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