Question

Find the absolute value of parameter $t$ for which the area of the triangle whose vertices are $A(-1,1,2)$; $B(1,2,3)$ and $C(t,1,1)$ is minimum.

Answer 1

Answer: 2

$\overrightarrow{AB}=2\hat{i}+\hat{j}+\hat{k},\overrightarrow{AC}=(t+1)\hat{i}+0\hat{j}-\hat{k}$
$\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& 1\\ t+1& 0& -1\end{array}\right|$
$=-\hat{i}+(t+3)\hat{j}-(t+1)\hat{k}$
$=\sqrt{1+(t+3{)}^2+(t+1{)}^2}$
$=\sqrt{2t^2+8t+11}$
Area of $\Delta ABC=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$
$=\frac{1}{2}\sqrt{2t^2+8t+1}$
let $f(t){\Delta }^2=\frac{1}{4}\left(2t^2+8t+1\right)$
$f^{\prime }(t)=0$
$\Rightarrow t=-2$
At $t=-2.f^{\prime \prime }(t)>0$
so $\Delta$ is minimum at $t=-2$