Finding a specific dot product from a cross product I am doing the no bullsh*t guide to linear algebra.  On page 161 problem 2.9 is:
Find a vector that is orthogonal to both $u_1 = (1,0,1)$ and $u_2 = (1,3,0)$ and whose dot product with the vector $v = (1,1,0)$ is equal to $8$.
I took the cross product of $u_1$ and $u_2$ and got $(-3,1,3)$.  From there I can see we need to find values such that $-3x + 1y = 8$.  However, I think I need another equation to solve this system.
I was able to draw the vectors $u_1, u_2$, their cross product, and figure out the solution $(12, -4, -12)$ by following the slope of the line made by the cross product vector, but I am not sure how to systematically solve this kind of problem.
Can someone explain the right way to get the answer?
 A: Your final answer is correct. The mistake in the first calculation was that you wrote that you must have $-3x+1y=8$ but it should've been $-3x+1x=8$.
You found that the cross product is $w=(-3,1,3)$ which means that any vector which is orthogonal to both $u_1$ and $u_2$ must lie on the line spanned by $w$, that is, must be of the form $\lambda w=(-3\lambda, \lambda, 3\lambda)$ for some $\lambda\in\mathbb{R}$. The dot product constraint gives
$$(-3\lambda,\lambda,3\lambda)\cdot(1,1,0)=8$$
so that $\lambda=-4$ so substituting this we get $(12,-4,-12)$ which is your solution.
A: Option:
1)Vector perpendicular to $(1,0,1)$ is $(a, b, -a)$.
2)$(a, b, - a) \cdot (1,3,0)=a+3b=0;$
We get $(-3b,b,3b).$
3)$b(-3,1,3)\cdot (1,1,0)=8;$
$b(-3+1)=8; b=-4;$
Finally we have $(12,-4,-12).$
A: I went the same way like @Snaw above:
Any multiple of vector  $\begin{pmatrix}-3 \\ 1 \\ 3 \end{pmatrix}$ fulfills the requirement to be orthogonal to the given vectors $u_1$ and $u_2$.
Let $k$ be the factor of that orthogonal vector. This finally leads to the equation:
$-3k\cdot 1+1k\cdot 1+3k\cdot 0 = 8$
which finally gives you the factor $k=-4$ for your solution.
A: Let $U = \begin{bmatrix}1 &0 &1\\1 & 3 &0\end{bmatrix}$ and $v = [1, 1, 0]$. Then you're being asked to find $x$ such that $Ux=0$ and $vx=8$. Note that $x$ is a three-dimensional column vector, while the $0$ in $Ux =0$ is a two-dimensional column vector. You can combine these two equations into one. Take $M$ to be a matrix with the first two rows being $U$ and the bottom row being $v$:
$M= \begin{bmatrix}1 &0 &1\\1 & 3 &0\\1&1&0\end{bmatrix}$
Then $Mx$ is a three-dimensional column vector where the first two entries are $0 $ (representing orthogonality) and the last entry is 8 (representing the dot product being $8$).
$\begin{bmatrix}1 &0 &1\\1 & 3 &0\\1&1&0\end{bmatrix} \begin{bmatrix}x_1\\x_2 \\x_3\end{bmatrix} =  \begin{bmatrix}0\\0\\8\end{bmatrix}$
The first two rows give a homogenous system. Remember that homogenous systems don't give unique solutions; any multiple of a homogenous system is also a solution. So once you've row-reduced the first two rows, you should have a solution expressed in terms of a free variable. You can then use the last row to solve for the value of that variable.
