〈b, a, b〉 and 〈a, b, a〉: For what non zero values of a and b are these two vectors parallel? Looking for a hint on how to solve this problem.
I know for the vectors to be parallel, the cross product must equal the zero vector, but I'm unsure on how to use that information to solve for values.
$\langle b, a, b\rangle \times\langle a, b, a\rangle$.
For what non zero values of $a$ and $b$ are these two vectors parallel?
 A: You don't really need to use the cross-product.
 If two vectors are parallel, then one is a scalar multiple of the other.
In this case, that means $b/a=a/b$, which implies $b^2=a^2$, which implies $b=\pm a$.
A: You have the right idea. Take the cross product and simplify the resulting expression. See if you can solve for $a$ in terms of $b$.
A: Cross product will work.  What is the cross product? $(a^2-b^2,0, b^2-a^2) = \bf 0$
For what values of $a,b$ does $a^2 - b^2 = 0$?
$a = \pm b$
Alternatively (and I think more simply)
The vectors are parallel if:
$(a,b,a) = \lambda (b,a,b)$
giving
$a  = \lambda b\\
b = \lambda a$
and
$\lambda = \pm 1$
A: You can use the cross product if you want to use a sledgehammer: the cross product is the formal determinant
$$
\det\begin{bmatrix} e_1 & e_2 & e_3 \\ b & a & b \\ a & b & a \end{bmatrix}
=(a^2-b^2)e_1+(b^2-a^2)e_3
$$
and so the condition is $a^2=b^2$.
You can also use the dot product, because parallelism is equivalent to
$$
|v\bullet w|=|v||w|
$$
that in your case becomes
$$
3|ab|=\sqrt{a^2+2b^2}\sqrt{2a^2+b^2}
$$
that becomes $2a^4+5a^2b^2+2b^4=9a^2b^2$ and so $2(a^2-b^2)=0$.
On the other hand, the problem can be more easily solved by observing that parallelism is the same as requiring
$$
\langle b,a,b\rangle=x\langle a,b,a\rangle
$$
so $b=xa$ and $a=xb$ and easily $x=\pm1$.
