Question Number 1016 by 123456 last updated on 14/May/15

[(x^→ ×a^→ )×b^→ ]×x^→ =c^→

Answered by rpatle69@gmail.com last updated on 14/May/15

[−b^→ ×(x^→  ×a^→ )]×x^→ =c^→   x^→ ×[b^→ ×(x^→ ×a^→ )]=c^→   x^→ ×[(b^→ .a^→ )x^→ −(b^→ .x^→ )a^→ ]=c^→

Commented byprakash jain last updated on 17/May/15

x^→ ×(b^→ ∙a^→ )x^→ −(b^→ .x^→ )(x^→ ×a^→ )=−(b^→ ∙x^→ )(x^→ ×a^→ )

Commented by123456 last updated on 17/May/15

−(b^→ ∙x^→ )(x^→ ×a^→ )∙a^→ =0=c^→ ∙a^→   −(b^→ ∙x^→ )(x^→ ×a^→ )∙x^→ =0=c^→ ∙x^→

Commented by123456 last updated on 17/May/15

x^→ =λ_1 a^→ +λ_2 (a^→ ×c^→ )  for some λ_1 ,λ_2 ∈R