Question bank/JEE Main/Mathematics

If the shortest distance between the
lines$x + 2\lambda = 2y = - 12z,x = y + 4\lambda = 6z - 12\lambda$ is
$4\sqrt{2}$units, then a value of $\lambda^{10}$ is:
$x + 2\lambda = 2y = - 12z,x = y + 4\lambda = 6z - 12\lambda$ds chp
y?kqÙke nwjh $4\sqrt{2}$bdkbZ gS] rks $\lambda^{10}$

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[IMAGE] $x + 2\lambda = 2y = - 12z$ & $x = y + 4\lambda = 6z - 12\lambda$
$\Rightarrow \frac{x + 2\lambda}{1} = \frac{y}{1/2} = \frac{z}{- 1/12}$
$\Rightarrow \frac{x}{1} = \frac{y + 4\lambda}{1} = \frac{z - 2\lambda}{1/6}$
$\Rightarrow \frac{x + 2\lambda}{12} = \frac{y}{6} = \frac{z}{- 1}$
$\Rightarrow \frac{x}{6} = \frac{y + 4\lambda}{6} = \frac{z - 2\lambda}{1}$
Vector normal to both lines nksuksa js
$= \left| \begin{matrix}
i & j & k \\
12 & 6 & - 1 \\
6 & 6 & 1
\end{matrix} \right|$
$= \overset{\hat{}}{i}(12) - \overset{\hat{}}{j}(18) + \overset{\hat{}}{k}(36) = 12\overset{\hat{}}{i} - 18\overset{\hat{}}{j} + 36\overset{\hat{}}{k}$
or vector parallel to normal to both
line$= 2\overset{\hat{}}{i} - 3\overset{\hat{}}{j} + 6\overset{\hat{}}{k}$
$\overset{\rightarrow}{AB} = (2\lambda)\overset{\hat{}}{i} - (4\lambda)\overset{\hat{}}{i} + (2\lambda)\overset{\hat{}}{k}$
Shortest distance will be projection of $\overset{\rightarrow}{AB}$on
vector parallel to the perpendicular to both lines
U;wure nwjh $\overset{\rightarrow}{AB}$dk nksuksa yEcor~
\Rightarrow 4\sqrt{2} = \left| \frac{\overset{\rightarrow}{AB} \cdot + \right| = \left| \right| = \left| = 4\lambda
$\lambda = \pm \sqrt{2}$

Source: Grass-G / M-21 Mathematics Paper 19 Dec Hindi Eng.docx