How to find a linearly independent vector?

Given two vectors $(1,2,8),(0,1,9)$ find a 3rd vector that is linearly independent from these two vectors.

I sort of have an idea how to go about solving the problem but I’m not 100% sure. I’m know we just want to find a vector that can’t be written as the sum of the two given vectors but how exactly do I go about finding one such vector?

Got it now many thanks to all the helpers some brilliant explanations.

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I like your idea about finding a vector that can’t be written as a sum of the two vectors above. Let’s take a look at what that would look like.

Every possible sum of these two vectors can be expressed as $c_{1}(1,2,8) + c_{2}(0,1,9)$ for some $c_{1}, c_{2}$ in $\Bbb R$. So, all possible sums can be expressed in the form $(c_{1}, 2c_{1} + c_{2}, 8c_{1} + 9c_{2})$.

We want to come up with a third vector $(v_{1}, v_{2}, v_{3})$ that can’t be expressed in the above form. Whatever $c_{1}$ and $c_{2}$ you pick for the linear combination above, we need that the third component $v_{3}$ is exactly $8c_{1} + 9c_{2}$. Let’s pick a vector whose third component is different from this (i.e., pick $c_{1}$ and $c_{2}$, and fill in the first two components of $(c_{1}, 2c_{1} + c_{2}, 8c_{1} + 9c_{2})$, but make the third component different from this).

So, even though you can pick $c_{1}$ and $c_{2}$ to be anything, I will pick $c_{1} = c_{2} = 1$. Then the vector I will construct will be:

$(1c_{1}, 2c_{1} + 1c_{2}, 11) = (1, 2 + 1, 11) = (1, 3, 11)$

Notice that I made the third component different from $8c_{1} + 9c_{2} = 8 + 9 = 17$. Then this new vector can’t be written as a linear combination of the previous two vectors, because that’s how we constructed it.

Here’s a simple method: you want to find a vector $(a,b,c)$ such that the system
$$x(1,2,8)+y(0,1,9)=(a,b,c)$$
has no solution. It’s a linear system because it can be written as
$$\begin{cases} x=a\\ 2x+y=b\\ 8x+9y=c \end{cases}$$
The matrix of this system is
$$\left[\!\begin{array}{cc|c} 1 & 0 & a \\ 2 & 1 & b \\ 8 & 9 & c \end{array}\!\right]$$
If we proceed with Gaussian elimination we get
\begin{align}
\left[\!\begin{array}{cc|c}
1 & 0 & a \\
2 & 1 & b \\
8 & 9 & c
\end{array}\!\right]
&\to
\left[\!\begin{array}{cc|c}
1 & 0 & a \\
0 & 1 & b-a \\
0 & 9 & c-8a
\end{array}\!\right]
\\&\to
\left[\!\begin{array}{cc|c}
1 & 0 & a \\
0 & 1 & b-a \\
0 & 0 & c-8a-9b+9a
\end{array}\!\right]
\end{align}
so we just need to have
$$a-9b+c\ne0$$
and we can choose whatever values of the parameters, so for example $a=1$, $b=0$ and $c=0$.

Of course infinitely many other choices are possible.

A different method is finding a non zero vector which is orthogonal to the two given vectors; if the vector is $(a,b,c)$ we get
$$\begin{cases} a+2b+8c=0\\ b+9c=0 \end{cases}$$
This gives $a=-2b-8c$ and $b=-9c$. So we can set $c=1$ and get $b=-9$, $a=10$.

With the first method we find all vectors that solve our problem, but your task is just finding one, so take your pick.