<p>If the shortest distance between the lines \(\dfrac{x-1}{\alpha} = \dfrac{y+1}{-1} = \dfrac{z}{1}\), \((\alpha \ne -1)\) and \(x + y + z + 1 = 0 = 2x - y + z + 3\) is \(\dfrac{1}{\sqrt{3}}\), then a value of \(\alpha\) is</p>
<p>\(-\dfrac{16}{19}\)</p>
<p>\(\dfrac{19}{16}\)</p>
<p>\(\dfrac{32}{19}\)</p>
<p>\(\dfrac{19}{32}\)</p>
Step-by-Step Solution
Key Concept: The second line is the intersection of two planes; find its direction vector via cross product of normal vectors, then use the shortest distance formula for skew lines: d = |((P₂-P₁)·(b₁×b₂))|/|b₁×b₂| and equate to 1/√3 to solve for α.
Step 1: Find the direction vector of the second line (intersection of planes x+y+z+1=0 and 2x-y+z+3=0). Direction vector: b_2 = (1,1,1)×(2,-1,1) = (1-(-1), 1-2, -1-2) = (2,-1,-3) Step 2: First line has direction b_1 = (α,-1,1) and passes through P_1(1,-1,0). Find a point on the second line: Set x=0 in both plane equations: y+z+1=0 and -y+z+3=0. Solving: z=2, y=-3, so P_2(0,-3,2). Step 3: Calculate (P_2-P_1) = (-1,-2,2) and b_1 × b_2 = (α,-1,1)×(2,-1,-3) = (3-(-1), -(−3α-2), -α-(-2)) = (4, 3α+2, 2-α) Step 4: Apply distance formula: d = |(-1,-2,2)·(4,3α+2,2-α)|/√(16+(3α+2)^2+(2-α)^2) = 1/√3 Numerator: |-4-2(3α+2)+2(2-α)| = |-4-6α-4+4-2α| = |-8α-4| = 4|2α+1| Step 5: Denominator: √(16+9α^2+12α+4+4-4α+α^2) = √(10α^2+8α+24) Setting up: 4|2α+1|/√(10α^2+8α+24) = 1/√3 ⟹ 16(2α+1)^2 = (10α^2+8α+24)/3 ⟹ 48(4α^2+4α+1) = 10α^2+8α+24 ⟹ 192α^2+192α+48 = 10α^2+8α+24 ⟹ 182α^2+184α+24 = 0 ⟹ 91α^2+92α+12 = 0 ⟹ (7α+4)(13α+3) = 0 ∴ α = -4/7 or α = -3/13
Correct Answer: C