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If \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=0, \text { show that } \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} . Interpret the result geometrically?

 

Answers (1)

Given that,

\begin{aligned} &\vec{a}+\vec{b}+\vec{c}=0\\ &\text { Find the value of } \overrightarrow{\mathrm{b}}\\ &\Rightarrow \overrightarrow{\mathrm{b}}=-\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}} \end{aligned}

\\ \text { Take } \vec{a} \times \vec{b} \\ \vec{a} \times \vec{b}=\vec{a} \times(-\vec{a}-\vec{c}) \\ \Rightarrow \vec{a} \times \vec{b}=(-\vec{a} \times \vec{a})-\vec{a} \times \vec{c} \\ \Rightarrow \vec{a} \times \vec{b}=0-\vec{a} \times \vec{c}

\\ {[\because \vec{a} \times \vec{a}=0]} \\ \Rightarrow \vec{a} \times \vec{b}=-\vec{a} \times \vec{c} \\ \Rightarrow \vec{a} \times \vec{b}=\vec{c} \times \vec{a}_{\ldots(i)}

\\ \text { [\% by anti-commutative law, }-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}_{1} \\ \text { Now, take } \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}} \text { . } \\ \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=(-\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{c}} \\ \Rightarrow \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{c}}

\\ \qquad\left[\because \overrightarrow{\mathrm{C}} \times \overrightarrow{\mathrm{c}}=0\right]. \\ \Rightarrow \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}-0 \\ \Rightarrow \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}} \\ \Rightarrow \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}...(ii)

\\ \begin{aligned} &\left[\because\right. \text { by anti-commutative law, }-\vec{a} \times \vec{c}=\vec{c} \times \vec{a}_{]}\\ &\text { From equations (i) and (ii), we have }\\ &\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}\\ &\Rightarrow \vec{a} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} \end{aligned}

Now, let us interpret the result graphically.

Let there be a parallelogram ABCD.

 

Here, https://gradeup-question-images.grdp.co/liveData/PROJ28891/1554370624968879.png  and https://gradeup-question-images.grdp.co/liveData/PROJ28891/1554370625718817.png .

And AB and AD sides are making angle θ between them.

Area of parallelogram is given by,

Area of parallelogram = Base × Height

So from the diagram, area of parallelogram ABCD can be written as,

\\ \begin{aligned} &\text { Area of parallelogram }=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \sin \theta\\ &\text { Or, }\\ &\text { Area of parallelogram }=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \end{aligned}

Since, parallelogram on the same base and between the same parallels are equal in area, so we have

\\ $$ |\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}|=|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}| $$\\ $This also implies that, $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}$\\ $ Thus, $ it is represented graphically.

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