If →b and →c are two non-collinear unit vectors and →a is any vector, then (→a⋅→b)→b+(→a⋅→c)→c+→a⋅(→b×→c)|→b×→c|2(→b×→c)=
Let →b=^i,→c=^j
∴|→b×→c|=|^k|=1
again, let a=a1^i+a2^j+a3^k
Now,
→a⋅→b=→a⋅^i=a1,→a⋅→c=→a⋅^j=a2
and →a⋅→b×→c|→b×→c|=→a⋅^k=a3
∴(→a⋅→b)→b+(→a⋅→c)→c+→a⋅→b×→c|→b×→c|(→b×→c)
=a1→b+a2→c+a3(→b×→c)=a1^i+a2^j+a3^k=→a