<p>Lines are \(\dfrac{x}{0} = \dfrac{y}{0} = \dfrac{z}{1} = \lambda\) (z axis) and \(x + y + 2z - 3 = 0,\ 2x + 3y + 4z - 4 = 0\). The shortest distance (S.D.) between the two lines is given by \(\text{S.D.} = \dfrac{|(\vec{c}-\vec{a})\cdot(\vec{b}\times\vec{d})|}{|\vec{b}\times\vec{d}|}\). Find the shortest distance.</p>
Step-by-Step Solution
Key Concept: The second line is the intersection of two planes; find its direction vector using the cross product of the plane normals, then apply the skew lines distance formula using a point on each line.
Step 1: Identify the first line: z-axis has point a⃗ = (0,0,0) and direction b⃗ = (0,0,1) Step 2: Find direction vector of second line (intersection of planes). Use normal vectors n⃗_1 = (1,1,2) and n⃗_2 = (2,3,4) :
d⃗ = n⃗_1 × n⃗_2 = |i j k|
1 1 2|
2 3 4|
= i(4-6) - j(4-4) + k(3-2) = (-2, 0, 1) Step 3: Find a point on the second line by setting y = 0 in both plane equations:
x + 2z = 3 and 2x + 4z = 4
From second: x + 2z = 2, contradicting first. Try x = 0:
y + 2z = 3 and 3y + 4z = 4
From these: 3(3-2z) + 4z = 4 → 9 - 6z + 4z = 4 → z = 5/2, y = -2
Point c⃗ = (0, -2, 5/2) Step 4: Calculate b⃗ × d⃗ = (0,0,1) × (-2,0,1) = (0·1 - 1·0, 1·(-2) - 0·1, 0·0 - 0·(-2)) = (0, -2, 0)
|b⃗ × d⃗| = 2 Step 5: Calculate c⃗ - a⃗ = (0, -2, 5/2)
(c⃗ - a⃗) · (b⃗ × d⃗) = (0, -2, 5/2) · (0, -2, 0) = 0 + 4 + 0 = 4 Step 6: S.D. = |4|/2 = 2 ∴ Answer: A
Correct Answer: A