Question

In a $△ABC,AB=ri+j,AC=si-j$ if the area of triangle is of unit magnitude, then

• A. $|r-s|=2$
• B. $|r+s|=1$
• C. $|r+s|=2$
• D. $|r-s|=1$

Answer 1

The correct option is C $|r+s|=2$

$\overrightarrow{AB}=r\hat{i}+\hat{j}$
$\overrightarrow{AC}=s\hat{i}-\hat{j}$
Area of $△ABC,A=\frac{1}{2}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ r& 1& 0\\ s& -1& 0\end{array}\right|$
$A=\frac{1}{2}\left[\hat{i}\right(0+0)-\hat{j}(0-0)+\hat{k}(-r-s\left)\right]$
$\Rightarrow A=\left(\frac{-r-s}{2}\right)\hat{k}$
Given, magnitude of area $|A|=1$
$\Rightarrow \left|\frac{-r-s}{2}\right|=1$
$\Rightarrow |r+s|=2$