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. 
 A: 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.
A: 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.
A: Sorry to add to an old question, but it came up for me and I think this method may be useful for others. The method is equivalent to @egreg's answer, but maybe a bit easier to implement programmatically when you need the last vector to make a complete basis - a square matrix of linearly independent rows and columns - as in this question.
A matrix with linearly independent rows has a non-zero determinant, so if you want to find the vector $(a,b,c)$ to make a complete basis, we need
$$
\begin{vmatrix}
a & b & c \\
1 & 2 & 8 \\
0 & 1 & 9 \\
\end{vmatrix} \neq 0
$$
The determinant can be computed with the elements in the top row and the minors (the Laplace expansion), so we get
$$
a 
\begin{vmatrix}
2 & 8 \\
1 & 9 \\
\end{vmatrix}
- b
\begin{vmatrix}
1 & 8 \\
0 & 9 \\
\end{vmatrix}
+ c
\begin{vmatrix}
1 & 2 \\
0 & 1 \\
\end{vmatrix}
\neq 0
$$
which again is
$$
10a - 9b + c \neq 0
$$
Any $(a,b,c)$ that satisfies that will be linearly independent. A really simple approach would be just to pick one of the elements with non-zero coefficients and set it to $1$, and set the other elements to zero. In this case none of the coefficients are zero, so $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ are all linearly independent with the first two vectors you gave.
