1.

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

Answer»

\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}}


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