90^(@)hatn` `=abhatn` where `hatn` is unit vector in direction of `veca xx vecb`. (6) Vector product for unit vector of cartesian co-ordinate system. In cartesian co-ordinate system `hati,HATJ and hatk` are the unit vectors in the direction of X, Y and Z-axis respectively, `therefore hatixxhati=|1||1|sin0^(@),hatn` `=vec0` `hatjxxhatj=|1||1|sin0^(@).hatn` `=vec0` and `hatkxxhatk=|1||1|sin0^(@).hatn` `vec0` and `hatixxhatj=|1||1|sin90^(@).hatn` `=1hatk` [`because hatk` is perpendicular unit vector of the plane `hatixxhatj`] `=1(hatk)` `=hatk` Similarly `hatjxxhatk=hatiandhatkxxhati=hatj` and `hatjxxhati=-hatk` `hatkxxhatj=-hatiandhatixxhatk=-hatj`  (7) Vector product of two vectors in cartesian co-ordinate system : Suppose `veca=a_(x)hati+a_(y)hatj+a_(z)hatk` and `vecb=b_(x)hati+b_(y)hatj+b_(z)hatk` then `vecaxxvecb=(a_(x)hati+a_(y)hatj.+a_(z)hatk)xx(b_(x)hati+b_(y)hatj+b_(z)hatk)` `=a_(x)b_(x)(hatixxhati)+a_(x)b_(y)(hatixxhatj)+a_(x)b_(z)(hatixxveck)` `+a_(y)b_(x)(hatjxxhati)+a_(y)b_(y)(hatjxxhatj)+a_(y)b_(z)(hatjxxhatk)` `+a_(z)b_(x)(hatkxxhati)+a_(z)b_(y)(hatkxxhatj)+a_(z)b_(z)(hatkxxhatk)` `therefore` From property (6) result, `vecaxxvecb=a_(x)b_(y)(hatk)+a_(x)b_(z)(-hatj)+a_(y)b_(x)(-hatk)+a_(y)b_(z)(hati)+a_(z)b_(x)(hatj)+a_(z)b_(y)(-hati)` `=a_(y)b_(z)hati-a_(z)b_(y)hati+a_(z)b_(x)hatj-a_(x)b_(z)hatj+a_(x)b_(y)hatk-a_(y)b_(x)hatk` `vecaxxvecb=|(hati,hatj,hatk),(a_(x),a_(y),a_z),(b_(x),b_(y),b_(z))|` |
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