JEE MAIN - Mathematics (2023 - 1st February Morning Shift - No. 21)
$$A(2,6,2), B(-4,0, \lambda), C(2,3,-1)$$ 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
11
Explanation
$$
\begin{aligned}
& \mathrm{A}(2,6,2) \quad \mathrm{B}(-4,0, \lambda), \mathrm{C}(2,3,-1) \mathrm{D}(4,5,0) \\\\
& \text { Area }=\frac{1}{2}|\overrightarrow{B D} \times \overrightarrow{A C}|=18 \\\\
& \overrightarrow{A C} \times \overrightarrow{B D}=\left|\begin{array}{ccc}
\hat{i} & j & k \\\\
0 & -3 & -3 \\\\
8 & 5 & -\lambda
\end{array}\right|
\end{aligned}
$$
$$ \begin{aligned} & =(3 \lambda+15) \hat{i}-\hat{j}(-24)+\hat{k}(-24) \\\\ & \overrightarrow{A C} \times \overrightarrow{B D}=(3 \lambda+15) \hat{i}+24 \hat{j}-24 \hat{k} \\\\ & =\sqrt{(3 \lambda+15)^2+(24)^2+(24)^2}=36 \\\\ & =\lambda^2+10 \lambda+9=0 \\\\ & =\lambda=-1,-9 \\\\ & |\lambda| \leq 5 \Rightarrow \lambda=-1 \\\\ & 5-6 \lambda=5-6(-1)=11 \end{aligned} $$
$$ \begin{aligned} & =(3 \lambda+15) \hat{i}-\hat{j}(-24)+\hat{k}(-24) \\\\ & \overrightarrow{A C} \times \overrightarrow{B D}=(3 \lambda+15) \hat{i}+24 \hat{j}-24 \hat{k} \\\\ & =\sqrt{(3 \lambda+15)^2+(24)^2+(24)^2}=36 \\\\ & =\lambda^2+10 \lambda+9=0 \\\\ & =\lambda=-1,-9 \\\\ & |\lambda| \leq 5 \Rightarrow \lambda=-1 \\\\ & 5-6 \lambda=5-6(-1)=11 \end{aligned} $$
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