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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 4 · Set 1 · Q35
Find the equation of a line (in vector and cartesian form) that passes through the point of intersection of lines (x -1)/2=(y -2)/3=(z -3)/4 and (x…
Find the equation of a line (in vector and cartesian form) that passes through the point of intersection of lines $\displaystyle \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{4}$ and $\displaystyle \frac{\mathrm{x}-4}{5}=\frac{\mathrm{y}-1}{2}=\mathrm{z}$ and is parallel to the vector $\displaystyle 3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}$.
Marking-scheme solution
Let $\displaystyle \dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=a$ & $\displaystyle \dfrac{x-4}{5}=\dfrac{y-1}{2}=z=b$
So any general point on these are $\displaystyle P(2 a+1,3 a+2,4 a+3)$ and $\displaystyle Q(5 b+4,2 b+1, b)$ respectively.
For intersection points: $\displaystyle 2 a+1=5 b+4, 4 a+3=b$
on solving we get, $\displaystyle a=b=-1$
Thus, required point of intesection is $\displaystyle (-1,-1,-1)$.
Required equation of line (cartesian form): $\displaystyle \dfrac{x+1}{3}=\dfrac{y+1}{2}=\dfrac{z+1}{-8}$
Required equation of line (Vector form): $\displaystyle \vec{r}=(-\hat{i}-\hat{j}-\hat{k})+\lambda(3 \hat{i}+2 \hat{j}-8 \hat{k})$
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.