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

Oct 6, 2017

The component is $= < \frac{6}{5} , \frac{3}{5} >$ and its length is $= \frac{3}{\sqrt{5}}$

#### Explanation:

The question is finding the projection of $\vec{B}$ onto $\vec{A}$

The projection of $\vec{B}$ onto $\vec{A}$ is

$p r o {j}_{A} B = \frac{\vec{A} . \vec{B}}{| \vec{A} {|}^{2}} \cdot \vec{A}$

$\vec{A} = < 2 , 1 >$

$\vec{B} = < 0 , 3 >$

Therefore,

$p r o {j}_{A} B = \frac{< 2 , 1 > . < 0 , 3 >}{| < 2 , 1 > {|}^{2}} \cdot < 2 , 1 >$

$= \frac{3}{5} \cdot < 2 , 1 >$

$= < \frac{6}{5} , \frac{3}{5} >$

So,

$| p r o {j}_{A} B | = | < \frac{6}{5} , \frac{3}{5} > | = \sqrt{{\left(\frac{6}{5}\right)}^{2} + {\left(\frac{3}{5}\right)}^{2}} = \sqrt{\frac{45}{25}}$

$= \frac{3}{\sqrt{5}}$