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,

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

$=3/sqrt5$