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If $\vec{a}, \vec{b}$ and $\vec{c}$ determine the vertices of a triangle, show that $\frac{1}{2}[\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}]$ gives the vector area of the triangle. Hence, deduce the condition that the three points $\vec{a}, \vec{b}$ and $\vec{c}$ are collinear. Also, find the unit vector normal to the plane of the triangle.
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If $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ are the vertices of a $\triangle \mathrm{ABC}$ Then
Area of $\triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|$
Now, $\overline{\mathrm{AB}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}$
$\therefore$ Area of $\triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|$ $=\frac{1}{2}|\vec{b} \times \vec{c}-\vec{b} \times \vec{a}-\vec{a} \times \vec{c}+\vec{a} \times \vec{a}|$
$=\frac{1}{2}|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}|$
Points are collinear.
$$
\begin{aligned}
&\therefore \frac{1}{2}[\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}]=0 \\
&\Rightarrow \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=0
\end{aligned}
$$
This is the required condition for collinearity of three points $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$.
Let $\hat{n}$ be the unit vector normal to the plane of the $\triangle A B C$. then $\hat{\mathrm{n}}=\frac{\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}}{|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|}=\frac{\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}|}$
Area of $\triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|$
Now, $\overline{\mathrm{AB}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}$
$\therefore$ Area of $\triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|$ $=\frac{1}{2}|\vec{b} \times \vec{c}-\vec{b} \times \vec{a}-\vec{a} \times \vec{c}+\vec{a} \times \vec{a}|$
$=\frac{1}{2}|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}|$
Points are collinear.
$$
\begin{aligned}
&\therefore \frac{1}{2}[\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}]=0 \\
&\Rightarrow \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=0
\end{aligned}
$$
This is the required condition for collinearity of three points $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$.
Let $\hat{n}$ be the unit vector normal to the plane of the $\triangle A B C$. then $\hat{\mathrm{n}}=\frac{\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}}{|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|}=\frac{\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}|}$
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