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If` veca,vecb,vecc` are three mutually perpendicular vectors, then the vector which is equally inclined to these vectors is (A) `veca+vecb+vecc` (B) `veca/|veca|+vecb/|vecb|+vec/|vecc|` (C) `veca/|veca|^2+vecb/|vecb|^2+vecc/|vecc|^2` (D) `|veca|veca-|vecb|vecb+|vecc|vecc`A. `veca+vecb+vecc`B. `(veca)/(|veca|)+(vecb)/(|vecb|)+(vecc)/(|vecc|)`C. `(veca)/(|veca|^(2))+(vecb)/(|vecb|^(2))+(vecc)/(|vecc|^(2))`D. |veca|veca-|vecb|vecb+|vecc|vecc|` |
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Answer» Correct Answer - B `veca.vecb.=vecb,vecc=vecc.veca=0` `"Let " vecr=(veca)/(|veca|)+(vecb)/(|vecb|)+(vecc)/(|vecc|)` `" cos "0 =(vecr.veca)/(|vecr||vecb|)=(1)/(|vecr|)` `" cos "phi = (vecr.vecb)/(|vecr||vecc|) =(1)/(|vecr|)` `" cos "Psi = (vecr.vecc)/(|vecr||vecc|) =(1)/(|vecr|)` `:. 0 = phi = Psi` |
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