Problem Statement
A structural steel cable follows path x = 2 + t , y = − 1 + 2 t , z = 3 − 2 t x = 2 + t, y = -1 + 2t, z = 3 - 2t x = 2 + t , y = − 1 + 2 t , z = 3 − 2 t M ( 5 , 5 , 2 ) M(5, 5, 2) M ( 5 , 5 , 2 )
Step-by-Step Mathematical Solution
1. Extract Cable Line Data
Reference point on cable (at t = 0 t=0 t = 0 P = ( 2 , − 1 , 3 ) P = (2, -1, 3) P = ( 2 , − 1 , 3 )
Direction vector of cable: v ⃗ = ⟨ 1 , 2 , − 2 ⟩ \vec{v} = \langle 1, 2, -2 \rangle v = ⟨ 1 , 2 , − 2 ⟩
2. Construct Displacement Vector P M ⃗ \vec{PM} P M
P M ⃗ = ⟨ 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} P M = ⟨ 5 − 2 , 5 − ( − 1 ) , 2 − 3 ⟩ = ⟨ 3 , 6 , − 1 ⟩ m 3. Compute Cross Product P M ⃗ × v ⃗ \vec{PM} \times \vec{v} P M × v
P M ⃗ × v ⃗ = ∣ i ^ j ^ k ^ 3 6 − 1 1 2 − 2 ∣ \vec{PM} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 6 & -1 \\ 1 & 2 & -2 \end{vmatrix} P M × v = i ^ 3 1 j ^ 6 2 k ^ − 1 − 2 P M ⃗ × v ⃗ = 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)) P M × v = i ^ ( 6 ( − 2 ) − ( − 1 ) ( 2 )) − j ^ ( 3 ( − 2 ) − ( − 1 ) ( 1 )) + k ^ ( 3 ( 2 ) − 6 ( 1 )) P M ⃗ × v ⃗ = i ^ ( − 12 + 2 ) − j ^ ( − 6 + 1 ) + k ^ ( 6 − 6 ) = − 10 i ^ + 5 j ^ + 0 k ^ = ⟨ − 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 P M × v = i ^ ( − 12 + 2 ) − j ^ ( − 6 + 1 ) + k ^ ( 6 − 6 ) = − 10 i ^ + 5 j ^ + 0 k ^ = ⟨ − 10 , 5 , 0 ⟩ 4. Calculate Vector Magnitudes
Magnitude of cross product:
∥ P M ⃗ × v ⃗ ∥ = ( − 10 ) 2 + 5 2 + 0 2 = 100 + 25 = 125 = 5 5 m 2 \|\vec{PM} \times \vec{v}\| = \sqrt{(-10)^2 + 5^2 + 0^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5}\text{ m}^2 ∥ P M × v ∥ = ( − 10 ) 2 + 5 2 + 0 2 = 100 + 25 = 125 = 5 5 m 2
Magnitude of direction vector v ⃗ \vec{v} v ∥ v ⃗ ∥ = 1 2 + 2 2 + ( − 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} ∥ v ∥ = 1 2 + 2 2 + ( − 2 ) 2 = 1 + 4 + 4 = 9 = 3 m
5. Calculate Shortest Clearance Distance d d d
d = ∥ P M ⃗ × v ⃗ ∥ ∥ v ⃗ ∥ = 5 5 3 ≈ 3.73 meters d = \frac{\|\vec{PM} \times \vec{v}\|}{\|\vec{v}\|} = \frac{5\sqrt{5}}{3} \approx 3.73\text{ meters} d = ∥ v ∥ ∥ P M × v ∥ = 3 5 5 ≈ 3.73 meters Real-Life Engineering Meaning & Application
Physical Interpretation: The calculated distance of 3.73 m 3.73\text{ m} 3.73 m 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} ≥ 3.0 m 3.73 m 3.73\text{ m} 3.73 m