यदि `vec(a)+vec(b)+vec(c)=0`, सिद्ध करें कि `(vec(a)xxvec(b))=(vec(b)xxvec(c))=(vec(c)xxvec(a))`.

Explanation

`vec(a)+vec(b)+vec(c)=vec(0)impliesvec(a)+vec(b)=-vec(c)`
`implies" "(vec(a)+vec(b))xxvec(b)=(-vec(c))xxvec(b)" "["दोनो तरफ "vec(b)" से सदिश गुणन लेने पर"]`
`implies" "(vec(a)xxvec(b))+(vec(b)xxvec(b))+(vec(b)xxvec(b))=(-vec(c))xxvec(b)" "["वितरण नियम से"]`
`implies" "(vec(a)xxvec(b))+vec(0)=(vec(b)xxvec(c))" "[because vec(b)xxvec(b)=vec(0)" तथा "(-vec(c))xxvec(b)=vec(b)xxvec(c)]`
`implies" "vec(a)xxvec(b)=vec(b)xxvec(c)" "...(i)`
पुन:, `vec(a)+vec(b)+vec(c)=0impliesvec(b)+vec(c)=-vec(a)`
`implies" "(vec(b)+vec(c))xxvec(c)=(-vec(a))xxvec(c)" "["वितरण नियम से"]`
`implies" "(vec(b)xxvec(c))+vec(0)=vec(c)xxvec(a)" "[because vec(c)xxvec(c)=vec(0)" तथा "(-vec(a))xxvec(c)=vec(c)xxvec(a)]`
`implies" "vec(b)xxvec(c)=vec(c)xxvec(a)" "...(2)`
(1) तथा (2) से हमें मिलता है, `vec(a)xxvec(b)=vec(b)xxvec(c)=vec(c)xxvec(a)`