How to plot Vector in 3d using two angles and initial point

Hi! Could you please help me

How to put 3d vector if i know initial point coordinates and two angles

I tries this one, but still could not understand where is my phi and theta on 3d according to matlab plotting

x0=1.5; %initial x position
y0=1.5; %initial y position
z0=3.0;  
r = sqrt(x0^2 + y0^2 + z0^2);
x1 = r * sin(Phi0) * cos(Theta0);
y1 = r * sin(Phi0) * sin(Theta0);
z1 = r * cos(Phi0);
line([x0 x1],[y0,y1],[z0,z1]); 
xlabel X; ylabel Y; zlabel Z
grid on; 

axis equal

%I also have tried this one 
%x1 =r * sind(theta) * cosd(phi); % !!!convert to cart mine last one 
%y1 = r* sind(theta) * sind(phi);
%z1 = r* cosd(theta);

Here his how my angles should be

Answer 1

[Askic V]

I must admit something is confusing me in your post.

If you have a point in 3D space and the second point is in origin, then you have all you need to represent a vector (two points are enough to determine its length and direction). So you can plot it just by using these two points (quiver3)

On the other hand, if you have az and el angles measured exactly the same way you posted on the picture then, you can calculate a vector axis components in the following way (assuming vector starts in origin and ends in points marked with star, with length of r).

The vector has there parts vx*i + vy*j + vz*k, where i, j and k are unit axis vectors.From your picture it is clear that projection to the xy plane is: r*sin(theta), because, prjection to the z axis is r*cos(theta).

In the xy plane, the x (i) component is r*sin(theta)*sin(phi), and y component is r*sin(theta)*cos(phi).

To me, much sense has to find a components of a vector given vector length and these two angles, for example:

[vx,vy,vz] = find_vector(2, pi/6, pi/30)

vx = 0.1045

vy = 0.1810

vz = 1.9890

function [vx,vy,vz] = find_vector(vec_length, azim, el)
    phi = azim;
    theta = el;
    r = vec_length;
    vx = r * sin(phi) * sin(theta);
    vy = r * cos(phi) * sin(theta);
    vz = r * cos(theta);    
    quiver3(0,0,0,vx,vy,vz,'off');
    view(45,20)
end

[Aknur]

@Askic V thank you for your clearification. My initial point is

x0=1.5; %initial x position

y0=1.5; %initial y position

z0=3.0;

The figure shows only angles poisition (Phi cover 0 -360, Theta 0-180), where is Phi and where is Theta.

I need to find second point to draw a vector.

Kindly ask about in this equation vx*i + vy*j + vz*k what I should take as vx, vy, vz? is there should be coordinates of second point?

[Askic V]

vx, vy, vz are coordinates of a vector for which length and two angles (azimuth and elevation) are known.

For your particular task, second point cannot be determined, because, you don't have a length, or radius as Walter said.

But since you need to have only a segment, then this code will probably do:

x0 = 1.5; %initial x position
y0 = 1.5; %initial y position
z0 = 3.0; %initial Z position 
[vx,vy,vz] = find_vector(1, pi/6, pi/30)

vx = 0.0523

vy = 0.0905

vz = 0.9945

quiver3(x0,y0,z0,x0+vx,y0+vy,z0+vz, 'off')
zlim([0 10])

function [vx,vy,vz] = find_vector(vec_length, azim, el)
    phi = azim;
    theta = el;
    r = vec_length;
    vx = r * sin(phi) * sin(theta);
    vy = r * cos(phi) * sin(theta);
    vz = r * cos(theta);    
end

[Aknur]

@Askic V thank you. But this vector goes up, and I need it ti go down. How I can do it. I tried use barycentrcin coordinates and it works for me almost. And one more question can I calculate without r. If I will multiply for r then my vector also will go up

x0=1.5; %initial x position

y0=1.5; %initial y position

z0=3.0;

theta = 30; %theta 30 60 45-30 60-85

phi=60; %phi does not work

r = sqrt(x0^2 + y0^2 + z0^2);

%uv = phitheta2uv([phi; theta])

u = r * sind(theta) * cosd(phi);

v = r * sind(theta) * sind(phi);

w = r * cosd(theta);

%quiver3(x0,y0,z0,r,u,v,w)

%[az,r]=cart2pol(x0,y0);

%[x1,y1,z1]=sph2cart(az,elev,r);

%x1 = sind(theta) * cosd(phi); % convert to cart mine

%y1 = sind(theta) * sind(phi);

%z1 = cosd(theta);

%x1 = cosd(theta) * cosd(phi); % convert to cart matlab

%y1 = cosd(theta) * cosd(phi);

%z1 = sind(theta);

line([x0 u],[y0 v],[z0 w]);

%xlabel X; ylabel Y; zlabel Z

%grid on;

axis equal

disp(theta);

disp(phi);

disp(u);

disp(v);

disp(w);

[Askic V]

Hello Ankur,

first we need to be sure whether x1 = sind(theta) * cosd(phi) or x1 = sind(theta) * sind(phi)?

According to your picture posted in the first post, projection of vector from origin to the xy plane is clearly r* sin(theta).

Now, based on the fact how x and y axes are labeled, it seems that y component is cos phi, and x component is sin(phi), therefore x = r*sin(theta)*sin(phi).

Do you see and agree with this?

Regarding the other part of the question about pointing downwards, I have made a little code snippet that should clarify the things:

clear
close all
x0 = 1.5; %initial x position
y0 = 1.5; %initial y position
z0 = 3.0; %initial Z position 
[vx,vy,vz] = find_vector(pi/6, pi/30);

v1 = [x0,y0,z0];
v2 = [vx,vy,vz]; %unit vector
% plot v1
quiver3(0,0,0,v1(1),v1(2),v1(3),'off');
text(v1(1),v1(2),v1(3),' \leftarrow v1')
hold on
% plot v2
quiver3 (0,0,0,v2(1),v2(2),v2(3),'off'); % unit vector calculated from angles)
text(v2(1),v2(2),v2(3),' \leftarrow v2')
% find v3 as v1-v2
v3 = v1-v2;
quiver3 (0,0,0,v3(1),v3(2),v3(3),'off');
text(v3(1),v3(2),v3(3),' \leftarrow v3');
% Now create and draw resultant vector that starts at x0,y0,z0 and 
% is anti collinear to v2 (point downwards)
quiver3 (x0,y0,z0,v3(1)-x0,v3(2)-y0,v3(3)-z0,'off', 'LineWidth',2);
view(-15,20)

function [vx,vy,vz] = find_vector(azim, el)
    phi = azim;
    theta = el;
    vx = sin(phi) * sin(theta);
    vy = cos(phi) * sin(theta);
    vz = cos(theta);    
end

If I understood your question, purple vector is what you're are looking for, right?

[Aknur]

Dear, @Askic V thank you for your valuable feedback. I understand your idea and the code almost. As I have point which is start point of vector and two angles. I tried to do it this way. To find unit vector which will be calculated accoridng these formulas

x1 = sind(theta) * sind(phi);

y1 = sind(theta) * cosd(phi);

z1 = cosd(theta);

But when I rotate ny cube to check if the angles show correct angles it does not work properly. For example when I put theta = 0, and phi = 30. The figure should show me that theta angle which is along z axis should be equal 0, but it shows like theta =30 and phi =45 according to the figure of matlab. Could you please to look on my code

I really became confused with angles and how to plot theta and phi. And dont know from where matlab default starts 0 degree of theta and phi

%figure

%pax = polaraxes;

Room = [

3 3 0;

0 3 0;

0 3 3;

3 3 3;

0 0 3;

3 0 3;

3 0 0;

0 0 0];

% edges by looking for all combinations of points on cube that

% differ by only one coordinate

sections_per_edge = 1;

weights = ((1:sections_per_edge-1) / sections_per_edge).';

edges = []; % indices into Room

points = [];

n = size(Room, 1);

for i = 1:n-1

pointA = Room(i, :);

for j = i+1:n

pointB = Room(j, :);

if nnz(pointA - pointB) == 1

edges = [edges; i, j];

points = [points; weights * pointA + (1 - weights) * pointB];

end

end

end

plot3(Room(:,1), Room(:,2), Room(:,3), '.b')

hold on

plot3(points(:,1), points(:,1), points(:,1), '.r', 'markersize', 10)

hold on

line([Room(edges(:,1), 1), Room(edges(:,2), 1)].', ...

[Room(edges(:,1), 2), Room(edges(:,2), 2)].', ...

[Room(edges(:,1), 3), Room(edges(:,2), 3)].', 'color', 'k')

plot3(1.5, 1.5, 3.0, '*', 'markersize', 15)

plot3(1.5, 0, 0, 'o', 'markersize', 7, 'MarkerFaceColor', 'black')

plot3(0, 1.5, 0, 'o', 'markersize', 7, 'MarkerFaceColor', 'black')

plot3(3, 1.5, 0,'o', 'markersize', 7, 'MarkerFaceColor', 'black')

plot3(1.5, 3, 0, 'o', 'markersize', 7, 'MarkerFaceColor', 'black')

theta = 0;

phi = 30;

%elev = (90 - Theta);

%az = 60;

x0=1.5; %initial x position

y0=1.5; %initial y position

z0=3.0;

x1 = sind(theta) * sind(phi);

y1 = sind(theta) * cosd(phi);

z1 = cosd(theta);

plot3 ([x0 x1],[y0 y1],[z0 z1]);

xlabel X; ylabel Y; zlabel Z

grid on;

axis equal

%axis=gca;

%ax.PhiLim = [0 360]

%Ax.PhiDir = 'clockwise';

%Ax.PhiZeroLocation = 'top';

%ax.ThetaLim = [0 180]

%Ax.ThetaDir = 'clockwise';

%Ax.ThetaZeroLocation = 'top';

This picture in general what I need, but confused with angles. They (the unit vector) do not show correct direction according to the figure. I hope I were able to explain

Thank you for time and effort

[Aknur]

@Askic V I am sorry for late answer

Answer 2

[Walter Roberson]

You have initial coordinates, and you calculate radius based on those coordinates. That is equivalent to saying that the initial point is on the surface of the sphere around the origin.

Then you calculate coordinates based on that radius and two angles. That is like asking "what point at the same radius as the initial point, is at this particular angle on that implied sphere?"

That is a valid thing to calculate, but is that what you want to calculate?

I suspect that what you are looking for is to find a vector at a particular spatial angle, that goes through the given point, and to sketch a segment of that vector,

[Aknur]

@Walter Roberson thank you for your feedback and explanation. Yes you are right. I want to find a vector which starts from initial position at a particular spatial angle, and sketch segment of that vector.

As I am right these two angles Phi and Theta gave me direction in 3d. And also I need to define x1, y1, z1 - not absolutely end point of vector (but one which will be in side of coordinates( 3,3,3) if possible