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Structural Civil Engineering (Conduit & Support Cable Clearance) Solution

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Problem Statement

A structural steel cable follows path x=2+t,y=−1+2t,z=3−2tx = 2 + t, y = -1 + 2t, z = 3 - 2t (meters). An electrical line passes near point M(5,5,2)M(5, 5, 2).

Step-by-Step Mathematical Solution

1. Extract Cable Line Data

  • Reference point on cable (at t=0t=0): P=(2,−1,3)P = (2, -1, 3)
  • Direction vector of cable: v⃗=⟨1,2,−2⟩\vec{v} = \langle 1, 2, -2 \rangle

2. Construct Displacement Vector PM⃗\vec{PM}

PM⃗=⟨5−2,5−(−1),2−3⟩=⟨3,6,−1⟩ m\vec{PM} = \langle 5 - 2, 5 - (-1), 2 - 3 \rangle = \langle 3, 6, -1 \rangle\text{ m}

3. Compute Cross Product PM⃗×v⃗\vec{PM} \times \vec{v}

PM⃗×v⃗=∣i^j^k^36−112−2∣\vec{PM} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 6 & -1 \\ 1 & 2 & -2 \end{vmatrix}
PM⃗×v⃗=i^(6(−2)−(−1)(2))−j^(3(−2)−(−1)(1))+k^(3(2)−6(1))\vec{PM} \times \vec{v} = \hat{i}(6(-2) - (-1)(2)) - \hat{j}(3(-2) - (-1)(1)) + \hat{k}(3(2) - 6(1))
PM⃗×v⃗=i^(−12+2)−j^(−6+1)+k^(6−6)=−10i^+5j^+0k^=⟨−10,5,0⟩\vec{PM} \times \vec{v} = \hat{i}(-12 + 2) - \hat{j}(-6 + 1) + \hat{k}(6 - 6) = -10\hat{i} + 5\hat{j} + 0\hat{k} = \langle -10, 5, 0 \rangle

4. Calculate Vector Magnitudes

  • Magnitude of cross product:
    ∥PM⃗×v⃗∥=(−10)2+52+02=100+25=125=55 m2\|\vec{PM} \times \vec{v}\| = \sqrt{(-10)^2 + 5^2 + 0^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5}\text{ m}^2
  • Magnitude of direction vector v⃗\vec{v}:
    ∥v⃗∥=12+22+(−2)2=1+4+4=9=3 m\|\vec{v}\| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3\text{ m}

5. Calculate Shortest Clearance Distance dd

d=∥PM⃗×v⃗∥∥v⃗∥=553≈3.73 metersd = \frac{\|\vec{PM} \times \vec{v}\|}{\|\vec{v}\|} = \frac{5\sqrt{5}}{3} \approx 3.73\text{ meters}

Real-Life Engineering Meaning & Application

  • Physical Interpretation: The calculated distance of 3.73 m3.73\text{ m} is the absolute minimum geometric air gap between the electrical conductor and the grounded steel structural cable.
  • Engineering Application: National Electrical Safety Code (NESC) and OSHA standards specify minimum flashover clearance distances around high-voltage transmission lines (typically ≥3.0 m\ge 3.0\text{ m} for medium voltage) to prevent electrical arcing under high humidity or windy cable sway conditions. A verified distance of 3.73 m3.73\text{ m} confirms compliance, allowing structural engineers to issue formal building certification.