Question

Find a unit vector perpendicular to the plane containing the vectors
a=2i+j+k and b=i+2j+k​

Answer 1

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

Given that

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

And

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

We have to find

$\rm :\longmapsto\:A \: unit \: vector \: \perp \: to \: both \: \vec{a} \: and \: \vec{b}$

$\rm :\longmapsto\:Let \: \vec{c} \: \: is \: \perp \: to \: both \: \vec{a} \: and \: \vec{b}$

Then,

$\rm :\longmapsto\:\hat{c} = \dfrac{\vec{a} \times \vec{b}}{ |\vec{a} \times \vec{b}| } - - - (1)$

Now,

Consider,

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

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

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

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

$\bf\implies \: \: \vec{a} \times \vec{b} \: = \: - \hat{i} - \hat{j} + 3\hat{k}$

Now,

Consider,

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

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

$\rm \: = \: \: \: \sqrt{ {( - 1)}^{2} + {( - 1)}^{2} + {(3)}^{2} }$

$\rm \: = \: \: \: \sqrt{1 + 1 + 9}$

$\rm \: = \: \: \: \sqrt{11}$

Hence,

On substituting the above values in equation (1), we get

$\rm :\longmapsto\:\hat{c} = \dfrac{ - \hat{i} - \hat{j} + 3\hat{k}}{ \sqrt{11} }$

$\bf\implies \:\hat{c} = \: - \dfrac{1}{ \sqrt{11} }\hat{i} - \dfrac{1}{ \sqrt{11} }\hat{j} + \dfrac{3}{ \sqrt{11} }\hat{k}$

Additional information :-

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

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

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

$\rm :\longmapsto\:\vec{c} = \vec{a} \times \vec{b} \implies \: \vec{c} \: \perp \: \vec{a} \: and \: \vec{c} \: \perp \: \vec{b}$