<p>If the length of the perpendicular from the point \((\beta, 0, \beta)\) (\(\beta \neq 0\)) to the line, \(\dfrac{x}{1} = \dfrac{y-1}{0} = \dfrac{z+1}{-1}\) is \(\sqrt{\dfrac{3}{2}}\), then \(|\beta|\) is equal to ______.</p>
Step-by-Step Solution
Key Concept: The perpendicular distance from a point to a line in 3D is given by ||(P₀-P)×d̂||, where P₀ is the point, P is any point on the line, and d̂ is the unit direction vector. Set this equal to √(3/2) and solve for β.
Step 1: Identify the line parameters. The line is x/1 = (y-1)/0 = (z+1)/(-1), so a point on the line is P(0, 1, -1) and the direction vector is d = (1, 0, -1). The magnitude is || d || = √2. Step 2: Form the vector from a point on the line to the given point. P_0P = (β, 0, β) - (0, 1, -1) = (β, -1, β+1). Step 3: Compute the cross product P_0P × d :
| i j k |
|β -1 β+1|
|1 0 -1|
= i [(-1)(-1) - (β+1)(0)] - j [β(-1) - (β+1)(1)] + k [β(0) - (-1)(1)]
= i (1) - j (-β - β - 1) + k (1)
= (1, 2β+1, 1) Step 4: Calculate the magnitude: || P_0P × d || = √(1 + (2β+1)^2 + 1) = √(2 + 4β^2 + 4β + 1) = √(4β^2 + 4β + 3). Step 5: Apply the distance formula: distance = || P_0P × d ||/|| d || = √(4β^2 + 4β + 3)/√2 = √(3/2). Step 6: Square both sides: (4β^2 + 4β + 3)/2 = 3/2, so 4β^2 + 4β + 3 = 3, giving 4β^2 + 4β = 0, thus 4β(β + 1) = 0. Step 7: Since β ≠ 0, we have β = -1, so |β| = 1 . ∴ Answer: 1
Correct Answer: 1