If the shortest distance between the linesand,then a value of    is : Option 1) Option 2) Option 3) Option 4)

V Vakul

As we learnt in

Shortest distance between two skew lines (Cartesian form) -

Shortest distance between

$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$and $\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}$is given by

$\left | \frac{\left ( \vec{b} \times \vec{b_{1}}\right )\cdot \left ( \vec{a} -\vec{a_{1}}\right )}{\left | \vec{b} \times \vec{b_{1}} \right |} \right |$ Where

$\vec{a}= x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k}$

$\vec{a_{1}}= x_{2}\hat{i}+y_{2}\hat{j}+z_{2}\hat{k}$

$\vec{b}= a_{1}\hat{i}+b_{1}\hat{j}+c_{1}\hat{k}$

$\vec{b_{1}}= a_{2}\hat{i}+b_{2}\hat{j}+c_{2}\hat{k}$

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Normal vector of line is $L_{2} is$

$\begin{vmatrix} \hat{i}& \hat{j} &\hat{k} \\ 1 & 1&1 \\ 2 & -1& 1 \end{vmatrix}=2\hat{i}+\hat{j}-3\hat{k}$

$b_{1}^{-1}\times b_{2}^{-1}=(\alpha\hat{i}-\hat{j}+\hat{k})\times (2\hat{i}+\hat{j}-\hat{3k})$

$=\begin{bmatrix} \hat{i} & \hat{j} &\hat{k} \\ \alpha &-1 & 1\\ 2 & 1 & -3 \end{bmatrix}=2\hat{i}+\hat{j}(3\alpha+2)+\hat{k}(\alpha+2)$

One point on $L_{2}$ is (0,1,-2) and on $L_{1}$ is (1,-1,0)

Distance =$\left | \frac{(\hat{i}-2\hat{j}+2\hat{k}). (2\hat{1}+(3\alpha+2)\hat{j})}{\sqrt{4+(3\alpha+2)^{2}+(\alpha+2)^{2}}} \right |=\frac{1}{\sqrt{3}}$

$3(2-6\alpha-4+2\alpha+4)^{2}=10\alpha^{2}+12+16\alpha$

$3\times 4(2\alpha-1)^{2}=2(5\alpha^{2}+8\alpha+6)$

$24\alpha^{2}+6-24\alpha=5\alpha^{2}+8\alpha+6$

$19\alpha^{2}-32\alpha=0\Rightarrow \alpha=\frac{32}{19}$

Option 1)

This option is incorrect

Option 2)

This option is incorrect

Option 3)

This option is correct

Option 4)

This option is incorrect

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