Question

If the area(in $\text{sq.units}$) of triangle whose vertices are $(2,1,3),(a,0,2)$ and $(-3,-1,0)$ is $\frac{\sqrt{14}}{2}.$ Then the integral value of $a=$

Answer 1

Let the vertices of triangle are $A(2,1,3),B(a,0,2)$ and $C(-3,-1,0)$
$\Rightarrow \overrightarrow{AB}=(a-2)\hat{i}-\hat{j}-\hat{k}\overrightarrow{AC}=-5\hat{i}-2\hat{j}-3\hat{k}\Rightarrow \overrightarrow{AB}\times \overrightarrow{AC}=\hat{i}+(3a-1)\hat{j}-\hat{k}(2a+1)$
So, Area of triangle $=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$
$\Rightarrow \frac{\sqrt{1+(3a-1{)}^2+(2a+1{)}^2}}{2}=\frac{\sqrt{14}}{2}$
squaring both sides and simplifying,

$\Rightarrow 13a^2-2a+3=14\Rightarrow 13a^2-2a-11=0\Rightarrow (13a+11)(a-1)=0\Rightarrow a=1,-\frac{11}{13}$