If shortest distance between the lines \( \vec{r}=\hat{\imath}-\hat{\jmath}+\lambda(2\hat{\imath}+\hat{k}) \) and \( \vec{r}=2\hat{\imath}-\hat{\jmath}+\mu(\hat{\imath}-\hat{\jmath}-\hat{k}) \) is \( \displaystyle\frac{\sqrt{14}}{14}m \) units, the value of m is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1
For lines \( \vec{r}=\vec{a}_1+\lambda\vec{b}_1 \) and \( \vec{r}=\vec{a}_2+\mu\vec{b}_2 \), the shortest distance is
\( \displaystyle d=\left|\frac{(\vec{b}_1\times\vec{b}_2)\cdot(\vec{a}_2-\vec{a}_1)}{|\vec{b}_1\times\vec{b}_2|}\right| \)
Here \( \vec{a}_1=\hat{\imath}-\hat{\jmath} \), \( \vec{b}_1=2\hat{\imath}+\hat{k} \), \( \vec{a}_2=2\hat{\imath}-\hat{\jmath} \), \( \vec{b}_2=\hat{\imath}-\hat{\jmath}-\hat{k} \).
\( \displaystyle \vec{b}_1\times\vec{b}_2=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\2 & 0 & 1\\1 & -1 & -1\end{vmatrix} \)
\( \displaystyle =\hat{\imath}(0+1)-\hat{\jmath}(-2-1)+\hat{k}(-2-0) \)
\( \displaystyle =\hat{\imath}+3\hat{\jmath}-2\hat{k} \)
\( \displaystyle |\vec{b}_1\times\vec{b}_2|=\sqrt{1+9+4}=\sqrt{14} \)
\( \displaystyle \vec{a}_2-\vec{a}_1=\hat{\imath} \)
\( \displaystyle (\vec{b}_1\times\vec{b}_2)\cdot(\vec{a}_2-\vec{a}_1)=1 \)
\( \displaystyle d=\frac{1}{\sqrt{14}} \)
\( \displaystyle d=\frac{\sqrt{14}}{14} \)
Comparing with \( \frac{\sqrt{14}}{14}m \), m = 1.