Question

Given that $\overrightarrow{a}=\hat{i}-\hat{j}$ and $\overrightarrow{b}=\hat{i}+2\hat{j}$ find a unit vector co-planar with $\hat{a}$ and $\hat{b}$ and perpendicular to $\hat{a}$

Answer 1

$Lettherequiredvector,$

$\overline{r}=a1\stackrel{\wedge }{i}+a2\stackrel{\wedge }{j}+a3\stackrel{\wedge }{k}$

$\overline{r}.\overline{a}=0implies(a1\stackrel{\wedge }{i}+a2\stackrel{\wedge }{j}+a3\stackrel{\wedge }{k})(\stackrel{\wedge }{i}-\stackrel{\wedge }{j})=0$

$⟹a1-a2=0impliesa1=a2$

$ \overset { \wedge }{ a } =\cfrac { \overset { \wedge }{ i } -\overset { \wedge }{ j } }{ \sqrt { 2 } } $

$\stackrel{\wedge }{b}=\frac{\stackrel{\wedge }{i}-\stackrel{\wedge }{j}}{\sqrt{5}}$

$\left|\begin{array}{ccc}a1& a2& a3\\ \frac{1}{\sqrt{2}}& \frac{-1}{\sqrt{2}}& 0\\ \frac{1}{\sqrt{5}}& \frac{2}{\sqrt{5}}& 0\end{array}\right|=0$

$⟹a1\left(0\right)-a2\left(0\right)+a3(\frac{2}{\sqrt{10}}+\frac{1}{\sqrt{10}})=0$

$⟹\frac{3a3}{\sqrt{10}}=0$

$⟹a3=0$

$\therefore \overline{r}=a1\stackrel{\wedge }{i}+a1\stackrel{\wedge }{j}=a1(\stackrel{\wedge }{i}+\stackrel{\wedge }{j})$

$\because \overline{r}isaunitvector.$

$⟹a1=\frac{1}{\sqrt{2}}$

$\therefore \overline{r}=\frac{\stackrel{\wedge }{i}+\stackrel{\wedge }{j}}{\sqrt{2}}$