Question

What is the unit vector in the direction of a=2i-2j+k is

Answer 1

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

Given that

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

We know that,

$\rm :\longmapsto\:Unit \: vector \: in \: direction \: of \: \vec{a} \: is \:$

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

Consider,

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

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

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

$\rm \: = \: \: \: \sqrt{9}$

$\rm \: = \: \: \:3$

Hence,

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

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

$\bf :\longmapsto\:\hat{a} = \dfrac{2}{3}\hat{i} - \dfrac{2}{3} \hat{j} + \dfrac{1}{3}\hat{k}$

Additional information :-

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

$\rm :\longmapsto\:\vec{a}.\vec{a} = { |\vec{a}| }^{2}$

$\rm :\longmapsto\:\vec{a} \times \vec{a} = 0$

$\rm :\longmapsto\:\vec{a} \times \vec{b} = - \: \vec{b} \times \vec{a}$

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

$\rm :\longmapsto\: |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}|sin \theta$

$\rm :\longmapsto\: {(\vec{a}.\vec{b})}^{2} + { |\vec{a} \times \vec{b}| }^{2} = { |\vec{a}| }^{2} { |\vec{b}| }^{2}$