Perpendicular Distance
Procedure
Given points l, m, and p, find the perpendicular distance between point p and line lm
$$d = \|\vec{pn}\| = \frac{\vec{pl}\cdot \vec{n}}{\|\vec{n}\|} = \vec{pl} \cdot \hat{n} $$
where \( \vec{n}\) is any vector normal to \(\vec{lm}\).
For a vector \( \mathbf{V} = [v_x\ ,\ v_y]\) the normal vector is
\[ \vec{n} = [-v_y\ ,\ v_x] \quad \mathrm{ or }\quad \vec{n} = [v_y\ ,\ -v_x] \]
For 2 dimensions, the dot product is \[\mathbf{A}\cdot\mathbf{B} = a_x \times b_x + a_y \times b_y\]
Useage
The perpendicular distance between a point and a line can often simplify sum of moments by eliminating the need to break the force into horizontal and vertical components. A secific example is the elevation cylinder on a hydraulic crane. Given the moment applied to the boom, the force in the clyinder is the moment divided by the perpedicular distance between the boom pivot and the cylinder.
Example
Points
| Point | X | Y |
|---|---|---|
| l | 30 | 20 |
| m | 250 | 200 |
| p | 80 | 220 |
Vectors
\(\mathbf{A} = \vec{pl} = [-50\ ,\ -200]\)
\(\mathbf{B} = \vec{lm} =[220\ ,\ 180]\)
\(\mathbf{n} = [180\ ,\ -220]\)
Vector Operations
\(\hat{n} = [180/284.25\ ,\ -220/284.25] = [.633\ ,\ -.774]\)
\(d = \mathbf{A}\cdot \hat{n} = -50\times .633+ \ -200\times - .774 = 123.15\)