101
0/318
Loading content...
In linear algebra and computational geometry, projecting a vector onto a line is a fundamental operation used extensively in graphics, physics simulations, and machine learning algorithms. Given a vector v and a line represented by a direction vector L, the orthogonal projection finds the point on the line L that is geometrically closest to the tip of vector v.
The projection formula leverages the dot product to decompose vector v into two orthogonal components: one parallel to L (the projection) and one perpendicular to L (the rejection). The projection is computed as:
$$\text{proj}_L(v) = \frac{v \cdot L}{L \cdot L} \cdot L$$
Where:
Geometric Interpretation: Imagine shining a light perpendicular to the line L onto vector v. The shadow cast by v onto L is exactly the orthogonal projection. This shadow represents the component of v that lies in the direction of L, with all perpendicular information discarded.
Your Task: Write a Python function that computes the orthogonal projection of a vector v onto a line defined by direction vector L. The function should return the projection vector with all components rounded to 3 decimal places.
v = [3, 4]
L = [1, 0][3.0, 0.0]The projection of [3, 4] onto the x-axis (represented by [1, 0]) drops the y-component entirely.
Step-by-step calculation:
Geometrically, projecting onto the x-axis means finding the x-coordinate while setting y to zero. The point (3, 0) is the closest point on the x-axis to the point (3, 4).
v = [2, 3]
L = [1, 1][2.5, 2.5]The line L = [1, 1] represents the 45-degree diagonal (y = x line). Projecting [2, 3] onto this diagonal finds the point equidistant in both x and y.
Step-by-step calculation:
The projection lies on the line y = x, confirming it's a valid point on the diagonal. The perpendicular distance from (2, 3) to (2.5, 2.5) is minimized.
v = [1, 2, 3]
L = [1, 0, 0][1.0, 0.0, 0.0]This is a 3D projection onto the x-axis. The direction vector [1, 0, 0] defines the x-axis in 3D space.
Step-by-step calculation:
This demonstrates that the formula works in any number of dimensions. Projecting onto the x-axis in 3D simply extracts the x-component while zeroing out y and z.
Constraints