List - I | List - 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) | \(\sqrt{\frac{2}{3}}\) |
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).
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).
Direction vectors:
d₁ = i + 2j + 3k
d₂ = 3i + 2j + 7k
Normal vector n̂ 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.
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.
y = 1.
(P) → (3)
(Q) → (4)
(R) → (1)
(S) → (5)