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

$\overrightarrow{ AB }=2 \hat{ i }+\hat{ j }+\hat{ k }, \overrightarrow{ AC }=( t +1) \hat{ i }+0 \hat{ j }-\hat{ k }$
$\overrightarrow{ AB } \times \overrightarrow{ AC }=\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 1 & 1 \\ t +1 & 0 & -1\end{vmatrix}$
$= -\hat{i}+(t+3) \hat{j}-(t+1) \hat{k} $
$ = \sqrt{1+(t+3)^{2}+(t+1)^{2}} $
$ = \sqrt{2 t^{2}+8 t+11} $
Area of $\Delta ABC =\frac{1}{2}|\overrightarrow{ AB } \times \overrightarrow{ AC }|$
$=\frac{1}{2} \sqrt{2 t^{2}+8 t+1}$
let $f(t) \Delta^{2}=\frac{1}{4}\left(2 t^{2}+8 t+1\right) $
$f'(t)=0 $
$\Rightarrow t=-2 $
At $ t=-2 . f''(t)>0$
so $\Delta$ is minimum at $t =-2$