Vectors: Find all values of a and b such that $(1,2,a)$ and $(2,(1-b),3)$ are perpendicular I know that, for the vectors to be perpendicular, their cross product must equal $0$.
i.e. $w = u \times v$, where $u = (1,2,a)$ and $v = (2,(1-b),3)$
From that, I got $w = ( (6-a+ab), (-3+2a), (-b-3) )$
When checking, $w \cdot u$ and $w \cdot v$ must equal $0$
This does occur, but I am no closer to finding what $a$ and $b$ are
 A: Noting that the dot product (which is a scalar, not the cross product, which is a vector) of any two perpendicular vectors must be $0$ (thanks to Macavity's comment on the question), we can solve for $a$ and $b$ using the definition of the dot product.
First of all, we know that, by the definition of the dot product, the dot product of any two vectors $(a_1, a_2, a_3)$ and $(b_1, b_2, b_3)$ $= a_1b_1 + a_2b_2 + a_3b_3$.
Here, substituting the respective values for $u_i$ and $v_i$ and simplifying the expression, we can find a "linear" equation representing all values for $a$ and $b$.
\begin{align}
u \cdot v &= u_1v_1 + u_2v_2 + u_3v_3 \\
          &= 1 \cdot 2 + 2(1-b) + a \cdot 3 \\
          &= 2 + (2-2b) + 3a \\
          &= 4 - 2b + 3a
\end{align}
As we know that the dot product of two perpendicular vectors is $0$, we can substitute and solve:
\begin{align}
      0 &= 4 - 2b + 3a \\
2b - 3a &= 4 \quad && \text{In standard form} \\
      a &= \frac{2b - 4} 3 \quad && \text{Solving for $a$} \\
      b &= \frac{3a + 4} 3 \quad && \text{Solving for $b$}
\end{align}
Therefore, given any value of $a$ or $b$, we should be able to solve for the other.
Note that there are, however, infinitely many solutions to the problem for $a$ and $b$ -  essentially, the values are represented by a linear model where $a$ varies accordingly to $b$ (and vice versa). Below would be a graph that contains values of $a$ and $b$, graphed by WolframAlpha:

A: Note that we can write 
$$u \cdot v = |u| |v| \cos \theta$$
where $\theta$ is the angle between $u$ and $v$ (where $u,v$ are vectors).  So, if we want the vectors to be perpendicular, we want $\theta = \pi/2$ so that the RHS is zero.  In particular, two vectors are perpendicular if
$u \cdot v = 0$.
For $u = (1,2,a)$ and $v = (2,(1-b),3)$ we have
$$ u \cdot v = 2 + 2(1-b) + 3a$$
so that, setting this equal to zero,
$$4 - 2b + 3a = 0 \implies 3a -2b = -4$$
for them to be perpendicular.
A: Scalar not cross/Vector product has to be zero for orthogonal vectors. Therefore the values of a and b has to satisfy $$ 3a-2b+4= 0$$
