Let →a, →b and →c be 3 mutually perpendicular unit vectors. If →d is a unit vector which is equally inclined to →a, →b and →c, then the value of |→a+→b+→c+→d|2 is p+√q then p+q is
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Solution
Since, the vector →d is a unit vector which is equally inclined to →a, →b and →c ⇒→d=→a+→b+→c√3