Question

In a $\Delta ABC$ , the sides $BC,CA$ and $AB$ are consecutive positive integers in increasing order. Let $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are position vectors of the vertices $A,B$ and $C$ respectively. If $\left(\overrightarrow{c}-\overrightarrow{a}\right)\cdot \left(\overrightarrow{b}-\overrightarrow{c}\right)=0$ , then the value of $\left|\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}\right|$ is equal to

Answer 1

Answer: 12

$\left(\overrightarrow{c}-\overrightarrow{a}\right)\cdot \left(\overrightarrow{b}-\overrightarrow{c}\right)=0\Rightarrow \overrightarrow{\mathrm{AC}}\cdot \overrightarrow{\mathrm{CB}}=0\Rightarrow \Delta ABC$ is right angle at $C.$

From diagram
$k^2+{\left(k+1\right)}^2={\left(k+2\right)}^2$
$k^2+k^2+2k+1=k^2+4k+4$
$k^2-2k-3=0\Rightarrow k=3,-1\left(\therefore k\ne -1\right)$
So, sides are $3,4,5$
Area of $\Delta ABC=\frac{1}{2}\left|\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}\right|=\frac{1}{2}\times 3\times 4$
$\Rightarrow \left|\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}\right|=12$