Verified Solution • 2025–26 Session moderate

QJEEM23Ma9d03: JEE Main 2023 Physics - Class 11 JEEM23Ma9d03 Solution: and are the vertices of a quadrilateral . If its area is 18…

A(2,6,2),B(−4,0,λ),C(2,3,−1)A(2,6,2), B(-4,0, \lambda), C(2,3,-1) and D(4,5,0),∣λ∣≤5D(4,5,0),|\lambda| \leq 5 are the vertices of a quadrilateral ABCDA B C D . If its area is 18 square units, then 5−6λ5-6 \lambda is equal to __________.

Step-by-Step Verified Solution

amp;A(2,6,2)B(−4,0,λ),C(2,3,−1)D(4,5,0)amp; Area =12∣BD→×AC→∣=18amp;AC→×BD→=∣i^amp;jamp;k0amp;−3amp;−38amp;5amp;−λ∣\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}

amp;=(3λ+15)i^−j^(−24)+k^(−24)amp;AC→×BD→=(3λ+15)i^+24j^−24k^amp;=(3λ+15)2+(24)2+(24)2=36amp;=λ2+10λ+9=0amp;=λ=−1,−9amp;∣λ∣≤5⇒λ=−1amp;5−6λ=5−6(−1)=11\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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