The correct option is B 2→a
Let →a=ax^i+ay^j+az^k
⇒^i×(→a×^i)=((^i⋅^i)→a−(^i⋅→a)^i)
=(→a−(→a⋅^i)¯i)=→a−ax^i
Similarly,
⇒^j×(→a×^j)=→a−ay^j
Similarly,
⇒^k×(→a×^k)=→a−az^k
Adding all above three cases, we get
=^i×(→a×^i)+^j×(→a×^j)+^k×(→a×^k)=3→a−(ax^i+ay^j+az^k)
=3→a−→a=2→a