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QJEEM24Ma7029: JEE Main 2024 Mathematics - Class 12 JEEM24Ma7029 Solution: If the shortest distance between the lines and is , then the…

If the shortest distance between the lines x−λ3=y−2−1=z−11\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1} and x+2−3=y+52=z−44\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4} is 4430\frac{44}{\sqrt{30}} , then the largest possible value of ∣λ∣|\lambda| is equal to _________.

Step-by-Step Verified Solution

amp;L1:x−λ3=y−2−1=z−11amp;L2:x+2−3=y+52=z−44\begin{aligned}& L_1: \frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1} \\& L_2: \frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}\end{aligned}

amp;n1×n2=∣i^amp;j^amp;k^3amp;−1amp;1−3amp;2amp;4∣amp;=−6i^−15j^+3k^amp;d=∣[(λ+2)i^+7j^−3k^][−6i^−15j^+3k^]∣−6i^−15j^+3k^∣∣=4430amp;∣−6λ−12−105−9270∣=4430amp;∣6λ+126∣=132amp;∣λ+21∣=22amp;λ+21=±22amp;∣λ∣max⁡=43\begin{aligned}& n_1 \times n_2=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\3 & -1 & 1 \\-3 & 2 & 4\end{array}\right| \\& =-6 \hat{i}-15 \hat{j}+3 \hat{k} \\& d=\left|\frac{[(\lambda+2) \hat{i}+7 \hat{j}-3 \hat{k}][-6 \hat{i}-15 \hat{j}+3 \hat{k}]}{|-6 \hat{i}-15 \hat{j}+3 \hat{k}|}\right|=\frac{44}{\sqrt{30}} \\& \left|\frac{-6 \lambda-12-105-9}{\sqrt{270}}\right|=\frac{44}{\sqrt{30}} \\& |6 \lambda+126|=132 \\& |\lambda+21|=22 \\& \lambda+21= \pm 22 \\& |\lambda|_{\max }=43\end{aligned}

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