If the shortest distance between the lines $I_1$ and $I_2$ given by $\vec{r}=a\hat{i}+2\hat{j}-\hat{k}+\lambda (2\hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=\hat{i}-\hat{j}+\hat{k}+\mu ( 2\hat{i}-\hat{j}+\hat{k})$ is $\sqrt{\frac{35}{6}}$ units, the values of 'a' can be :
Answer & explanation
Correct answer: option 1
The correct answer is option (1) → 0, -8
$l_1:\vec{r}=a\hat{i}+2\hat{j}-\hat{k}+\lambda (2\hat{i}-\hat{j}+\hat{k})$
$l_2:\vec{r}=\hat{i}-\hat{j}+\hat{k}+\mu ( 2\hat{i}-\hat{j}+\hat{k})$
these are parallel lines
$\vec{a_1}=a\hat i+2\hat j-\hat k$
$\vec{a_2}=a\hat i+2\hat j-\hat k$
$\vec{a_2}-\vec{a_1}=(1-a)\hat i-3\hat j+2\hat k$
$\vec b=2\hat i-\hat j+\hat k$
shortest distance
$=\frac{|\vec b×(\vec{a_2}-\vec{a_1})|}{|\vec b|}⇒\vec b×(\vec{a_2}-\vec{a_1})=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-1&1\\1-a&-3&2\end{vmatrix}$
$⇒\frac{|-\hat i+(a+3)\hat j+(a+5)\hat k|}{\sqrt{6}}=\frac{\sqrt{35}}{\sqrt{6}}$
so squaring both sides
$(-1)^2+(a+3)^2+(a+5)^2=35$
$a^2+6a+a^2+10a+25+9=34$
$a^2+8a=0$
$a=0,-8$