How to find the x,y elements of a 3D vector when origin, direction, z are known

[dsoltyka]

I believe this is the correct place to post this as I believe I'm going to need to solve this as a linear system, however I suppose it might be solvable using trig as well. However, I've been at it for a while and I'm out of ideas and I think I'm missing something silly.

Consider the following:

Origin = (-1.1258, 100.8336, 2489.9998)
Direction = (-0.1115, 0.0826, -0.9903)

I need to use that information to find a final point in 3D space at an arbitrary Z coordinate, -512. That final point must lie on a line parallel to a line drawn from the origin in the given direction

I had originally tried to treat it like a right triangle and solving for the hypotenuse length, and multiplying that by my direction to get the final location vector, however either that won't work or I did it wrong.

Any ideas?

 

[dsoltyka]

Nevermind, I was in fact over thinking the solution. It was simple to solve when I used the formula for a 3D line.

 

[Rasalhague]

Let p denote the vector you call "origin", and let q denote the vector you call "direction". Let r denote your "final point" whose z-coordinate is -512. Then there's a number $\lambda$ such that

$$\mathbf{p}+\lambda \mathbf{q}=\mathbf{r},$$

or expressed as matrices,

$$\begin{bmatrix}p_1\\ p_2\\ p_3\end{bmatrix}+ \lambda \begin{bmatrix}q_1\\ q_2\\ q_3\end{bmatrix} = \begin{bmatrix}r_1\\ r_2\\ r_3\end{bmatrix}.$$

Addition works componentwise, so we have

$$p_3 + \lambda q_3 = -512.$$

Just substitute the numerical values of the third components of p and q, and solve for $\lambda$.

(EDIT: Oh, I see you got there already!)

 

[dsoltyka]

Let p denote the vector you call "origin", and let q denote the vector you call "direction". Let r denote your "final point" whose z-coordinate is -512. Then there's a number $\lambda$ such that

$$\mathbf{p}+\lambda \mathbf{q}=\mathbf{r},$$

or expressed as matrices,

$$\begin{bmatrix}p_1\\ p_2\\ p_3\end{bmatrix}+ \lambda \begin{bmatrix}q_1\\ q_2\\ q_3\end{bmatrix} = \begin{bmatrix}r_1\\ r_2\\ r_3\end{bmatrix}.$$

Addition works componentwise, so we have

$$p_3 + \lambda q_3 = -512.$$

Just substitute the numerical values of the third components of p and q, and solve for $\lambda$.

(EDIT: Oh, I see you got there already!)

I did, but the elegance of your explanation would have made it simpler for sure. Thank you :)

 

[blue_raver22]

I would approach this problem by first understanding the concept of a 3D vector. A 3D vector is a mathematical representation of a point in 3D space, consisting of three components (x, y, z) that represent the coordinates of the point. In this case, the origin and direction given are also represented as 3D vectors.

To find the final point at an arbitrary z coordinate, we can use the concept of vector addition. We know that the final point must lie on a line parallel to the given direction vector, so we can add the z component of the origin vector to the z component of the direction vector to get the final z coordinate. In this case, it would be 2489.9998 + (-512) = 1977.9998.

Next, we can use the concept of scalar multiplication to find the x and y components of the final point. Since we know the direction vector, we can multiply it by a scalar value to get a new vector that has the same direction but a different magnitude. This scalar value can be calculated by dividing the z component of the final point by the z component of the direction vector. In this case, it would be 1977.9998 / (-0.9903) = -1995.882.

Finally, we can use the x and y components calculated to find the final point by adding them to the x and y components of the origin vector. This would give us a final point of (-1.1258 + (-1995.882 * (-0.1115)), 100.8336 + (-1995.882 * 0.0826), 1977.9998) = (-221.3116, -156.5928, 1977.9998).

In summary, to find the x and y elements of a 3D vector when the origin, direction, and z coordinate are known, we can use the concepts of vector addition and scalar multiplication. This approach is based on the idea that the direction vector remains constant and can be scaled to find the final point at an arbitrary z coordinate. I hope this helps in solving the problem.