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The length of the longer diagonal of the parallelogram constructed on ` 5veca + 2vecb and veca - 3vecb, ` if it is given that ` |veca|=2sqrt2, |vecb|=3 and veca. Vecb= pi/4` isA. 15B. `sqrt3`C. `sqrt593`D. `sqrt369` |
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Answer» Correct Answer - C The diagonals of the parallelogram are ` vecalpha = 5veca + 2vecb + veca -3vecb = 6veca - vecb and beta = +- ( 4 veca + 5vecb)` Now, ` |vecalpha|= | 6 veca - vecb|` ` Rightarrow |vecalpha| = sqrt(36 |veca|^(2) + |vecb|^(2) -12 (veca .vecb))` `|vecalpha|=sqrt(36xx8+9-12xx 2sqrt2xx3xx1/sqrt2) =15` and, ` |vecbeta|=|4veca +5vecb|` `|vecbeta|=sqrt(16|veca|^(2)+25|vecb|^(2)+40(veca.vecb))` `|vecbeta|=sqrt(16xx8+25xx9 +40xx2sqrt2xx3xx1/sqrt2)=sqrt593` Clearly, `|vecbeta| gt |vecalpha|` Hence, the length of the longer diagonal is `sqrt593` |
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