The two adjacent sides of a parallelogram are $2\hat{\mathrm{i}}-4\hat{\mathrm{j}}-5\hat{\mathrm{k}}\text{ and }2\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}$ Find the two-unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram.

OR

Find the area of the triangle whose vertices are $\mathrm{P}(-1,2,-1),\mathrm{Q}(3,-1,2)$ and $\mathrm{R}(2,3,-1)$

Let ABCD be the given parallelogram with
$\overrightarrow{\mathrm{AB}}=2\hat{\mathrm{i}}-4\hat{\mathrm{j}}-5\hat{\mathrm{k}}\text{ and }\overrightarrow{\mathrm{AD}}=2\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}$
Clearly, the diagonal $\overrightarrow{\mathrm{AC}}\text{ is given by }\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AD}}$
$=4\hat{\mathrm{i}}-2\hat{\mathrm{j}}-2\hat{\mathrm{k}}$
and the diagonal $\overrightarrow{\mathrm{BD}}\text{ is given by }\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{BA}}$
$=\overrightarrow{\mathrm{AD}}-\overrightarrow{\mathrm{AB}}=6\hat{\mathrm{j}}+8\hat{\mathrm{k}}$
[using parallelogram law of addition]

Now, the unit vector along $\overrightarrow{\mathrm{AC}}$ is given by
$\begin{array}{l}\frac{\overrightarrow{\mathrm{AC}}}{\left|\overrightarrow{\mathrm{AC}}\right|}=\frac{4\widehat{\mathrm{i}}-2\widehat{\mathrm{j}}-2\widehat{\mathrm{k}}}{\sqrt{16+4+4}}\\ =\frac{4\widehat{\mathrm{i}}-2\widehat{\mathrm{j}}-2\widehat{\mathrm{k}}}{\sqrt{24}}\\ =\frac{4\widehat{\mathrm{i}}-2\widehat{\mathrm{j}}-2\widehat{\mathrm{k}}}{2\sqrt{6}}\\ =\frac{1}{\sqrt{6}}(2\widehat{\mathrm{i}}-\widehat{\mathrm{j}}-\widehat{\mathrm{k}})\end{array}$
and the unit vector along $\overrightarrow{\mathrm{BD}}$ is given by
$\frac{\overrightarrow{\mathrm{BD}}}{\left|\overrightarrow{\mathrm{BD}}\right|}=\frac{6\widehat{\mathrm{j}}+8\widehat{\mathrm{k}}}{\sqrt{36+64}}=\frac{6\widehat{\mathrm{j}}+8\widehat{\mathrm{k}}}{10}=\frac{1}{5}(3\widehat{\mathrm{j}}+4\widehat{\mathrm{k}})$
Now, area of parallelogram ABCD
$\begin{array}{r}=\frac{1}{2}|\overrightarrow{\mathrm{AC}}\times \overrightarrow{\mathrm{BD}}|\\ \mathrm{Here}, \overrightarrow{\mathrm{AC}}\times \overrightarrow{\mathrm{BD}}=\left|\begin{array}{ccc}\widehat{\mathrm{i}}& \widehat{\mathrm{j}}& \widehat{\mathrm{k}}\\ 4& -2& -2\\ 0& 6& 8\end{array}\right|\end{array}$
$\begin{array}{l}=\widehat{\mathrm{i}}(-16+12)-\widehat{\mathrm{j}}(32-0)+\widehat{\mathrm{k}}(24-0)\\ =-4\widehat{\mathrm{i}}-32\widehat{\mathrm{j}}+24\widehat{\mathrm{k}}\\ \text{ and }|\overrightarrow{\mathrm{AC}}\times \overrightarrow{\mathrm{BD}}|=\sqrt{(-4{)}^2+(-32{)}^2+(24{)}^2}\\ =\sqrt{4^2\left(1+8^2+6^2\right)}\\ =4\sqrt{1+64+36}=4\sqrt{101}\end{array}$
$\therefore$ Area of parallelogram ABCD $$\begin{gathered} =\frac{1}{2}\times 4\sqrt{101} \\ =2\sqrt{101}\text{ sq units } \end{gathered}$$

OR

Let $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \mathrm{and} \overrightarrow{\mathrm{c}}$ be the position vectors of points P, Q and R, respectively.
Then, $\overrightarrow{\mathrm{a}}=-\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}-\widehat{\mathrm{k}},\overrightarrow{\mathrm{b}}=3\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+2\widehat{\mathrm{k}}\text{ and }\overrightarrow{\mathrm{c}}=2\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}-\widehat{\mathrm{k}}$
Clearly, the area of $\mathrm{\Delta PQR}=\frac{1}{2}|\overrightarrow{\mathrm{PQ}}\times \overrightarrow{\mathrm{PR}}|$
Now, $\overrightarrow{\mathrm{PQ}}=$ Position vector of Q - Position vector of P
$\begin{array}{l}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}=(3\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+2\widehat{\mathrm{k}})-(-\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}-\widehat{\mathrm{k}})\\ =4\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}+3\widehat{\mathrm{k}}\end{array}$
$\overrightarrow{\mathrm{PR}}=$ Position vector of R - Position vector of P
$=\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}=(2\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}-\widehat{\mathrm{k}})-(-\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}-\widehat{\mathrm{k}})=3\widehat{\mathrm{i}}+\widehat{\mathrm{j}}$
$\begin{array}{l}\therefore \overrightarrow{\mathrm{PQ}}\times \overrightarrow{\mathrm{PR}}=\left|\begin{array}{ccc}\overrightarrow{\mathrm{i}}& \widehat{\mathrm{j}}& \widehat{\mathrm{k}}\\ 4& -3& 3\\ 3& 1& 0\end{array}\right|\\ =(0-3)\widehat{\mathrm{i}}-(0-9)\widehat{\mathrm{j}}+(4+9)\widehat{\mathrm{k}}\\ =-3\widehat{\mathrm{i}}+9\widehat{\mathrm{j}}+13\widehat{\mathrm{k}}\\ \text{ and }|\overrightarrow{\mathrm{PQ}}\times \overrightarrow{\mathrm{PR}}|=\sqrt{(-3{)}^2+(9{)}^2+(13{)}^2}\\ =\sqrt{9+81+169}\\ =\sqrt{259}\end{array}$
$\begin{array}{r}So area of \mathrm{\Delta PQR}=\frac{1}{2}|\overrightarrow{\mathrm{PQ}}\times \overrightarrow{\mathrm{PR}}|\\ =\frac{1}{2}\sqrt{259}\text{ sq units }\end{array}$