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Find the vector components of `veca= 2hati+ 3hatj` along the directions of `hati+hatj`. |
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Answer» Correct Answer - `(5)/(2) (hati+hatj)` Vector component of `veca` in the direction of `vecb(hati+hatj)` `" "=(acos theta ) vecb= ((abcos theta )/(b))(vecb)/(b) = (veca*vecb)/(|vecb|^(2))vecb` `= ((2hati+3hatj)*(hati+hatj))/([sqrt(1^(2)+1^(2))]^(2))(hati+hatj)` `= (2xx1+3xx1)/(2) (hati+hatj)= (5)/(2) (hati+hatj)` |
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