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Vectors `3veca-5vecb and 2veca + vecb` are mutually perpendicular. If `veca + 4 vecb and vecb - veca` are also mutually perpendicular, then the cosine of the angle between `veca nad vecb` isA. `19/(5sqrt43)`B. `19/(3sqrt43)`C. `19/(sqrt45)`D. `19/(6sqrt43)` |
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Answer» Correct Answer - a `(3veca-5vecb).(2veca+vecb)=0` `or 6|veca|^(2)-5|vecb|^(2)= 7 veca.vecb` `Also , (veca+4vecb).(vecb-veca)=0` `or -|veca|^(2)+ 4|vecb|^(2)= 3 veca.vecb` `or 6/7 |veca|^(2)-5/7|vecb|^(2)= -1/3 |veca|^(2)+ 4/3|vecb|^(2)` ` or 25|veca|^(2) = 43 |vecb|^(2)` ` Rightarrow 3 veca. vecb= - |veca|^92) + 4|vecb|^(2)= 57/25 |vecb|^(2)` `or 3 |veca||vecb| cos theta = 57/25 |vecb|^(2)` `or 3sqrt(43/25) |vecb|^(2) cos theta = 57/25 |vecb|^(2)` ` or cos theta = 19/(5sqrt43)` |
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