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JEE Main 27-Jan-2024 Paper - Shift 1 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 27-Jan-2024 Paper - Shift 14 Marks
MCQ
If the shortest distance between the line $\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and $\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$ is $\frac{6}{\sqrt{5}}$, then the sum of all possible values of $\lambda$ is :
  • A
    5
  • B
    8
  • C
    7
  • D
    10
✓

Answer

$\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$
$\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$
the shortest distance between the lines
$=\left|\frac{(\vec{a}-\vec{b}) \cdot\left(\overrightarrow{d_1} \times \overrightarrow{d_2}\right)}{\left|\overrightarrow{d_1} \times \overrightarrow{d_2}\right|}\right| $
$=\left|\frac{\left|\begin{array}{ccc} \lambda-4 & 0 & 2 \\1 & 2 & -3 \\ 2 & 4 & -5 \end{array}\right|}{\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{array}\right|}\right| $
$=\left|\frac{(\lambda-4)(-10+12)-0+2(4-4)}{|2 \hat{i}-1 \hat{j}+0 \hat{k}|}\right| \\$
$\frac{6}{\sqrt{5}}=\left|\frac{2(\lambda-4)}{\sqrt{5}}\right|$
$3=|\lambda-4|$
$\lambda-4= \pm 3$
$\lambda=7,1$
Sum of all possible values of $\lambda$ is $=8$

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