The sum of all values of , for which the shortest distance between the lines (x+1)/( )=(y-…

Question

The sum of all values of $\alpha $, for which the shortest distance between the lines $\frac{x+1}{\alpha }=\frac{y-2}{-1}=\frac{z-4}{-\alpha }$ and $\frac{x}{\alpha }=\frac{y-1}{2}=\frac{z-1}{2\alpha }$ is $\sqrt{2}$, is

Accepted Answer

Lines into vector form

$\begin{gathered} {L}_{1}:\vec{r}=-\hat{i}+2\hat{j}+4\hat{k}+\lambda (\alpha \hat{i}-\hat{j}-\alpha \hat{k}) \ {L}_{2}:\vec{r}=\hat{j}+\hat{k}+\mu (\alpha \hat{i}+2\hat{j}+2\alpha \hat{k}) \end{gathered}$

Lines are not parallel because direction ratios are not proportional

$\frac{\alpha }{\alpha }
e \frac{-1}{2}$

So lines are skew lines.

Minimum distance between two skew lines $\vec{r}=\vec{a}+\lambda \vec{b}$ and $\vec{r}=\vec{c}+\mu \vec{d}$ is given as

$d=|\frac{(\vec{a}-\vec{c})\cdot (\vec{b}	imes \vec{d})}{|\vec{b}	imes \vec{d}|}|$

$\vec{a}=-\hat{i}+2\hat{j}+4\hat{k},\, \vec{c}=\hat{j}+\hat{k}$

$\vec{b}=\alpha \hat{i}-\hat{j}-\alpha \hat{k},\, \vec{d}=\alpha \hat{i}+2\hat{j}+2\alpha \hat{k}$

$\vec{b}	imes \vec{d}=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ \alpha  \quad  -1 \quad  -\alpha \ \alpha  \quad  2 \quad  2\alpha \end{vmatrix}$

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

$	herefore $ $\vec{b}	imes \vec{d}=-3{\alpha }^{2}\hat{j}+3\alpha \hat{k}$

$\Rightarrow$ $|\vec{b}	imes \vec{d}|=\sqrt{9{\alpha }^{4}+9{\alpha }^{2}}$

$\Rightarrow$ $\vec{a}-\vec{c}=-\hat{i}+\hat{j}+3\hat{k}$

$\Rightarrow$ $|\frac{(\vec{a}-\vec{c})\cdot (\vec{b}	imes \vec{d})}{|\vec{b}	imes \vec{d}|}|=\sqrt{2}$

$\Rightarrow$ $|\frac{(-\hat{i}+\hat{j}+3\hat{k})\cdot (-3{\alpha }^{2}\hat{j}+3\alpha \hat{k})}{\sqrt{9{\alpha }^{4}+9{\alpha }^{2}}}|=\sqrt{2}$

$\Rightarrow$ $|\frac{-3{\alpha }^{2}+9\alpha }{3\sqrt{{\alpha }^{4}+{\alpha }^{2}}}|=\sqrt{2}$

$\Rightarrow$ $|\frac{-3\alpha (\alpha -3)}{3\alpha \sqrt{{\alpha }^{2}+1}}|=\sqrt{2}$

$\Rightarrow$ $|\frac{\alpha -3}{\sqrt{1+{\alpha }^{2}}}|=\sqrt{2}$

squaring

$\Rightarrow$ $\frac{(\alpha -3{)}^{2}}{1+{\alpha }^{2}}=2$

$\Rightarrow$ ${\alpha }^{2}+9-6\alpha =2+2{\alpha }^{2}$

$\Rightarrow$ ${\alpha }^{2}+6\alpha -7=0$

This equation has two roots say ${\alpha }_{1}$ and ${\alpha }_{2}$

So, sum of roots $=-\frac{b}{a}=-\frac{6}{1}=-6$

$	herefore $ ${\alpha }_{1}+{\alpha }_{2}=-6$

Solutions · 1