The shortest distance between the lines whose vector equations are \( \vec{r}=\left(6\hat{\imath}+2\hat{\jmath}+2\hat{k}\right)+\lambda\left(\hat{\imath}-2\hat{\jmath}+2\hat{k}\right) \) and \( \vec{r}=\left(-4\hat{\imath}-\hat{k}\right)+\mu\left(3\hat{\imath}-2\hat{\jmath}-2\hat{k}\right) \)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 9 units
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
\( \displaystyle d=\left|\frac{(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)}{|\vec{b}_1\times\vec{b}_2|}\right| \)
Here \( \vec{a}_1=6\hat{\imath}+2\hat{\jmath}+2\hat{k} \), \( \vec{b}_1=\hat{\imath}-2\hat{\jmath}+2\hat{k} \), \( \vec{a}_2=-4\hat{\imath}-\hat{k} \), \( \vec{b}_2=3\hat{\imath}-2\hat{\jmath}-2\hat{k} \).
\( \displaystyle \vec{a}_2-\vec{a}_1=-10\hat{\imath}-2\hat{\jmath}-3\hat{k} \)
\( \displaystyle \vec{b}_1\times\vec{b}_2=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\1 & -2 & 2\\3 & -2 & -2\end{vmatrix} \)
\( \displaystyle =\hat{\imath}(4+4)-\hat{\jmath}(-2-6)+\hat{k}(-2+6) \)
\( \displaystyle =8\hat{\imath}+8\hat{\jmath}+4\hat{k} \)
\( \displaystyle |\vec{b}_1\times\vec{b}_2|=\sqrt{64+64+16} \)
\( \displaystyle =\sqrt{144}=12 \)
\( \displaystyle (\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)=-80-16-12=-108 \)
\( \displaystyle d=\left|\frac{-108}{12}\right|=9\ \text{units} \)
NCERT (2026-27 edition):
Three Dimensional Geometry
