✓ Board-verified
Mathematics · 2026 · 5 marks
CBSE 2026 · Region 2 · Set 1 · Q35
A line passing through the points A(1, 2, 3) and B(5, 8, 11) intersects the line r =4 i + j +λ(5 i +2 j + k ). Find the co-ordinates of the point of…
A line passing through the points A($\displaystyle 1$, $\displaystyle 2$, $\displaystyle 3$) and B($\displaystyle 5$, $\displaystyle 8$, $\displaystyle 11$) intersects the line $\displaystyle \vec{r}=4 \hat{i}+\hat{j}+\lambda(5 \hat{i}+2 \hat{j}+\hat{k})$. Find the co-ordinates of the point of intersection. Hence, write the equation of a line passing through the point of intersection and perpendicular to both the lines.
Marking-scheme solution
The line passing through the points $\displaystyle A(1,2,3)$ and $\displaystyle B(5,8,11)$ is given by
$\displaystyle l_{1}: \dfrac{x-1}{5-1}=\dfrac{y-2}{8-2}=\dfrac{z-3}{11-3}$ i.e. $\displaystyle \dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\mu$
Also the given line in its cartesian form is, $\displaystyle l_{2}: \dfrac{x-4}{5}=\dfrac{y-1}{2}=\dfrac{z}{1}=\lambda$
Any point on $\displaystyle l_{1}$ is $\displaystyle P(2\mu+1, 3\mu+2, 4\mu+3)$
Any point on $\displaystyle l_{2}$ is $\displaystyle Q(5\lambda+4, 2\lambda+1, \lambda)$
When the lines intersect, the points $\displaystyle P$ and $\displaystyle Q$ must coincide.
$\displaystyle \therefore 2\mu+1=5\lambda+4, 3\mu+2=2\lambda+1$ and $\displaystyle 4\mu+3=\lambda$
$\displaystyle \Rightarrow 2\mu-5\lambda=3$ ..... (i) , $\displaystyle 3\mu-2\lambda=-1$ ...... (ii) and $\displaystyle 4\mu-\lambda=-3$ ...... (iii)
solving any two above equations, we get $\displaystyle \mu=\lambda=-1$
The point of intersection is $\displaystyle (-1,-1,-1)$.
Let the required line $\displaystyle (l)$ passing through $\displaystyle (-1,-1,-1)$ be $\displaystyle \dfrac{x+1}{a}=\dfrac{y+1}{b}=\dfrac{z+1}{c}$
Since $\displaystyle l \perp l_{1}$ and $\displaystyle l \perp l_{2}$,
$\displaystyle 2a+3b+4c=0, 5a+2b+c=0$
$\displaystyle \Rightarrow \dfrac{a}{3-8}=\dfrac{b}{20-2}=\dfrac{c}{4-15}$ i.e. $\displaystyle \dfrac{a}{-5}=\dfrac{b}{18}=\dfrac{c}{-11}$
Hence, the required line is $\displaystyle \dfrac{x+1}{-5}=\dfrac{y+1}{18}=\dfrac{z+1}{-11}$ or $\displaystyle \dfrac{x+1}{5}=\dfrac{y+1}{-18}=\dfrac{z+1}{11}$
All Three Dimensional Geometry questions2026 question paperNCERT solutions: Three Dimensional GeometryPractice Three Dimensional Geometry →
More from Three Dimensional Geometry
- Find the image A^′ of the point A(1,6,3) in the line x/1=(y-1)/2=(z-2)/3. Also, find the equation of the line joining A and A^′. OR Find a point P on…2025
- Find the values of λ, for which the distance of point (2,1, λ) from plane 3 x+5 y+ 4 z=11 is 2 √2 units.2022
- Find the value of λ if the following lines are perpendicular to each other: l_1: (1-x)/-3=(3 y-2)/(2 λ)=(z-3)/3; l_2: (x-1)/(3 λ)=(1-y)/1=(2 z-5)/32025
- The value of λ for which the angle between the lines r= i+ j+ k+p(2 i+ j+2 k) and r=(1+q) i+(1+q λ) j+(1+q) k is (π)/2 is:2023
- If a line makes an angle of (π)/4 with the positive directions of both x -axis and z -axis, then the angle which it makes with the positive direction…2024
- Assertion: If a line makes angles α, β, γ with positive direction of the coordinate axes, then sin^2 α+ sin^2 β+ sin^2 γ=2. Reason (R): The sum of…2023
- 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. OR Find the…2026
- Find the equation of the plane passing through points (2, 1, 0), (3, -2, -2) and (1, 1, -7). Also, obtain its distance from the origin. OR Find the…2022
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.