### Vector Algebra Question 21

#### Question 21 - 01 February - Shift 1

$A(2,6,2), B(-4,0, \lambda), C(2,3,-1)$ 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

## Show Answer

#### Answer: 11

#### Solution:

#### Formula: Vector products of two vectors; Area of quadrilateral.

$A(2,6,2) \quad B(-4,0, \lambda), C(2,3,-1) D(4,5,0)$

Area $=\frac{1}{2}|\overrightarrow{{}B D} \times \overrightarrow{{}A C}|=18$

$\overrightarrow{{}A C} \times \overrightarrow{{}B D}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -3 & -3 \\ 8 & 5 & -\lambda\end{vmatrix} $

$=(3 \lambda+15) \hat{i}-\hat{j}(-24)+\hat{k}(-24)$

$\overrightarrow{{}A C} \times \overrightarrow{{}B D}=(3 \lambda+15) \hat{i}+24 \hat{j}-24 \hat{k}$

$=\sqrt{(3 \lambda+15)^{2}+(24)^{2}+(24)^{2}}=36$

$=\lambda^{2}+10 \lambda+9=0$

$=\lambda=-1,-9$

$|\lambda| \leq 5 \Rightarrow \lambda=-1$

$5-6 \lambda=5-6(-1)=11$