If $\overrightarrow {\rm{a}} ,\overrightarrow {\rm{b}}$ and $\overrightarrow {\rm{c}}$ determine the vertices of a triangle, show that $\frac{1}{2}[\overrightarrow {\rm{b}} \times \overrightarrow {\rm{c}} + \overrightarrow {\rm{c}} \times \overrightarrow {\rm{a}} + \overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} ]$ gives the vector area of the triangle. Hence, deduce the condition that the three points $\overrightarrow {\rm{a}} ,\overrightarrow {\rm{b}}$ and $\overrightarrow {\rm{c}}$ are collinear. Also, find the unit vector normal to the plane of the triangle.

(i) If $\overrightarrow {\rm{a}} ,\overrightarrow {\rm{b}}$ and $\overrightarrow {\rm{c}}$ are

collinear, then the area of the triangle formed by the vectors will be zero.

(ii) As we know, $\overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} = |\overrightarrow {\rm{a}} ||\overrightarrow {\rm{b}} |\sin \theta \widehat {\rm{n}}$

Since, $\overrightarrow {\rm{a}} ,\overrightarrow {\rm{b}}$ and $\overrightarrow {\rm{c}}$ are the vertices of a $\Delta ABC$ as shown.

$\therefore$ Area of $\Delta ABC = \frac{1}{2}|\overrightarrow {AB} \times \overrightarrow {AC} |$

Now, $\overrightarrow {AB} = \vec b - \vec a$ and $\overrightarrow {AC} = \vec c - \vec a$

$\therefore$ Area of $\Delta ABC = \frac{1}{2}|\vec b - \vec a \times \vec c - \vec a|$

$= \frac{1}{2}|\overrightarrow {\rm{b}} \times \overrightarrow {\rm{c}} - \overrightarrow {\rm{b}} \times \overrightarrow {\rm{a}} - \overrightarrow {\rm{a}} \times \overrightarrow {\rm{c}} + \overrightarrow {\rm{a}} \times \overrightarrow {\rm{a}} |$

$= \frac{1}{2}|\overrightarrow {\rm{b}} \times \overrightarrow {\rm{c}} + \overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} + \overrightarrow {\rm{c}} \times \overrightarrow {\rm{a}} + \vec 0|$

$= \frac{1}{2}|\overrightarrow {\rm{b}} \times \overrightarrow {\rm{c}} + \overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} + \overrightarrow {\rm{c}} \times \overrightarrow {\rm{a}} |$ ……(i).

For three points to be collinear, area of the $\Delta ABC$ should be equal to zero.

$\Rightarrow$ $\frac{1}{2}[\overrightarrow {\rm{b}} \times \overrightarrow {\rm{c}} + \overrightarrow {\rm{c}} \times \overrightarrow {\rm{a}} + \overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} ] = 0$

$\Rightarrow$ $\overrightarrow {\rm{b}} \times \overrightarrow {\rm{c}} + \overrightarrow {\rm{c}} \times \overrightarrow {\rm{a}} + \overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} = 0$ …..(ii)

This is the required condition for collinearity of three points $\overrightarrow {\rm{a}} ,\overrightarrow {\rm{b}}$ and $\overrightarrow {\rm{c}}$.

Let $\hat n$ be the unit vector normal to the plane of the $\Delta ABC$.
$\therefore$ $\widehat {\rm{n}} = \frac{{\overrightarrow {{\rm{AB}}} \times \overrightarrow {{\rm{AC}}} }}{{|\overrightarrow {{\rm{AB}}} \times \overrightarrow {{\rm{AC}}} |}}$

$= \frac{{\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a}}{{|\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a|}}$