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Say we have a vector $\vec{v}$ that defines the chemical bond between two atoms, and whose components are known

$$ \vec{v} = \begin{bmatrix}v_1 \\ v_2 \\ v_3 \end{bmatrix} $$

Lets define another vector $\vec{u}$ that is collinear to $\vec{v}$, which represents the same chemical bond but with a different length, and whose length is known

$$ \vec{u} = c \vec{v} $$

From

$$ \lvert \vec{u} \rvert = \sqrt{(cv_1)^2 + (cv_2)^3 + (cv_3)^2} $$

I was able to derive the following relationship (which I suppose for those who do linear algebra is a well-known quality of collinear vectors)

$$ \lvert \vec{u} \rvert = c \lvert \vec{v} \rvert $$

Since the magnitudes of both vectors are known, we can find $c$ easily (the initial bond distance is of course known, and we are free to define the new bond distance as we please).

However, and perhaps this is trivial, what I want is to obtain a quantity that I can add to the specific atomic coordinates in my molecule. Say my system consists of two molecules, and I want to adjust the distance between these two molecules, by translating one of the molecules in a direction defined by the vector between one atom of molecule 1 and one atom of molecule $2$.

To implement this in my code, I suppose I have to work with the coordinates themselves (since a vector on its own could be anywhere in space). So we have three sets of coordinates: atom $A$, initial atom $B$, and final atom $B$

$$ \begin{align} A &= (A_x, A_y, A_z)\\[3mm] B_i &= (B_{x,i}, B_{y,i}, B_{z,i})\\[3mm] B_f &= (B_{x,f}, B_{y,f}, B_{z,f}) \end{align} $$

Intuitively I want to set up a set of three linear equations, one for each coordinate, but I am unsure how to proceed. Additionally I sort of want to end up using the result $\lvert \vec{u} \rvert = c \lvert \vec{v} \rvert$ (because collinearity is included here....... and because I derived it, lol).

Any thoughts on how to solve this (I think) simple problem?


If I understood Narlin's answer correctly, we should have the following.

Given the unit vector

$$ \vec{w} = \frac{\vec{v}}{\lvert \vec{v} \rvert} = $$

and

$$ c = \frac{\lvert \vec{u} \rvert }{\lvert \vec{v} \rvert} $$

we should have that

$$ (B_{f,x}, B_{f,y}, B_{f_z}) = (A_x, A_y, A_z) + \frac{\lvert \vec{u} \rvert }{\lvert \vec{v} \rvert} \cdot \begin{bmatrix}w_1 \\ w_2 \\ w_3 \end{bmatrix} $$

Or, if we separate the coordinates into their own equations

$$ \begin{align} B_{f,x} &= A_x + \frac{\lvert \vec{u} \rvert }{\lvert \vec{v} \rvert} w_1 \\[3mm] B_{f,y} &= A_y + \frac{\lvert \vec{u} \rvert }{\lvert \vec{v} \rvert} w_2 \\[3mm] B_{f,z} &= A_z + \frac{\lvert \vec{u} \rvert }{\lvert \vec{v} \rvert} w_3 \end{align} $$

Numerical example

Defining the points $A$ and $B_i$

$$ \begin{align} A &= (0,0,0) \\[3mm] B_i &= (0,2,4) \end{align} $$

we have that the vector $\vec{AB_i}$

$$ \vec{AB_i} = \begin{bmatrix}0 \\ 2 \\ 4 \end{bmatrix} $$

and $c=2$, we have that $\lvert \vec{AB_i} \rvert = 2\sqrt{5}$ and that $\lvert \vec{AB_f} \rvert = 4\sqrt{5}$. The unit vector $\vec{w}$ becomes

$$ \vec{w} = \begin{bmatrix}0 \\ 5^{-1/2} \\ 2\cdot 5^{-1/2} \end{bmatrix} $$

We then have everything to find the coordinates of point $B_f$.

Using the formula from Narlin's answer, I get that

$$ \begin{align} B_{f,x} &= 0 + 2 \cdot 0 &= 0 \\[3mm] B_{f,y} &= 0 + 2 \cdot 5^{-1/2} &= \frac{2}{\sqrt{5}} \\[3mm] B_{f,z} &= 0 + 2 \cdot 2 \cdot 5^{-1/2} &= \frac{4}{\sqrt{5}} \end{align} $$

which is clearly wrong. The correct answer should be $B_f = (0,4,8)$.

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  • $\begingroup$ Is $B_f$ the point $C$ in your question title? $\endgroup$ – Narlin Oct 12 '18 at 21:07
  • $\begingroup$ Yes, that is correct. $\endgroup$ – Yoda Oct 12 '18 at 21:16
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In this answer, variable $c$ represents the length of the new vector. This is a change of variable from the ones used in the question. Here, $c=\Vert u \Vert$. The OP used $\Vert u \Vert$ to represent the new vector length. $$\\$$You know coordinates for A and B. You know the length between A and B. You also know the length $c$ of the new vector. Define a unit vector as follows: $$w=(B-A)/|(B-A)|$$ The new point $C = A + c\cdot w$. Create a numeric example and this should work for you. $$\\$$(edit) You know coordinates for A and B. You know the length between A and B. In your example, the length between A & B is $s=\Vert(B-A)\Vert=2\sqrt{5}$
You also know the length $c$ of the new vector. From your example, even though you said $c=2$. What you really want is for $c$ to double the length of vector $s$. That is, you want $c=2|B-A|=4\sqrt{5}$ Define a unit vector as follows: $$\mathbf{w}=(B-A)/\Vert(B-A)\Vert$$ The new point $C = A + c\cdot \mathbf{w}$. Defining $c$ this way is important in order to move away from having point A at $(0,0,0)$ $$\mathbf{w}=\frac{B-A}{\Vert B-A\Vert}=\left(\begin{array}{c}0\\\frac{\sqrt{5}}{5}\\\frac{\sqrt{5}}{5}\end{array}\right)$$ $$c=2|B-A|=4\sqrt{5}$$ $$\left(\begin{array}{c} 0\\ 4\\ 8 \end{array}\right)=\left(\begin{array}{c} 0\\ 0\\ 0 \end{array}\right)+c\cdot\left(\begin{array}{c} 0\\ \frac{\sqrt{5}}{5}\\ \frac{\sqrt{5}}{5} \end{array}\right)$$ That is, $$C=A+c\mathbf{w}$$

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  • $\begingroup$ It is a bit unclear in your answer which quantities are vectors and which scalars, and which are points. Could you edit? $\endgroup$ – Yoda Oct 12 '18 at 21:57
  • $\begingroup$ A and B are points but are treated as vector points. That is, they have 3 components and are subtracted, multiplied, etc as if they were vectors. $w$ will turn out to be a vector. point A is treated as a vector - again 3 components. Only $c$ is a scalar. $\endgroup$ – Narlin Oct 12 '18 at 22:14
  • $\begingroup$ I am not able to make your solution work. See my updated question. $\endgroup$ – Yoda Oct 14 '18 at 13:22
  • $\begingroup$ I see what you are doing. I will update my answer in a few minutes. It really won't change any, but will explain how to get your value. $\endgroup$ – Narlin Oct 14 '18 at 17:24
  • $\begingroup$ I'm pretty bad at LaTex. Takes a few tries. $\endgroup$ – Narlin Oct 14 '18 at 18:21

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