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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 3 · Set 1 · Q35
(a) Represent the equations of lines l_1 and l_2 in vector form and check whether they are intersecting or not. l_1: (x +3)/-3=(y -1)/1=(z -5)/5;…
Represent the equations of lines $\displaystyle l_{1}$ and $\displaystyle l_{2}$ in vector form and check whether they are intersecting or not. \[\begin{aligned} & l_{1}: \frac{\mathrm{x}+3}{-3}=\frac{\mathrm{y}-1}{1}=\frac{\mathrm{z}-5}{5} \\ & l_{2}: \frac{\mathrm{x}+1}{-1}=\frac{2-\mathrm{y}}{-2}=\frac{\mathrm{z}-5}{5} \end{aligned} \]
Opposite sides of a square are along the lines : \[\begin{aligned} & \vec{r}=\hat{\mathrm{i}}+2 \hat{j}-4 \hat{\mathrm{k}}+\lambda(2 \hat{\mathrm{i}}+3 \hat{j}+6 \hat{\mathrm{k}}) \\ & \vec{r}=3 \hat{\mathrm{i}}+3 \hat{j}-5 \hat{\mathrm{k}}+\mu(2 \hat{\mathrm{i}}+3 \hat{j}+6 \hat{\mathrm{k}}) \end{aligned} \] Find the area of the square if direction ratios of other pair of opposite sides of the square are given by $\displaystyle <-3, 6, p>$. Also, find the value of p.
Marking-scheme solution
$\displaystyle l_{1}: \vec{r}=(-3 \hat{i}+\hat{j}+5 \hat{k})+\lambda(-3 \hat{i}+\hat{j}+5 \hat{k})$
$\displaystyle l_{2}: \vec{r}=(-\hat{i}+2 \hat{j}+5 \hat{k})+\mu(-\hat{i}+2 \hat{j}+5 \hat{k})$
$\displaystyle \overrightarrow{a_{1}}=-3 \hat{i}+\hat{j}+5 \hat{k} ; \quad \overrightarrow{b_{1}}=-3 \hat{i}+\hat{j}+5 \hat{k}$
$\displaystyle \overrightarrow{a_{2}}=-\hat{i}+2 \hat{j}+5 \hat{k} ; \quad \overrightarrow{b_{2}}=-\hat{i}+2 \hat{j}+5 \hat{k}$
$\displaystyle \overrightarrow{a_{2}}-\overrightarrow{a_{1}}=2 \hat{i}+\hat{j}$
$\displaystyle \overrightarrow{b_{1}} \times \overrightarrow{b_{2}}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ -3 & 1 & 5 \\ -1 & 2 & 5\end{vmatrix}=-5 \hat{i}+10 \hat{j}-5 \hat{k}$
$\displaystyle \because\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=0$
$\displaystyle \therefore$ Both lines are intersecting.
$\displaystyle \overrightarrow{a_{1}}=\hat{i}+2 \hat{j}-4 \hat{k} ; \quad \overrightarrow{a_{2}}=3 \hat{i}+3 \hat{j}-5 \hat{k} ; \quad \vec{b}=2 \hat{i}+3 \hat{j}+6 \hat{k}$
$\displaystyle \overrightarrow{a_{2}}-\overrightarrow{a_{1}}=2 \hat{i}+\hat{j}-\hat{k}$
$\displaystyle \left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \times \vec{b}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 2 & 3 & 6\end{vmatrix}=9 \hat{i}-14 \hat{j}+4 \hat{k}$
$\displaystyle \left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \times \vec{b}\right|=\sqrt{293}$
Distance between given parallel lines $\displaystyle =\dfrac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \times \vec{b}\right|}{|\vec{b}|}=\dfrac{\sqrt{293}}{7}$
$\displaystyle \therefore$ Length of side of square $\displaystyle =$ distance between parallel lines $\displaystyle =\dfrac{\sqrt{293}}{7}$ units
Area of square $\displaystyle =\dfrac{293}{49}$ square units
$\displaystyle \because$ Adjacent sides of square are perpendicular to each other.
$\displaystyle \therefore-3 \times 2+6 \times 3+6 p=0 \Rightarrow p=-2$
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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.