Given that
(→a.→b)→b+(→a.→c)→c+→a(→b×→c)∣∣∣→b×→c∣∣∣(→b×→c)
And →b⊥→c hence, →b×→c=∣∣∣→b∣∣∣∣∣→c∣∣
=(→a.→b)→b+(→a.→c)→c+→a(∣∣∣→b∣∣∣∣∣→c∣∣)∣∣∣∣∣∣→b∣∣∣∣∣→c∣∣∣∣∣∣∣∣→b∣∣∣∣∣→c∣∣
Given that vector →b,→c are unit vector hence ,
∣∣∣→b∣∣∣=1,∣∣→c∣∣=1
=(→a.→b)→b+(→a.→c)→c+→a(1.1)|1.1|1.1
=(→a.→b)→b+(→a.→c)→c+→a