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The vector `(vec(a)+3vec(b))` is perpendicular to `(7 vec(a)-5vec(b))` and `(vec(a)-4vec(b))` is perpendicular to `(7vec(a)-2vec(b))`. The angle between `vec(a)` and `vec(b)` is :A. `30^(@)`B. `45^(@)`C. `60^(@)`D. None of these |
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Answer» Correct Answer - C `(vec(a)+3vec(b)).(7vec(a)-5vec(b))=0` `rArr 7a^(2)-15b^(2)+16 vec(a). vec(b)=0` …(i) and, `(vec(a)-4vec(b)).(7vec(a)-2vec(b))=0` `rArr 7a^(2)+8b^(2)-30 vec(a).vec(b)=0` …(ii) By adding (i) and (ii) `rArr -23b^(2)+46vec(a).vec(b)=0 rArr 2vec(a).vec(b)=b^(2)` So `7a^(2)-15b^(2)+8b^(2)=0 rArr a^(2)=b^(2)` `rArr 2ab cos theta=b^(2) rArr 2 cos theta=1` `rArr theta=cos^(-1) (1//2)=60^(@)` |
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