[SOLVED] Draw line with a point and an angle

[saejox]

Hi,

I like to draw a line using a point such as Vector3(1,2,5) and an angle. Angle is in Vector3 format -- Vector3(0,1,0), Vector3(1,0,1) etc.. .--

I know how to draw line between two points, but cant figure out how to draw it with an angle.

Thanks

[omniter]

By angle, do you mean euler angles? If so, what do these euler angles describe? Please give a more detailed explanation of the desired output and its relation to the given inputs.

[saejox]

By angle, do you mean euler angles? If so, what do these euler angles describe? Please give a more detailed explanation of the desired output and its relation to the given inputs.

i want my casted ray to be drawn. i have a starting point and a vector3 direction.

given line length is 10,
start point 5,4,-2

for direction 0,1,0 ; drawline( 5,4,-2 , 5,14,-2)
for direction 1,0,0 ; drawline( 5,4,-2 , 15,4,-2)
for direction -1,0,0 ; drawline (5,4,-2 , -5,14,-2)

[saejox]

i'm sure most of you already draw raycast line for debug purposes. thats what i want to do.

[saejox]

found it!
never had to deal with vectors and points since freshman.

here is what did,


step 1: normalize direction vector first
step 2: magnitude is a scalar. like 5, 10. its length of the line
step 3: just call => drawline( (origin.x, origin.y, origin.z) , (origin.x + direction.x * magnitude, origin.y + direction.y * magnitude, origin.z + direction.z * magnitude) )


this way i can draw ray's with any magnitude i want.

thanks.

[omniter]

If you already know how to draw a line between two points, then this is just one extra step. If you have point A, a direction V, and a length t, then you can get point B with this:
B = A + t*norm(V)

EDIT:
Seems like you found a solution already. It looks a bit verbose though, considering you're working with Vector3s. The code should be just as simple as the equation I gave.

[saejox]

If you already know how to draw a line between two points, then this is just one extra step. If you have point A, a direction V, and a length t, then you can get point B with this:
B = A + t*norm(V)

EDIT:
Seems like you found a solution already. It looks a bit verbose though, considering you're working with Vector3s. The code should be just as simple as the equation I gave.

it the same equation, mine looks long as a function call.

thank you for your reply.

[lingfors]

step 1: normalize direction vector first
step 2: magnitude is a scalar. like 5, 10. its length of the line
step 3: just call => drawline( (origin.x, origin.y, origin.z) , (origin.x + direction.x * magnitude, origin.y + direction.y * magnitude, origin.z + direction.z * magnitude) )

You know operators + and * are overloaded for vectors, right? So you could just calculate end point as

Code: Select all

Vector3 endPoint = origin + magnitude * direction;