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If ` vecalpha+ vecbeta+ vecgamma=a vecdeltaa n d vecbeta+ vecgamma+ vecdelta=b vecalpha, vecalphaa n d vecdelta`are non-colliner, then ` vecalpha+ vecbeta+ vecgamma+ vecdelta`equalsa. `a vecalpha`b. `b vecdelta`c. `0`d. `(a+b) vecgamma`A. `a vecalpha `B. `b vecdelta`C. `0`D. `(a+ b)vecgamma` |
Answer» Correct Answer - C Given `vecalpha + vecbeta + vecgamma = a vecdelta" "` (i) `" "vecbeta + vecgamma + vecdelta = b vecalpha" "` (ii) From (i), ` vecalpha + vecbeta + vecdelta = (a+1) vecdelta" "` (iii) From (ii), `vecalpha + vecbeta + vecgamma + vecdelta = (b+1)vecalpha " "` (iv) From (iii) and (iv), we get `" "(a+1) vecdelta = (b+1) vecalpha " "` (v) Since `vecalpha` is not parallel to `vecdelta`, From (v), `a=1 =0 and b+1 =0` From (iii), `vecalpha + vecbeta + vecgamma + vecdelta =0` |
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