Question

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

Answer 1

$\left(\overset{ \rightarrow }{c} - \overset{ \rightarrow }{a}\right) \cdot \left(\overset{ \rightarrow }{b} - \overset{ \rightarrow }{c}\right) = 0 \Rightarrow \overset{ \rightarrow }{A C} \cdot \overset{ \rightarrow }{C B} = 0 \Rightarrow \Delta A B C$ is right angle at $C.$

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