यदि `vec(a)xxvec(b)=vec(b)xxvec(c)ne vec(0)` तो दिखाएँ कि `vec(a)+vec(c)=k vec(b)`जहाँ k एक अदिश है |

Explanation

`vec(a)xxvec(b)=vec(b)xxvec(c)`
`implies" "vec(a)xxvec(b)-vec(b)xxvec(c)=0`
`implies" "(vec(a)xxvec(b))+(vec(c)xxvec(b))=vec(0)" "[because (vec(b)xxvec(c))=-(vec(c)xxvec(b))]`
`implies" "(vec(a)+vec(c))xxvec(b)=vec(0)" "["वितरण नियम से"]`
`implies" "(vec(a)+vec(c))||vec(b)`
`implies" "(vec(a)+vec(c))=k vec(b)`, जहाँ k एक अदिश है |