How to rotate a line around one of its vertexes

Question

I am making my first raycasting engine, and would like to rotate a line over an angle θ

How does one do this? Would it be possible to show me some basic C++ code or some pseudocode?

This image describes my problem: enter image description here

I decided to make all of this in graphics.h, because it is the simplest graphics header for C/C++.

Also interested, because cross product works only in 3 dimensions — , Jan 03 '19 at 10:43
If you are doing linear algebra, pick a linear algebra library and express it using values of appropriate types. Then you can write code that looks like the maths `B = P + M * (A - P);` *can be* valid C++ — Caleth, Jan 03 '19 at 12:08
This website can be of help to you: http://geomalgorithms.com/ — kvantour, Jan 03 '19 at 12:08

score 4 · Accepted Answer · answered Jan 03 '19 at 10:53

You want:

B = P + M * (A - P)

Where M is a 2D rotation matrix:

M = |  cos(ϴ)  -sin(ϴ) |
    |  sin(ϴ)   cos(ϴ) |

In C++ it could be written as:

float c = cos(theta), s = sin(theta);
float dx = ax - px, dy = ay - py;
float bx = px + c * dx - s * dy;
float by = py + s * dx + c * dy;

score 2 · Answer 2 · answered Jan 03 '19 at 10:51

2

One simple algorithm:

All these three steps can be done using one 3x3 matrix multiplication.

answered Jan 03 '19 at 10:51

Maxim Egorushkin

YSC · Answer 3 · 2019-01-03T13:42:35.793

The scalar product of two vectors have the following property:

vec(PA) . vec(PB) = rho cos theta

Taking the definition of our two vectors:

vec(PA) = (x_a-x_p, y_a-y_p)
vec(PB) = (x_b-x_p, y_b-y_p)

We can get:

(x_a-x_p)(x_b-x_p) + (y_a-y_p)(y_b-y_p) = rho cos theta (1)

Since PA=PB, we also have:

(x_a-x_p)^2 + (y_a-y_p)^2 = (x_b-x_p)^2 + (y_b-y_p)^2 (2)

From (1) and (2) you can derive x_band y_b with some arithmetic autopilot.

3 Answers3