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If the vectors `3vecP+vecq, 5vecP - 3vecq and 2 vecp+ vecq, 4 vecp - 2vecq` are pairs of mutually perpendicular vectors, the find the angle between vectors `vecp and vecq`. |
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Answer» `3 vecp + vecq and 5vecp-3vecq` are perpendicular. thereforem, `(3vecP +vecq). (5vecp -3vecq)=0` `15vecp^(2)-3vecq^(2)=4vecp.vecq` `2vecp +vecq and 4vecp -2vecq` are perpendicular , therefore, `(2vecp +vecq).(4vecp - 2vecq)=0` `8vecp^(2)=2vecq^(2)` `vecq^(2)=4vecp^(2)` Now, `cos theta= (vecp.vecq)/(|vecp||vecq|)` Substituting `vecq^(2)=4vecp^(2) "in" (i), 3vecp^(2)=4vecp.vecq` `cos theta = 3/4.vecp^(2)/(|vecp|2|vecp|)=3/8` or ` theta=cos^(-1)""3/8` |
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