Question

Find the vector area of a triangle $OAB$ where $\overrightarrow{OA}=a,\overrightarrow{OB}=b$ and they are inclined at an angle $\theta$. Also, find the vector area of a triangle whose vertices are the points $A,B$ and $C$.

• A. $\frac{1}{2}(a\times b)$
• B. $\frac{1}{2}(b\times a)$
• C. $\frac{1}{2}(a\times b+b\times c+c\times a)$
  
• D. None of these

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{1}{2}(a\times b)$
C $\frac{1}{2}(a\times b+b\times c+c\times a)$

We know that $a\times b=ab\sin {\theta }_{\hat{n}}$

Area of $△OAB=\frac{1}{2}OA.OB\sin \theta$

$=\frac{1}{2}ab\sin {\theta }_{\hat{n}}$

$\therefore a\times b=2\Delta$, where $\Delta$ is area of $△ABC$

Hence vector area of triangle $OAB$ is $\frac{1}{2}(a\times b)$

(area of parallelogram whose adjacent sides are given by a and b)

Now referred to $O$ as origin,

Let the position vectors of $A,B,C$ be $a,b$ and $c$ respectively.

Then $\overrightarrow{BC}=c-b,\overrightarrow{BA}=a-b.$

$\therefore$ Vectors area of $△ABC$ is $\frac{1}{2}\overrightarrow{BC}\times \overrightarrow{BA}$

$=\frac{1}{2}(c-b)\times (a-b)$ $=\frac{1}{2}(c\times a-b\times a-c\times b+b\times b)$ $=\frac{1}{2}(a\times b+b\times c+c\times a)$
$(\because b\times b=0$ and $-c\times b=b\times c).$