In the university assignment we were given 3 vectors which can form linear combination which is expressed in another vector. They told that there some guessing involved but it gets easier with practice, but for the last 13 and a half hour i can't figure anything matching those vectors left and right. I don't want to go against university rules, but it looks very absurd because no way in hell this is guesswork.

I try to systematically get 1 in every axis to get the required variable, and then calculate another vector which has 0 in that axis and 1 in the following one, but there are no such possibilities as far as i see.

I know that if i post the assignment here people will solve it in no time, but that's not really the problem. If that's really the case of guesswork i should reconsider my choice of study as i clearly don't have the intuition for it.

P.S. googling for a solution algorithm doesn't help - or i wasn't lucky at finding anything.

  • $\begingroup$ @DietrichBurde i sadly cannot - given that's an assignment which will be graded. I was hoping that there is some universal way which yields the solution 100% if followed correctly. They all have 3 axis, so i guess i can try all of their combinations and combinations of their combinations with common factors from like 1 to 10, but i doubt that will be fast and efficent. $\endgroup$ Commented Nov 29, 2021 at 20:01
  • $\begingroup$ well that's what i tried - i not always can bring it to echelon form that the issue. I know there is a way but i don't see it. Isn't gaussian elimination heavily relies on guessing and trial and error? $\endgroup$ Commented Nov 29, 2021 at 20:03
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    $\begingroup$ First do a "toy example" with, say $2$ vectors in $\Bbb R^3$. Then look at your example. The method of "linear algebra" always works, without "guessing". $\endgroup$ Commented Nov 29, 2021 at 20:05
  • $\begingroup$ I'm not quite sure I understand the question, do you mean that you're given a vector and you have to express it as a linear combination of three other vectors, or that you have to find three vectors such that any vector can be written as a linear combination of the three vectors? Or do you mean something else? $\endgroup$ Commented Nov 29, 2021 at 20:23
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    $\begingroup$ Let me get this right. You have a vector $\mathbf{u}=(u_x,u_y,u_z)$, and you want to express it as a linear combination of 3 basis vectors, $\mathbf{v}^i=(v_x^i,v_y^i,v_z^i)$ with $i=1,2,3$. So you want to find $x,y,z$ in $\mathbf{u}=x\mathbf{v}^1+y\mathbf{v}^2+z\mathbf{v}^3$ (the superscripts are labels, not exponents). Writing this equation for each component of $\bf{u}$ results in a system of 3 equations with 3 unknowns, which can be solved exactly. Is that what you need? $\endgroup$
    – Luismi98
    Commented Nov 29, 2021 at 20:28

1 Answer 1


Write your problem as a system of linear equations, put it in matrix form and then solve it. You can always bring it into echeleon form or prove that it does not have a solution without any guesswork; there are numerous tutorials online that explain how to do this step by step.

Or, since I assume that you are dealing with vectors in $\mathbb R^n$ and have a scalar product, you can always express any vector $\vec v$ in terms of any (Edit: orthogonal) basis $\{\vec x_1,\vec x_2,\vec x_3\}$ by projecting onto each individual basis element:

\begin{equation} \vec v = \frac{\vec v \cdot \vec x_1}{\|\vec x_1\|}\vec x_1 + \frac{\vec v \cdot \vec x_2}{\|\vec x_2\|}\vec x_2 + \frac{\vec v \cdot \vec x_3}{\|\vec x_3\|}\vec x_3. \end{equation}

Mathematical intuition is not about random guesswork. You will make your life a lot easier by assuming that every problem in your university classes has some smart way of solving it and by trying to find the trick. No solution just falls from the sky.

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    $\begingroup$ this assumes $\{x_1,x_2,x_3\}$ is an orthogonal basis of $\Bbb{R}^3$. $\endgroup$
    – peek-a-boo
    Commented Nov 29, 2021 at 20:37
  • $\begingroup$ @peek-a-boo Ah yes, true, my bad. $\endgroup$ Commented Nov 29, 2021 at 21:07

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