a vector has amagnitude of ten along negative x axices then it is represented as
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The resultant of the given two vectors can easily be figured out just by resolving the vectors into x and y components
Resolving A vector, we get 10cos(30°) and 10sin(30°) , which is equal to 5√3 and 5 along x and y axis respectively
Now resolving B vector, we get -20cos(30°)[carries negative sign as it is along negative X-axis] and 20sin(30°), which is equal to -10√3 and 10 along X and Y axis respectively
Equating the resolved vectors that are along x axis, we get -5√3 ,which can be assigned to Rx (just for our convenience)
Equating the resolved vectors that are along Y axis, we get 15,which can be taken as Ry
Thus the X and Y components of the resultant vector has been found out
Futher Pythagoras theorem can be used to calculate the resultant vector ‘R’ using its X and Y components ( i.e Rx and Ry)
Therefore ‘R’ vector =√(Rx^2 + Ry^2)
=√(5√3^2 + 15^2) = √(225+75) = √300
=10√3 = 17.32
Resolving A vector, we get 10cos(30°) and 10sin(30°) , which is equal to 5√3 and 5 along x and y axis respectively
Now resolving B vector, we get -20cos(30°)[carries negative sign as it is along negative X-axis] and 20sin(30°), which is equal to -10√3 and 10 along X and Y axis respectively
Equating the resolved vectors that are along x axis, we get -5√3 ,which can be assigned to Rx (just for our convenience)
Equating the resolved vectors that are along Y axis, we get 15,which can be taken as Ry
Thus the X and Y components of the resultant vector has been found out
Futher Pythagoras theorem can be used to calculate the resultant vector ‘R’ using its X and Y components ( i.e Rx and Ry)
Therefore ‘R’ vector =√(Rx^2 + Ry^2)
=√(5√3^2 + 15^2) = √(225+75) = √300
=10√3 = 17.32
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