If the sum of two-unit vectors $\overrightarrow{\mathrm{a}}\mathrm{ }\mathrm{and}\mathrm{ }\overrightarrow{\mathrm{b}}$ is a unit vector, show that the magnitude of their difference is $\sqrt{3}$

OR

Show that area of the parallelogram whose diagonal are given by $\overrightarrow{\mathrm{a}}\mathrm{ }\mathrm{and}\mathrm{ }\overrightarrow{\mathrm{b}}$ is $\frac{\left|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\right|}{2}$ . Also find the area of the parallelogram whose diagonals are $2\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+\widehat{\mathrm{k}}\mathrm{ }\mathrm{and}\mathrm{ }\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}-\widehat{\mathrm{k}}$.

Assume that,

$\overrightarrow{c}=\hat{a}+\hat{b}$

Then, according to given condition $\overrightarrow{c}$ is a unit vector, i.e. $\left|\overrightarrow{c}\right|=1$.

To show $|\hat{a}-\hat{b}|=\sqrt{3}$

Consider, $\overrightarrow{c}=\hat{a}+\hat{b}$

$$\begin{gathered} \Rightarrow \mathrm{ }\left|\overrightarrow{c}\right|=|\hat{a}+\hat{b}| \\ \Rightarrow \mathrm{ }1=|\hat{a}+\hat{b}| \\ \Rightarrow |\hat{a}+\hat{b}{|}^2=1 \end{gathered}$$

$$\begin{gathered} \Rightarrow \mathrm{ }(\stackrel{\mathrm{ˆ}}{\mathrm{a}}+\stackrel{\mathrm{ˆ}}{\mathrm{b}})\cdot (\stackrel{\mathrm{ˆ}}{\mathrm{a}}+\stackrel{\mathrm{ˆ}}{\mathrm{b}})=1 \Rightarrow |\stackrel{\mathrm{ˆ}}{\mathrm{a}}{|}^2+2\stackrel{\mathrm{ˆ}}{\mathrm{a}}\cdot \stackrel{\mathrm{ˆ}}{\mathrm{b}}+|\stackrel{\mathrm{ˆ}}{\mathrm{b}}{|}^2=1 \\ \Rightarrow 1+2\stackrel{\mathrm{ˆ}}{\mathrm{a}}\cdot \stackrel{\mathrm{ˆ}}{\mathrm{b}}+1=1 \\ \Rightarrow \mathrm{ }2\stackrel{\mathrm{ˆ}}{\mathrm{a}}\cdot \stackrel{\mathrm{ˆ}}{\mathrm{b}}=-1\dots \left(\mathrm{i}\right) \end{gathered}$$

Now consider,  $|\hat{a}-\hat{b}{|}^2=(\hat{a}-\hat{b})\cdot (\hat{a}-\hat{b})$

$$\begin{gathered} =|\hat{a}{|}^2-2\hat{a}\cdot \hat{b}+|\hat{b}{|}^2 \\ =1-(-1)+1=3 \end{gathered}$$

$\Rightarrow \mathrm{ }|\hat{a}-\hat{b}|=\sqrt{3}$

Which have proven.

OR

Assume that ABCD be a parallelogram such that,

$\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{p}},\overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{q}}\Rightarrow \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{q}}$

$\therefore \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{a}}\mathrm{ }\dots .$(i) [By triangle law of vectors]

And $\overrightarrow{\mathrm{BD}}=\overrightarrow{\mathrm{BA}}+\overrightarrow{\mathrm{AD}}=-\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{b}}$

Adding (i) and (ii), we get

$\bar{a}+\bar{b}=2\bar{q}$

Subtracting (ii) from Eq. (i), we get

$$\begin{gathered} \overrightarrow{a}-\overrightarrow{b}=2\overrightarrow{p} \\ \Rightarrow \overrightarrow{p}=\frac{1}{2}(\overrightarrow{a}-\overrightarrow{b}) \end{gathered}$$

Now, $\overrightarrow{\mathrm{p}}\times \overrightarrow{\mathrm{q}}=\frac{1}{4}(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})\times (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})$

$=\frac{1}{4}(\bar{a}\times \bar{a}+\bar{a}\times \bar{b}-\overrightarrow{b}\times \bar{a}-\bar{b}\times \bar{b})$

So, the  area of a parallelogram $\mathrm{ABCD}=|\overrightarrow{p}\times \overrightarrow{q}|=\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|$

Now, area of a parallelogram, whose diagonals are $2\mathrm{i}-\mathrm{j}+\mathrm{k} \mathrm{and} \mathrm{i}+3\mathrm{j}-\dot{\mathrm{k}}$

$=\frac{1}{2}\left|\right(2\mathrm{i}-\mathrm{j}+\mathrm{k})\times (\mathrm{i}+3\mathrm{j}-\mathrm{k}\left)\right|$

$=\frac{1}{2}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -1& 1\\ 1& 3& -1\end{array}\right|$

$=\frac{1}{2}\left|i\right(1-3)-j(-2-1)+k(6+1\left)\right|$

$=\frac{1}{2}|-2\stackrel{\mathrm{ˆ}}{\mathrm{i}}+3\stackrel{\mathrm{ˆ}}{\mathrm{j}}+7\stackrel{\mathrm{ˆ}}{\mathrm{k}}|=\frac{1}{2}\sqrt{4+9+49}=\frac{1}{2}\sqrt{62} \mathrm{sq}.\mathrm{units}$Therefore, the area of parallelogram ABCD is $\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|$ have proven and the Area of parallelogram is $\frac{1}{2}\sqrt{62}\mathrm{unit}{\mathrm{ }}^2$