Question

$A(2,6,2),B(-4,0,\lambda ),C(2,3,-1)$ and $D(4,5,0),|\lambda |\le 5$ are the vertices of a quadrilateral $ABCD$. If its area is $18$ square units, then $5-6\lambda$ is equal to________

Answer 1

Answer: 11

$A(2,6,2)B(-4,0,\lambda ),C(2,3,-1)D(4,5,0)$
$\text{ Area }=\frac{1}{2}|\overrightarrow{BD}\times \overrightarrow{AC}|=18$
$\overrightarrow{AC}\times \overrightarrow{BD}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& -3& -3\\ 8& 5& -\lambda \end{array}\right|$
$=(3\lambda +15)\hat{i}-\hat{j}(-24)+\hat{k}(-24)$
$\overrightarrow{AC}\times \overrightarrow{BD}=(3\lambda +15)\hat{i}+24\hat{j}-24\hat{k}$
$=\sqrt{(3\lambda +15{)}^2+(24{)}^2+(24{)}^2}=36$
$={\lambda }^2+10\lambda +9=0$
$=\lambda =-1,-9$
$|\lambda |\le 5\Rightarrow \lambda =-1$
$5-6\lambda =5-6(-1)=11$