CBSE 2023 · Region 5 · Set 1 · Q31 · 3 marks
Find the distance between the lines: \[\begin{aligned} & \vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k}) \\ & \vec{r}=(3 \hat{i}+3 \hat{j}-5 \hat{k})+\mu(4 \hat{i}+6 \hat{j}+12 \hat{k}) \end{aligned} \]
Marking-scheme solution
Here
$$
\begin{aligned}
& \overrightarrow{a_{1}}=\hat{i}+2 \hat{j}-4 \hat{k}, \overrightarrow{\mathrm{b}_{1}}=2 \hat{i}+3 \hat{j}+6 \hat{k}
& \overrightarrow{a_{2}}=3 \hat{i}+3 \hat{j}-5 \hat{k}, \overrightarrow{\mathrm{b}_{2}}=4 \hat{i}+6 \hat{j}+12 \hat{k}
\end{aligned}Here, $\displaystyle \overrightarrow{\mathrm{b}}_{1}$ and $\displaystyle \overrightarrow{\mathrm{b}}_{2}$ are parallel vectors.
Three Dimensional GeometryShortest Distance between Two LinesApplyshort_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.