Question

if â and b are the unit vectors along a and b respectively , then what is the projection of b on â?​

Answer 1

$\large\underline{\sf{Solution-}}$

We know that

$\rm :\longmapsto\:Projection \: of \: \vec{a} \: on \: \vec{b} \: = \dfrac{\vec{a}.\vec{b}}{ |\vec{b}| }$

Since,

$\rm :\longmapsto\:\hat{a} \: is \: unit \: vector.$

So,

$\rm :\longmapsto\: |\hat{a}| = 1$

Now,

$\rm :\longmapsto\:Projection \: of \: \vec{b} \: on \: \hat{a} \:$

$\rm \:  =  \:  \: \dfrac{\vec{b}.\hat{a}}{ |\hat{a}| }$

$\rm \:  =  \:  \: \vec{b}.\hat{a}$

Hence,

$\rm :\longmapsto\:Projection \: of \: \vec{b} \: on \: \hat{a} \: = \vec{b}.\hat{a}$

Proof of projection of vector a on vector b

Let assume that

$\rm :\longmapsto\:\vec{a} \: and \: \vec{b} \: inclined \: each \: other \: at \: p$

So,

$\rm :\longmapsto\:\vec{a}.\vec{b} = |\vec{a}| \: |\vec{b}| \: cosp$

Now, In triangle OAB

$\rm :\longmapsto\:cosp = \dfrac{OB}{OA}$

$\rm :\longmapsto\:OB = OA \: cosp$

$\rm :\longmapsto\:OB = |\hat{a}| \: cosp$

$\rm :\longmapsto\:OB = \dfrac{ |\vec{a}| \: |\vec{b}| \: cosp}{ |\vec{b}| }$

$\rm :\longmapsto\:OB = \dfrac{ \vec{a} \: . \:\vec{b}}{ |\vec{b}| }$