If |→a|=3,∣∣→b∣∣=4,|→c|=5,→a⊥(→b+→c),→b⊥(→c+→a) and →c⊥(→a+→b) then √2∣∣→a+→b+→c∣∣ is equal to
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Solution
Given, →a⊥(→b+→c),→b⊥(→c+→a) and →c⊥(→a+→b) ⇒→a,→b,→c are perpendicular to each other.....(i) ∴√2|→a+→b+→c|=√2√|→a|2+|→b|2+|→c|2 [ using (i) ] =√2⋅√32+42+52=10