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What is the component of `(3hati+4hatj)` along `(hati+hatj)` ?A. `(1)/(2)(hatj+hati)`B. `(3)/(2) (hatj+hati)`C. `(5)/(2)(hatj+hati)`D. `(7)/(2)(hatj+hati)` |
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Answer» Correct Answer - 4 Component of `vecA` along `vecB = A cos theta = (vecA*vecB)/(B)` in vector form `= ((vecA*vecB)/(B))hatB= ((veca*vecB)/(B^(2)))vecB` `vecA*vecB= (3hati+ 4hatj)*(hati+hatj)=7` `B^(2) = (sqrt(1^(2)+ 1^(2)))^(2)= 2` thus required component is `= (7)/(2) (hati+hatj)` |
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