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| Answer» N1)question1)|A|=3 ,|B|=4,|c|=6a) find max. value and minimum values of |a+b-c|solution:-Here ,it's very IMPORTANT to, understand,that,mod only shows the MAGNITUDE,....if |A|=3....then it means,that the magnitude of that vector is eqaul to 3.Whether it is positive or negative is not fixed..If we take,VectorA and VectorB as negativethen|A+B-C|=|-3-4-6|=13this will be the maximum value,coz,it is the sum of all three......now, let's think on minimum valueIf vectorA and Vector B will be taken as positivethen|A+B-C|=|3+4-6|=1|A+B-C|=|3+4-6|=1This will be the minimum value....b)If vectorA ,vectorB,vectorC are mutually perpendicular,then find|A+B-2C|Solution:-If the three vectors(X,y,z) are perpendicular then the sum of their magnitude=(x^2+y^2+z^2)^1/2herex=A=3y=B=4,z=-2c=-2×6=12(-)magnitude=(9+16+14)^1/2=(169)^1/2=13c)can(|A+2B+C|=0?solution:-minimum value which can be possible|A+2B+C|=|-3+4×2-6|=1 minimum value will be1 minimum value will be1 HENCE 0 can't be possible | |