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

 

Answers (1)

We have,

Let ABCD be a parallelogram.

In ABCD,

\\ \mathrm{AB}=\overrightarrow{\mathrm{p}} \\ \mathrm{AD}=\overrightarrow{\mathrm{q}} $$

And since, AD || BC
$\mathrm{So}, \mathrm{BC}=\overrightarrow{\mathrm{q}}$
We need to show that,

\\ $ Area of parallelogram $ \mathrm{ABCD}=\frac{|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|}{2}

\\$Where, \overrightarrow{\mathrm{a}} and \overrightarrow{\mathrm{b}} \text{are diagonals of the parallelogram} \mathrm{ABCD}\\ \text{Now, by triangle law of addition, we get} \\\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}

 

 

\\ \Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}} \\ \Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{a}}(\mathrm{say})_{\ldots(\mathrm{i})} \\ \Rightarrow \overrightarrow{\mathrm{BD}}=\overrightarrow{\mathrm{BA}}+\overrightarrow{\mathrm{AD}} \\ \text { Similarly, } \\ \Rightarrow \overrightarrow{\mathrm{BD}}=-\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}

\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{BD}}=\overrightarrow{\mathrm{b}}(\mathrm{say}) \ldots(\mathrm{ii})\\ &\text { Adding equations (i) and (ii), we get }\\ &\vec{a}+\vec{b}=(\vec{p}+\vec{q})+(-\vec{p}+\vec{q}) \end{aligned}

\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{q}}\\ &\Rightarrow \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=2 \overrightarrow{\mathrm{q}}\\ &\Rightarrow \overrightarrow{\mathrm{q}}=\frac{1}{2}(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\\ &\text { And, }\\ &\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}=(\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}})-(-\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}) \end{aligned}

\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{q}}\\ &\Rightarrow \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}=2 \overrightarrow{\mathrm{p}}\\ &\Rightarrow \overrightarrow{\mathrm{p}}=\frac{1}{2}(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})\\ &\text { Now, } \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}} \text { can be written as, } \end{aligned}

\\ \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left(\frac{1}{2}(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})\right) \times\left(\frac{1}{2}(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\right) \\ \Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\frac{1}{4}(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \\ \Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\frac{1}{4}(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{b}})

\\ \\ \quad\left[\because \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{b}}=0_{\text {and }}-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\right]

\begin{array}{l} \Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\frac{2}{4}(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \\ \Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\frac{1}{2}(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \end{array}

We know that,Vector area of parallelogram ABCD is given by,
 

Area of parallelogram ABCD \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}

\\ =\frac{1}{2}|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \\ =\frac{|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|}{2}

Hence, shown.

Now, we need to find the area of parallelogram whose diagonals are  2 \hat{\imath}-\hat{\jmath}+\hat{k}_{\text {and }} \hat{\imath}+3 \hat{\jmath}-\hat{k}

We have already derived the relationship between area of parallelogram and diagonals of parallelogram, which is

\\ \begin{aligned} &\text { Area of parallelogram }=\frac{|\vec{a} \times \vec{b}|}{2}\\ &\text { Here, } \vec{a}=2 \hat{\imath}-\hat{\jmath}+\hat{k}\\ &\text { And, } \overrightarrow{\mathrm{b}}=\hat{\imath}+3 \hat{\jmath}-\hat{\mathrm{k}}\\ &\Rightarrow \text { Area of parallelogram }=\frac{|(2 \hat{i}-\hat{j}+\hat{k}) \times(\hat{i}+3 \hat{\jmath}-\hat{k})|}{2} \end{aligned}

\begin{array}{l} =\frac{1}{2}|| \begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{\mathrm{k}} \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{array}|| \\ =\frac{1}{2}|[\hat{\imath}((-1)(-1)-(1)(3))-\hat{\jmath}((2)(-1)-(1)(1))+\hat{\mathrm{k}}((2)(3)-(-1)(1))]| \\ =\frac{1}{2}|[\hat{\imath}(1-3)-\hat{\jmath}(-2-1)+\hat{\mathrm{k}}(6+1)]| \\ =\frac{1}{2}|-2 \hat{\imath}+3 \hat{\jmath}+7 \hat{\mathrm{k}}| \end{array}

\\ \begin{aligned} &=\frac{1}{2} \sqrt{(-2)^{2}+3^{2}+7^{2}}\\ &=\frac{1}{2} \sqrt{4+9+49}\\ &=\frac{1}{2} \sqrt{62}\\ &\text { Thus, area of required parallelogram is } \frac{1}{2} \sqrt{62} \text { sq units } \end{aligned}

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