Question

Find the unit vector perpendicular to each of the vectors i-2j - 2k and 2i+j-2k.​

Answer 1

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

Let assume that

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

and

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

Consider,

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

$\rm \:  =  \:  \: \rm \: = \: \: \: \begin{gathered}\sf \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\1& - 2& - 2\\2&1& - 2\end{array}\right | \end{gathered}$

$\rm \:  =  \:  \: (4 + 4)\hat{i} - \hat{j}( - 2 + 4) + \hat{k}(1 + 4)$

$\rm \:  =  \:  \: 8\hat{i} - 2\hat{j} + 5\hat{k}$

$\bf\implies \:\vec{a} \times \vec{b} = 8\hat{i} - 2\hat{j} + 5\hat{k}$

Consider,

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

$\rm \:  =  \:  \: |8\hat{i} - 2\hat{j} + 5\hat{k}|$

$\rm \:  =  \:  \: \sqrt{ {(8)}^{2} + {( - 2)}^{2} + {(5)}^{2} }$

$\rm \:  =  \:  \: \sqrt{64 + 4 + 25}$

$\rm \:  =  \:  \: \sqrt{93}$

$\bf\implies \: |\vec{a} \times \vec{b}| = \sqrt{93}$

Thus,

$\rm :\longmapsto\:Unit \: vector \: perpendicular \: to \: \vec{a} \: and \: \vec{b} \: is \:$

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

$\rm \:  =  \:  \: \dfrac{8\hat{i} - 2\hat{j} + 5\hat{k}}{ \sqrt{93} }$

$\rm \:  =  \:  \: \dfrac{8}{ \sqrt{93} } \hat{i} - \dfrac{2}{ \sqrt{93} } \hat{j} + \dfrac{5}{ \sqrt{93} } \hat{k}$

Additional information :-

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

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

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

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

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

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

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