Question
If $\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+\hat{\text{k}},\vec{\text{c}}=2\hat{\text{j}}-\hat{\text{k}}$are three vectors, find the aera of the parallelogram having diagonals $\big(\vec{\text{a}}+\vec{\text{b}}\big)$ and $\big(\vec{\text{b}}+\vec{\text{c}}\big).$

Answer

It is given that $\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+\hat{\text{k}},\vec{\text{c}}=2\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{a}}+\vec{\text{b}}=\big(2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}\big)+\big(-\hat{\text{i}}+\hat{\text{k}}\big)=\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{b}}+\vec{\text{c}}=\big(-\hat{\text{i}}+\hat{\text{k}}\big)+\big(2\hat{\text{j}}-\hat{\text{k}}\big)=-\hat{\text{i}}+2\hat{\text{j}}$
WE know that the area of parallelogram $\frac{1}{2}\big|\vec{\text{d}}_1\times\vec{\text{d}}_2\big|$ where $\vec{\text{d}}_1$ and $\vec{\text{d}}_2$ are the diagonal vectors.
Now,
$\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{b}}+\vec{\text{c}}\big)=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-3&2\\-1&2&0 \end{vmatrix}=-4\hat{\text{i}}-2\hat{\text{j}}-\hat {\text{k}}$
$\therefore$ Area of the parallelogram having diagonals $\big(\vec{\text{a}}+\vec{\text{b}}\big)$ and $\big(\vec{\text{b}}+\vec{\text{c}}\big)$
$=\frac{1}{2}\big|\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{b}}+\vec{\text{c}}\big)\big|=\frac{1}{2}\big|-4\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}}\big|$
$=\frac{1}{2}\sqrt{(-4)^2+(-2)^2+(-1)^2}$
$=\frac{\sqrt{21}}{2}\text{square units}$

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