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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)(hati+hati)`D. `(7)/(2)(hatj+hati)`

Answer» Correct Answer - D
Component of `vecA` along `vecB=A cos theta=(vecA.vecB)/(B)`
`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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