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The resultant of `vecA` and `vecB` is perpendicular to `vecA`. What is the angle between `vecA` and `vecB` ?A. `cos ^(-1)((A)/(B))`B. `cos^(-1)(-(A)/(B))`C. `sin^(-1)((A)/(B))`D. `sin^(-1)(-(A)/(B))` |
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Answer» Correct Answer - B `tan alpha=(B sin theta)/(A+B cos theta)=tan90^(@)=oo` thus, `A+B cos theta=0rArrcos theta=-A//B` |
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