CBSE 2026 · Region 1 · Set 3 · Q32 · 5 marks
Show that line AB passing through points $\displaystyle \mathrm{A}(0,4,1), \mathrm{B}(2,3,-1)$ and the line CD passing through points $\displaystyle \mathrm{C}(4,5,0), \mathrm{D}(2,6,2)$ are parallel. Also, find distance between them.
Marking-scheme solution
Drs of AB are $\displaystyle < 2, -1, -2 >$
Drs of CD are $\displaystyle < -2, 1, 2 >$
These are proportional. Hence, the lines are parallel.
Lines in vector form are
$\displaystyle \vec{r} = 4\hat{\jmath} + \hat{k} + \lambda(2\hat{\imath} - \hat{\jmath} - 2\hat{k})$
$\displaystyle \vec{r} = 4\hat{\imath} + 5\hat{\jmath} + \mu(2\hat{\imath} - \hat{\jmath} - 2\hat{k})$
$\displaystyle \overrightarrow{a_2} - \overrightarrow{a_1} = 4\hat{\imath} + \hat{\jmath} - \hat{k}, \vec{b} = 2\hat{\imath} - \hat{\jmath} - 2\hat{k}$
$\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1}) \times \vec{b} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 4 & 1 & -1 \\ 2 & -1 & -2 \end{vmatrix} = -3\hat{\imath} + 6\hat{\jmath} - 6\hat{k}$
Distance between these parallel lines $\displaystyle = \dfrac{\left|(\overrightarrow{a_2}-\overrightarrow{a_1})\times\vec{b}\right|}{|\vec{b}|} = \dfrac{\sqrt{81}}{\sqrt{9}} = 3$
Three Dimensional GeometryShortest Distance between Two LinesApplylong_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.