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  • If a +b +c =0, show that a x b =b xc =c xa . Interpret the result geometrically?

Q: 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)

Solutiom
Given that,

$$
\vec{a}+\vec{b}+\vec{c}=0
$$


Find the value of $\vec{b}$

$$
\Rightarrow \vec{b}=-\vec{a}-\vec{c}
$$


Take $\vec{a} \times \vec{b}$

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

[\% by anti-commutative law, $-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}_1$ Now, take $\vec{b} \times \vec{c}$.

$$
\begin{aligned}
& \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=(-\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{c}} \\
& \Rightarrow \overrightarrow{\mathrm{~b}} \times \overrightarrow{\mathrm{c}}=-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{c}} \\
& {[\because \overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{c}}=0]} \\
& \Rightarrow \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}-0 \\
& \Rightarrow \overrightarrow{\mathrm{~b}} \times \overrightarrow{\mathrm{c}}=-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}} \\
& \Rightarrow \overrightarrow{\mathrm{~b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} \ldots(i i)
\end{aligned}
$$
 

$[\because$ by anti-commutative law, $-\vec{a} \times \vec{c}=\vec{c} \times \vec{a}]$ From equations (i) and (ii), we have

$$
\begin{aligned}
& \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} \\
& \Rightarrow \vec{a} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}
\end{aligned}
$$
 

Now, let us interpret the result graphically.

Let there be a parallelogram ABCD.

Here, $A B=\vec{a}$ and $A D=\vec{b}$.
And $A B$ and $A D$ sides are making angle $\theta$ between them.
Area of parallelogram is given by,
Area of parallelogram $=$ Base $\times$ Height
So from the diagram, area of parallelogram ABCD can be written as,
Area of parallelogram $=|\vec{a}||\vec{b}| \sin \theta$
Or,
Area of parallelogram $=|\vec{a} \times \vec{b}|$

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, $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{a} \times \vec{c}$
Thus, itis represented graphically.

Posted by

infoexpert22

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