我想在处理中将矢量投影到矢量a和矢量c上。 在我的草图矢量中，a是红色，c是蓝色，我希望c垂直于b，但这是我遇到很多麻烦的地方。我正在使用JAMA库试图让这更容易。对此有任何帮助非常感谢，因为我已经被困了大约一个星期了。
float X=200; // Origin : Note we have now centred the origin in the
X-direction float Y=350; float ax=150; // Vector a resolved into
components float ay=-50; float bx=0; // Vector b resolved into
components float by=-150; float cx=150; float cy=200;
Matrix a; Matrix b; Matrix c;
void setup() {
size(400,400); // Create a drawing window
strokeWeight(3); // Make pen 3 pixels wide for all lines
double [][] anums = {{ax},
{ay}};
double [][] bnums = {{bx},
{by}};
double [][] cnums = {{-cy},
{cx}};
a = new Matrix(anums);
b = new Matrix(bnums);
c = new Matrix(cnums); }
void draw() {
background(255); // Clear screen
// Evaluate equation (1.5)
// STEP1: Insert code here that computes a_unit (i.e. the unit vector in the
// direction of a
double length = a.norm2();
Matrix a_unit= a.times(1/length);
// STEP2: Insert code here to compute the dot product of b and a_unit
Matrix a_unit_T = a_unit.transpose();
Matrix projection = a_unit_T.times(b);
double lp = projection.get(0,0);
// STEP3 Insert code here to compute the vector p using equation 1.5 above Matrix p = a_unit.times(lp);
float px = (float)p.get(0,0);
float py = (float)p.get(1,0);
float ax = (float)a.get(0,0);
float ay = (float)a.get(1,0);
float bx = (float)b.get(0,0);
float by = (float)b.get(1,0);
float cx = (float)c.get(0,0);
float cy = (float)c.get(1,0);
// Draw the projection of b onto a
stroke(0,0,0); // Use a black pen
ellipse(X+px,Y+py,10,10); // point where b projects onto a
line(X+px,Y+py,X+bx,Y+by); // line from a to point of projection on b
stroke(255,0,0); // Make pen red
arrow(X,Y,X+ax,Y+ay); // Draw vector a starting at (X,Y)
//stroke(0,0,255);
//arrow(X,Y,X-ax,Y+ay);
stroke(0,255,0); // Make pen green
arrow(X,Y,X+bx,Y+by); // Draw vector b starting at (X,Y)
// STEP 4. Insert code here to add a new vector at 90 degrees to the vector a
stroke(0,0,255);
arrow(X,Y,X+cx,Y+cy);
// STEP 5. Insert code here to compute and draw the projection of b onto c
double length1 = c.norm2();
Matrix c_unit= c.times(1/length1);
// STEP2: Insert code here to compute the dot product of b and a_unit
Matrix c_unit_T = c_unit.transpose();
Matrix projection1 = c_unit_T.times(b);
double lp1 = projection.get(0,0);
// STEP3 Insert code here to compute the vector p using equation 1.5 above
Matrix r = c_unit.times(lp1);
float rx = (float)r.get(0,0);
float ry = (float)r.get(1,0);
stroke(0,0,0); // Use a black pen
ellipse(X+rx,Y+ry,10,10); // point where b projects onto a
line(X+rx,Y+ry,X+bx,Y+by); // line from a to point of projection on b
if (mouseButton == RIGHT)
{
a.set(0,0,(double)mouseX-X);
a.set(1,0,(double)mouseY-Y);
}
if (mouseButton == LEFT)
{
b.set(0,0,(double)mouseX-X);
b.set(1,0,(double)mouseY-Y);
} } // Draw an arrow from (x1,y1) to (x2,y2) void arrow(float x1, float y1, float x2, float y2) { line(x1, y1, x2, y2);
pushMatrix(); translate(x2, y2); float a = atan2(x1-x2, y2-y1);
rotate(a); line(0, 0, -8, -8); line(0, 0, 8, -8); popMatrix(); }
答案 0 :(得分：0)
这是代码配对，
float X=200; // Origin : Note we have now centred the origin in the X-direction
float Y=350;
float ax=300; // Vector a resolved into components
float ay=-100;
float bx=0; // Vector b resolved into components
float by=-300;
Matrix a;
Matrix b;
void setup()
{
size(400,400); // Create a drawing window
strokeWeight(3); // Make pen 3 pixels wide for all lines
double [][] anums = {{ax},
{ay}};
double [][] bnums = {{bx},
{by}};
a = new Matrix(anums);
b = new Matrix(bnums);
}
void draw()
{
background(255); // Clear screen
// Evaluate equation (1.5)
// STEP1: Insert code here that computes a_unit (i.e. the unit vector in the
// direction of a
double length = a.norm2();
Matrix a_unit = a.times(1/length);
// STEP2: Insert code here to compute the dot product of b and a_unit
Matrix a_unit_T = a_unit.transpose();
Matrix projection = a_unit_T.times(b);
double lp = projection.get(0,0);
// STEP3: Insert code here to compute the vector p using equation 1.5 above
Matrix p = a_unit.times(lp);
float px = (float)p.get(0,0);
float py = (float)p.get(1,0);
float ax = (float)a.get(0,0);
float ay = (float)a.get(1,0);
float bx = (float)b.get(0,0);
float by = (float)b.get(1,0);
// Draw the projection of b onto a
stroke(0,0,0); // Use a black pen
ellipse(X+px,Y+py,10,10); // point where b projects onto a
line(X+px,Y+py,X+bx,Y+by); // line from a to point of projection on b
stroke(255,0,0); // Make pen red
arrow(X,Y,X+ax,Y+ay); // Draw vector a starting at (X,Y)
stroke(0,255,0); // Make pen green
arrow(X,Y,X+bx,Y+by); // Draw vector b starting at (X,Y)
// STEP 4. Insert code here to add a new vector at 90 degrees to the vector a
double [][] cnums = {{ay},
{-ax}};
Matrix c = new Matrix(cnums);
float cx = (float)c.get(0,0);
float cy = (float)c.get(1,0);
stroke(0,0,255);
arrow(X,Y,X+cx,Y+cy);
// STEP 5. Insert code here to compute and draw the projection of b onto c
double length1 = c.norm2();
Matrix c_unit= c.times(1/length1);
Matrix c_unit_T = c_unit.transpose();
Matrix projection1 = c_unit_T.times(b);
double lp1 = projection1.get(0,0);
Matrix r = c_unit.times(lp1);
float rx = (float)r.get(0,0);
float ry = (float)r.get(1,0);
stroke(0,0,0); // Use a black pen
ellipse(X+rx,Y+ry,10,10); // point where b projects onto a
line(X+rx,Y+ry,X+bx,Y+by); // line from a to point of projection on b
if (mouseButton == RIGHT)
{
a.set(0,0,(double)mouseX-X);
a.set(1,0,(double)mouseY-Y);
}
if (mouseButton == LEFT)
{
b.set(0,0,(double)mouseX-X);
b.set(1,0,(double)mouseY-Y);
}
}
// Draw an arrow from (x1,y1) to (x2,y2)
void arrow(float x1, float y1, float x2, float y2)
{
line(x1, y1, x2, y2);
pushMatrix();
translate(x2, y2);
float a = atan2(x1-x2, y2-y1);
rotate(a);
line(0, 0, -8, -8);
line(0, 0, 8, -8);
popMatrix();
}