यदि `vec(a)cdot vec(b)=vec(a)cdot vec(c), vec(a)xxvec(b)=vec(a)xxvec(c)" तथा "vec(a)ne vec(0)`, तो सिद्ध करें कि `vec(b)=vec(c)`.

Explanation

दिया है : `vec(a)cdot vec(b)=vec(a)cdot vec(c)" तथा "vec(a)ne vec(0)`
`implies" "vec(a)cdot vec(b)-vec(a)cdot vec(c)=vec(0)" तथा "vec(a)ne vec(0)`
`implies" "vec(a)cdot (vec(b)-vec(c))=0" तथा "vec(a) ne vec(0)`
`implies" "vec(b)-vec(c)=vec(0)" या "vec(a)_|_(vec(b)-vec(c))`
`implies" "vec(b)=vec(c)" या "vec(a)_|_(vec(b)-vec(c))" "...(1)`
पुन: दिया है, `vec(a)xxvec(b)=vec(a)xxvec(c)" तथा "vec(a)ne vec(0)`
`implies" "(vec(a)xxvec(b))-(vec(a)xxvec(c))=vec(0)" तथा "vec(a)ne vec(0)`
`implies" "vec(a)xx(vec(b)-vec(c))=vec(0)" तथा "vec(a)ne vec(0)`
`implies" "(vec(b)-vec(c))=vec(0)" या "vec(a)||(vec(b)-vec(c))`
`implies" "vec(b)=vec(c)" या "vec(a)||(vec(b)-vec(c))" "...(2)`
(1) तथा (2) से हमें मिलता है, `vec(b)=vec(c)`
[`because vec(a)_|_(vec(b)-vec(c))" तथा "vec(a)||(vec(b)-vec(c))` दोनों एक साथ सत्य नहीं हो सकते]