Question:

Let y ∈ R be such that the lines \(L_1:\frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}\) and \(L_2:\frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma}\) intersect. Let R1 be the point of intersection of L1 and L2. Let O = (0, 0 ,0), and \(\hat{n}\) denote a unit normal vector to the plane containing both the lines L1 and L2.
Match each entry in List-I to the correct entry in List-II.
List - IList - II
(P)γ equals(1)\(-\hat{i}-\hat{j}+\hat{k}\)
(Q)A possible choice for \(\hat{n}\) is(2)\(\sqrt{\frac{3}{2}}\)
(R)\(\overrightarrow{OR_1}\) equals(3)1
(S)A possible value of \(\overrightarrow{OR_1}.\hat{n}\) is(4)\(\frac{1}{\sqrt6}\hat{i}-\frac{2}{\sqrt6}\hat{j}+\frac{1}{\sqrt6}\hat{k}\)
  (5)\(\sqrt{\frac{2}{3}}\)
The correct option is

Updated On: Mar 7, 2025
  • (P) → (3) (Q) → (4) (R) → (1) (S) → (2)
  • (P) → (5) (Q) → (4) (R) → (1) (S) → (2)
  • (P) → (3) (Q) → (4) (R) → (1) (S) → (5)
  • (P) → (3) (Q) → (1) (R) → (4) (S) → (5)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

Intersection of Lines and Plane Equation

1. Parametric Equations for L₁ and L₂:

For L₁:

(x + 11)/1 = (y + 21)/2 = (z + 29)/3

Rewriting in parametric form:

P(λ - 11, 2λ - 21, 3λ - 29).

For L₂:

(x + 16)/3 = (y + 11)/2 = (z + 4)/7

Rewriting in parametric form:

Q(3μ - 16, 2μ - 11, 7μ - 4).

2. Equating Coordinates of L₁ and L₂:

Setting the parametric equations equal to each other:

λ - 11 = 3μ - 16

2λ - 21 = 2μ - 11

3λ - 29 = 7μ - 4

Solving for λ and μ:

λ = 5, μ = 10.

The point of intersection:

P = Q = (-1, -1, 1).

3. Equation of the Plane Containing L₁ and L₂:

Direction vectors:

d₁ = i + 2j + 3k

d₂ = 3i + 2j + 7k

Normal vector is given by the cross product:

n̂ = d₁ × d₂

Computing the determinant:

n̂ = | i j k |
       | 1 2 3 |
       | 3 2 7 |

Expanding:

n̂ = (-i - 2j + k) / √6.

4. Dot Product of OR₁ and n̂:

Vector OR₁:

OR₁ = -i - j + k

Computing dot product:

(-1)(-1) + (-1)(-2) + (1)(1) = 1 + 2 + 1 = 4.

Normalizing:

n̂ = (-i - 2j + k) / √6.

Final dot product:

OR₁ · n̂ = (-1 + 2 + 1) / √6 = √2/3.

5. Solving for y-coordinate:

y = 1.

Final Matching:

(P) → (3)

(Q) → (4)

(R) → (1)

(S) → (5)

Was this answer helpful?
0
0

Top Questions on Three Dimensional Geometry

View More Questions

Questions Asked in JEE Advanced exam

View More Questions