Mathematics - Vector Question with Solution | TestHub

D.R. 's of ${\overrightarrow{\mathrm{a}}}_1=\left(-1,0,3\right)$

D.R. 's of ${\overrightarrow{\mathrm{a}}}_2=\left(0,-1,2\right)$

D.R. 's of ${\overrightarrow{\mathrm{b}}}_1=\left(1,-\mathrm{a},0\right)$ 

D.R. 's of ${\overrightarrow{\mathrm{b}}}_2=\left(1,-1,1\right)$ 

Now, ${\overrightarrow{\mathrm{a}}}_2-{\overrightarrow{\mathrm{a}}}_1=\hat{i}-\hat{j}-\hat{k}$

and ${\overrightarrow{\mathrm{b}}}_1\times {\overrightarrow{\mathrm{b}}}_2=\left|\begin{array}{ccc}\hat{\mathit{i}}& \hat{\mathit{j}}& \hat{\mathit{k}}\\ 1& -\mathrm{a}& 0\\ 1& -1& 1\end{array}\right|$

${\overrightarrow{\mathrm{b}}}_1\times \overrightarrow{{\mathrm{b}}_2}=\hat{i}\left(-a\right)-\hat{j}+\hat{k}\left(a-1\right)$

i.e. $\left|\overrightarrow{{\mathrm{b}}_1}\times \overrightarrow{{\mathrm{b}}_2}\right|=\sqrt{a^2+1+{\left(a-1\right)}^2}$

So, $\left({\overrightarrow{\mathrm{a}}}_2-{\overrightarrow{\mathrm{a}}}_1\right)·\left({\overrightarrow{\mathrm{b}}}_1\times {\overrightarrow{\mathrm{b}}}_2\right)=2-2a$

Shortest distance between the lines, $\frac{2\left(1-a\right)}{\sqrt{a^2+1+{\left(a-1\right)}^2}}=\sqrt{\frac{2}{3}}$

Squaring on both the sides, we get,

$3{\left(1-a\right)}^2=a^2-a+1$

 i.e. $a=2,\frac{1}{2}$