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A vector sqrt(3) hati+hatj rotates about its fail through an angle 300 in clock wise direction then the new vector is |
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Answer» Solution :The magnitude of `SQRT(3)hati+hatj` is `sqrt(3+1)=2` the ANGLE made by the vectorwith x-axis is `"tan" theta=(A_(y))/(A_(x))=(1)/(sqrt(3))` `:. theta=30^(@)`  When the given vector rotates 30 in clock wise its direction changes ALONG x - axis but its magnitude does not change. `:.` The new VETOR is `2hati` 
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