Vector Algebra — Class 12 Maths Solution

Since, $\quad \vec a + \vec b + \vec c = 0$

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

Now, $\vec a \times \vec b = \vec a \times ( - \vec c - \vec a)$

$= \vec a \times ( - \vec c) + \vec a \times ( - \vec a) = - \vec a \times \vec c$

$\Rightarrow$ $\quad \vec a \times \vec b = \vec c \times \vec a$ ……..(i)
Also, $\overrightarrow {\rm{b}} \times \overrightarrow {\rm{c}} = ( - \overrightarrow {\rm{c}} - \overrightarrow {\rm{a}} ) \times \overrightarrow {\rm{c}}$

$= ( - \vec c \times \vec c) + ( - \vec a \times \vec c) = - \vec a \times \vec c$

$\Rightarrow$ $\overrightarrow {\rm{b}} \times \overrightarrow {\rm{c}} = \overrightarrow {\rm{c}} \times \overrightarrow {\rm{a}}$ ……….(ii)

From Eqs. (i) and (ii), $\quad \vec a \times \vec b = \vec b \times \vec c = \vec c \times \vec a$

Geometrical interpretation of the result

If $ABCD$ is a parallelogram such that $\overrightarrow {AB}$

$= \vec a$

and $\overrightarrow {AD} = \vec b$

and these adjacent sides are making angle $\theta$ between each other,

Area of parallelogram $ABCD = |\overrightarrow {\rm{a}} ||\overrightarrow {\rm{b}} ||\sin \theta | = |\overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} |$

Since, parallelogram on the same base and between the same parallels are equal in area.

We can say that, $\quad |\overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} | = |\overrightarrow {\rm{a}} \times \overrightarrow {\rm{c}} | = |\overrightarrow {\rm{b}} \times \overrightarrow {\rm{c}} |$

This further implies that, $\overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} = \overrightarrow {\rm{a}} \times \overrightarrow {\rm{c}} = \overrightarrow {\rm{b}} \times \overrightarrow {\rm{c}}$

So, area of the parallelograms formed by taking any two sides represented by $\overrightarrow {\rm{a}} ,\overrightarrow {\rm{b}}$ and $\overrightarrow {\rm{c}}$

as adjacent are equal.