Question

Find the projection of the vector [2i-3j+6k] on the vector [i+2j+2k].

Answer 1

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

$\rm :\longmapsto\:Let \: \vec{a} = 2\hat{i} - 3\hat{j} + 6\hat{k}$

and

$\rm :\longmapsto\:Let \: \vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}$

We know,

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

Consider,

$\rm :\longmapsto\:\vec{a}.\vec{b}$

$\:  \: \rm  =  \:  \: (2\hat{i} - 3\hat{j} + 6\hat{k}).(\hat{i} + 2\hat{j} + 2\hat{k})$

$\:  \: \rm  =  \:  \: 2 - 6 + 12$

$\:  \: \rm  =  \:  \: 8$

$\boxed{\bf\implies \:\vec{a}.\vec{b} = 8}$

Consider,

$\rm :\longmapsto\: |\vec{b}|$

$\:  \: \rm  =  \:  \: \sqrt{ {(1)}^{2} + {(2)}^{2} + {(2)}^{2} }$

$\:  \: \rm  =  \:  \: \sqrt{1 + 4 + 4}$

$\:  \: \rm  =  \:  \: \sqrt{9}$

$\:  \: \rm  =  \:  \: 3$

$\boxed{\bf\implies \: |\vec{b}| = 3}$

Hence,

${\bf :\longmapsto\:Projection \: of \: \vec{a} \: on \: \vec{b} = \dfrac{\vec{a}.\vec{b}}{ |\vec{b}| } = \dfrac{8}{3} }$

Additional Information :-

$\boxed{ \sf \: \vec{a}.\vec{b} = |\vec{a}| |\vec{b}| cos \theta}$

$\boxed{ \sf \: |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| sin\theta}$

$\boxed{ \sf \: \vec{a}.\vec{b} = 0 \implies \: \vec{a} \: \perp \: \vec{b}}$

$\boxed{ \sf \: \vec{a} \times \vec{b} = 0 \implies \: \vec{a} \: \parallel \: \vec{b}}$

$\boxed{ \sf \: \vec{a}.\vec{a} = { |\vec{a}| }^{2} }$

$\boxed{ \sf \: \vec{a} \times \vec{a} = 0}$

$\boxed{ \sf \: \vec{a}.\vec{b} = \vec{b}.\vec{a}}$

$\boxed{ \sf \: \vec{a} \times \vec{b} = - \vec{b} \times \vec{a}}$