Problem Statement
A robotic arm axis line is defined by x − 1 2 = y − 2 3 = z − 1 2 \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-1}{2} 2 x − 1 = 3 y − 2 = 2 z − 1 M ( 4 , 3 , 7 ) M(4, 3, 7) M ( 4 , 3 , 7 )
Step-by-Step Mathematical Solution
1. Extract Line Geometry
From the symmetric equation:
Point on the arm line: P = ( 1 , 2 , 1 ) P = (1, 2, 1) P = ( 1 , 2 , 1 )
Direction vector of the arm segment: v ⃗ = ⟨ 2 , 3 , 2 ⟩ \vec{v} = \langle 2, 3, 2 \rangle v = ⟨ 2 , 3 , 2 ⟩
2. Construct Displacement Vector P M ⃗ \vec{PM} P M
P M ⃗ = ⟨ 4 − 1 , 3 − 2 , 7 − 1 ⟩ = ⟨ 3 , 1 , 6 ⟩ dm \vec{PM} = \langle 4 - 1, 3 - 2, 7 - 1 \rangle = \langle 3, 1, 6 \rangle\text{ dm} P M = ⟨ 4 − 1 , 3 − 2 , 7 − 1 ⟩ = ⟨ 3 , 1 , 6 ⟩ dm 3. Compute Cross Product P M ⃗ × v ⃗ \vec{PM} \times \vec{v} P M × v
P M ⃗ × v ⃗ = ∣ i ^ j ^ k ^ 3 1 6 2 3 2 ∣ \vec{PM} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 6 \\ 2 & 3 & 2 \end{vmatrix} P M × v = i ^ 3 2 j ^ 1 3 k ^ 6 2 P M ⃗ × v ⃗ = i ^ ( 1 ⋅ 2 − 6 ⋅ 3 ) − j ^ ( 3 ⋅ 2 − 6 ⋅ 2 ) + k ^ ( 3 ⋅ 3 − 1 ⋅ 2 ) \vec{PM} \times \vec{v} = \hat{i}(1\cdot 2 - 6\cdot 3) - \hat{j}(3\cdot 2 - 6\cdot 2) + \hat{k}(3\cdot 3 - 1\cdot 2) P M × v = i ^ ( 1 ⋅ 2 − 6 ⋅ 3 ) − j ^ ( 3 ⋅ 2 − 6 ⋅ 2 ) + k ^ ( 3 ⋅ 3 − 1 ⋅ 2 ) P M ⃗ × v ⃗ = i ^ ( 2 − 18 ) − j ^ ( 6 − 12 ) + k ^ ( 9 − 2 ) = − 16 i ^ + 6 j ^ + 7 k ^ = ⟨ − 16 , 6 , 7 ⟩ \vec{PM} \times \vec{v} = \hat{i}(2 - 18) - \hat{j}(6 - 12) + \hat{k}(9 - 2) = -16\hat{i} + 6\hat{j} + 7\hat{k} = \langle -16, 6, 7 \rangle P M × v = i ^ ( 2 − 18 ) − j ^ ( 6 − 12 ) + k ^ ( 9 − 2 ) = − 16 i ^ + 6 j ^ + 7 k ^ = ⟨ − 16 , 6 , 7 ⟩ 4. Calculate Vector Magnitudes
Magnitude of cross product:
∥ P M ⃗ × v ⃗ ∥ = ( − 16 ) 2 + 6 2 + 7 2 = 256 + 36 + 49 = 341 ≈ 18.466 \|\vec{PM} \times \vec{v}\| = \sqrt{(-16)^2 + 6^2 + 7^2} = \sqrt{256 + 36 + 49} = \sqrt{341} \approx 18.466 ∥ P M × v ∥ = ( − 16 ) 2 + 6 2 + 7 2 = 256 + 36 + 49 = 341 ≈ 18.466
Magnitude of direction vector v ⃗ \vec{v} v ∥ v ⃗ ∥ = 2 2 + 3 2 + 2 2 = 4 + 9 + 4 = 17 ≈ 4.123 \|\vec{v}\| = \sqrt{2^2 + 3^2 + 2^2} = \sqrt{4 + 9 + 4} = \sqrt{17} \approx 4.123 ∥ v ∥ = 2 2 + 3 2 + 2 2 = 4 + 9 + 4 = 17 ≈ 4.123
5. Calculate Minimum Clearance Distance d d d
d = ∥ P M ⃗ × v ⃗ ∥ ∥ v ⃗ ∥ = 341 17 = 341 17 = 20.0588 ≈ 4.48 dm ( 44.8 cm ) d = \frac{\|\vec{PM} \times \vec{v}\|}{\|\vec{v}\|} = \frac{\sqrt{341}}{\sqrt{17}} = \sqrt{\frac{341}{17}} = \sqrt{20.0588} \approx 4.48\text{ dm} \quad (44.8\text{ cm}) d = ∥ v ∥ ∥ P M × v ∥ = 17 341 = 17 341 = 20.0588 ≈ 4.48 dm ( 44.8 cm ) Real-Life Engineering Meaning & Application
Physical Interpretation: The perpendicular distance of 4.48 dm 4.48\text{ dm} 4.48 dm 44.8 cm 44.8\text{ cm} 44.8 cm Engineering Application: In robotics software motion planners (such as ROS MoveIt), safety envelopes are established around all rigid links. If the robot's physical arm casing radius (15 cm 15\text{ cm} 15 cm 20 cm 20\text{ cm} 20 cm 10 cm 10\text{ cm} 10 cm 45 cm 45\text{ cm} 45 cm 44.8 cm 44.8\text{ cm} 44.8 cm