Question

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

Answer 1

Correct Answer: 11

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

Answer 2

Using the area formula for a quadrilateral in 3D: \[ \text{Area} = \frac{1}{2} \left| \overrightarrow{BD} \times \overrightarrow{AC} \right| \] Computing cross product: \[ \overrightarrow{AC} \times \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 
0 & -3 & -3 
8 & 5 & -\lambda \end{vmatrix} \] Expanding determinant: \[ = (3\lambda + 15) \hat{i} - (-24) \hat{j} + (-24) \hat{k} \] \[ = (3\lambda + 15) \hat{i} + 24\hat{j} - 24\hat{k} \] \[ \left| (3\lambda + 15)^2 + 24^2 + (-24)^2 \right| = 36 \] \[ \lambda^2 + 10\lambda + 9 = 0 \] Solving for \( \lambda \): \[ \lambda = -1, -9 \] Since \( |\lambda| \leq 5 \), we take \( \lambda = -1 \), \[ 5 - 6(-1) = 11 \]