The shortest distance between lines whose vector equations are: \( \vec{r}=(1-t)\hat{\imath}+(t-2)\hat{\jmath}+(3-2t)\hat{k} \) and \( \vec{r}=(s+1)\hat{\imath}+(2s-1)\hat{\jmath}-(2s+1)\hat{k} \) , is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{8}{\sqrt{29}} \) units
Separating the parameter, each line takes the form \( \vec{r}=\vec{a}+\lambda\vec{b} \):
\( \displaystyle \vec{r}=(\hat{\imath}-2\hat{\jmath}+3\hat{k})+t(-\hat{\imath}+\hat{\jmath}-2\hat{k}) \)
\( \displaystyle \vec{r}=(\hat{\imath}-\hat{\jmath}-\hat{k})+s(\hat{\imath}+2\hat{\jmath}-2\hat{k}) \)
So \( \vec{a}_1=\hat{\imath}-2\hat{\jmath}+3\hat{k} \), \( \vec{b}_1=-\hat{\imath}+\hat{\jmath}-2\hat{k} \), \( \vec{a}_2=\hat{\imath}-\hat{\jmath}-\hat{k} \), \( \vec{b}_2=\hat{\imath}+2\hat{\jmath}-2\hat{k} \).
The shortest distance between skew lines is
\( \displaystyle d=\left|\frac{(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)}{|\vec{b}_1\times\vec{b}_2|}\right| \)
\( \displaystyle \vec{a}_2-\vec{a}_1=\hat{\jmath}-4\hat{k} \)
\( \displaystyle \vec{b}_1\times\vec{b}_2=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\-1 & 1 & -2\\1 & 2 & -2\end{vmatrix} \)
\( \displaystyle =(-2+4)\hat{\imath}-(2+2)\hat{\jmath}+(-2-1)\hat{k} \)
\( \displaystyle =2\hat{\imath}-4\hat{\jmath}-3\hat{k} \)
\( \displaystyle |\vec{b}_1\times\vec{b}_2|=\sqrt{4+16+9}=\sqrt{29} \)
\( \displaystyle (\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)=0-4+12=8 \)
\( \displaystyle d=\frac{8}{\sqrt{29}}\ \text{units} \)
NCERT (2026-27 edition):
Shortest Distance Between Two Lines
