Given, diagram can be projected on xy -plane as shown here. |OA|=R=|OB|=|OC| OB has projected as AB′ and OC is projected as B′C′ . Hence, |AB′|=|BC′|=R and ∠BB′C′=90° . Hence, the resultant of OA,OB and OC is OC′ OC′=OB+BC′ ⇒|OC′|=|OB|+|BC′| ...(i) In ΔBB′C′ , using Pythagoras |BC′|=√R2+R2=√2R From Eq. (i), we get |OC′|=R+√2R=(√2+1)R