Given vector A=2i + 1j and vector B=3j, how do you find the component of B in the direction of A?

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Answer 1 · 2017-10-06T06:16:08

The component is #=<6/5,3/5># and its length is #=3/sqrt5#

The question is finding the projection of #vecB# onto #vecA#

The projection of #vecB# onto #vecA# is

#proj_A B=(vecA.vecB)/(|vecA|^2)*vecA#

#vecA = <2,1>#

#vecB=<0,3>#

Therefore,

#proj_A B=(<2,1>. <0,3>)/(|<2,1>|^2)*<2,1>#

#=(3)/(5)*<2,1>#

#=<6/5,3/5>#

So,

#|proj_A B|=|<6/5,3/5>|= sqrt((6/5)^2+(3/5)^2)=sqrt(45/25)#

#=3/sqrt5#