CBSE 2026 · Region 4 · Set 2 · Q35 · 5 marks
Find the vector and cartesian equations of the line passing through the point of intersection of the lines $\displaystyle \vec{\mathrm{r}}=(\hat{\mathrm{i}}+\hat{j}-\hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}-\hat{j})$ and $\displaystyle \overrightarrow{\mathrm{r}}=(4 \hat{\mathrm{i}}-\hat{\mathrm{k}})+\mu(2 \hat{\mathrm{i}}+3 \hat{\mathrm{k}})$ and parallel to the line $\displaystyle \frac{\mathrm{x}-1}{-2}=\frac{7-\mathrm{y}}{-3}=\mathrm{z}$.
Marking-scheme solution
Lines are $\displaystyle \vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(3 \hat{i}-\hat{j})$ and $\displaystyle \vec{r}=(4 \hat{i}-\hat{k})+\mu(2 \hat{i}+3 \hat{k})$
So any general point on these lines are $\displaystyle P(3 \lambda+1,-\lambda+1,-1)$ & $\displaystyle Q(2 \mu+4,0,3 \mu-1)$ respectively.
For intersection points: $\displaystyle -\lambda+1=0,3 \mu-1=-1$ & $\displaystyle 3 \lambda+1=2 \mu+4$
on solving we get, $\displaystyle \lambda=1, \mu=0$
Thus, required point of intesection is $\displaystyle (4,0,-1)$.
Required equation of line (cartesian form): $\displaystyle \dfrac{x-4}{-2}=\dfrac{y}{3}=\dfrac{z+1}{1}$
Required equation of line (Vector form): $\displaystyle \vec{r}=(4 \hat{i}-\hat{k})+\omega(-2 \hat{i}+3 \hat{j}+\hat{k})$
Three Dimensional GeometryEquation of a Line in SpaceApplylong_answerhard
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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.