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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 3 · Set 1 · Q25
(a) If the lines (x -3)/1=(1- y)/1=(z +2)/p and (2- x)/3=(y +1)/5=(z +56)/(2 p) are perpendicular to each other, then find the value(s) of p.
(a)
If the lines $\displaystyle \frac{\mathrm{x}-3}{1}=\frac{1-\mathrm{y}}{1}=\frac{\mathrm{z}+2}{\mathrm{p}}$ and $\displaystyle \frac{2-\mathrm{x}}{3}=\frac{\mathrm{y}+1}{5}=\frac{\mathrm{z}+56}{2 \mathrm{p}}$ are perpendicular to each other, then find the value(s) of p.
OR
(b)
Find the vector equation of a line passing through the origin and perpendicular to both the lines $\displaystyle \vec{\mathrm{r}}=2 \hat{i}-\hat{j}+2 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}+2 \hat{k})$ and $\displaystyle \vec{\mathrm{r}}=\mu(\hat{i}-\hat{j}+\hat{k})$.
Marking-scheme solution
Direction Ratios of line $\displaystyle \dfrac{x-3}{1}=\dfrac{1-y}{1}=\dfrac{z+2}{p}$ are $\displaystyle <1,-1, p>$
Direction Ratios of line $\displaystyle \dfrac{2-x}{3}=\dfrac{y+1}{5}=\dfrac{z+56}{2 p}$ are $\displaystyle <-3,5,2 p>$
$\displaystyle \because$ Given both lines are perpendicular to each other.
$\displaystyle \therefore 1 \times(-3)+(-1) \times 5+p \times 2 p=0$
$\displaystyle \Rightarrow p= \pm 2$
The vector parallel to required line is given by $\displaystyle \vec{b}=(3 \hat{i}+4 \hat{j}+2 \hat{k}) \times(\hat{i}-\hat{j}+\hat{k})=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 2 \\ 1 & -1 & 1\end{vmatrix}=6 \hat{i}-\hat{j}-7 \hat{k}$
Equation of required line is given by $\displaystyle \vec{r}=\delta(6 \hat{i}-\hat{j}-7 \hat{k})$
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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.