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} $
$=(3 \lambda+15) \hat{i}-\hat{j}(-24)+\hat{k}(-24)$
$\overrightarrow{{}A C} \times \overrightarrow{{}B D}=(3 \lambda+15) \hat{i}+24 \hat{j}-24 \hat{k}$
$=\sqrt{(3 \lambda+15)^{2}+(24)^{2}+(24)^{2}}=36$
$=\lambda^{2}+10 \lambda+9=0$
$=\lambda=-1,-9$
$|\lambda| \leq 5 \Rightarrow \lambda=-1$
$5-6 \lambda=5-6(-1)=11$
