Question

Find a unit vector perpendicular to each of the vector a +b and a - b , where
ā= 3î + 29 +2k and =î +2j-2k .​

Answer 1

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

Given that

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

and

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

Now,

Let assume that,

$\rm :\longmapsto\:\vec{c} = \vec{a} + \vec{b}$

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

$\rm \:  =  \:  \: 4\hat{i} + 4\hat{j}$

Let assume that,

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

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

$\rm \:  =  \:  \: 2\hat{i} + 4\hat{k}$

Consider,

$\rm :\longmapsto\:\vec{c} \times \vec{d}$

$\rm \: = \: \: \: \begin{gathered}\sf \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\4&4&0\\2&0& 4\end{array}\right | \end{gathered}$

$\rm \:  =  \:  \: \hat{i}(16 - 0) - \hat{j}(16 - 0) + \hat{k}(0 - 8)$

$\rm \:  =  \:  \: 16\hat{i} -16 \hat{j} - 8 \hat{k}$

Consider,

$\rm :\longmapsto\: |\vec{c} \times \vec{d}|$

$\rm \:  =  \:  \: |16\hat{i} - 16\hat{j} - 8\hat{k}|$

$\rm \:  =  \:  \: \sqrt{ {(16)}^{2} + {( - 16)}^{2} + {( - 8)}^{2} }$

$\rm \:  =  \:  \: \sqrt{256 + 256 + 64}$

$\rm \:  =  \:  \: \sqrt{576}$

$\rm \:  =  \:  \: 24$

Now,

$\rm :\longmapsto\:Unit \: vector \: \perp \: to \: both \: \vec{c} \: and \: \vec{d} \: is \:$

$\rm \:  =  \:  \: \dfrac{\vec{c} \times \vec{d}}{ |\vec{c} \times \vec{d}| }$

$\rm \:  =  \:  \: \dfrac{16\hat{i} - 16\hat{j} - 8\hat{k}}{24}$

$\rm \:  =  \:  \: \dfrac{2}{3}\hat{i}- \dfrac{2}{3}\hat{j} - \dfrac{1}{3}\hat{k}$

Additional Information :-

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

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

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

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

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

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

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