# 〈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?

• Which two vectors? $\langle b,a,b \rangle$ and $\langle a,b,a \rangle$ ? If so, what does "cross" mean? – GEdgar Jan 28 at 22:46
• Welcome to Mathematics Stack Exchange. To be parallel, $b/a=a/b\implies b^2=a^2\implies b=\pm a$ – J. W. Tanner Jan 28 at 22:46
• See my edits to get a hint on how we format mathematics. – John Hughes Jan 28 at 23:03

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 $$b=\pm a$$.

You have the right idea. Take the cross product and simplify the resulting expression. See if you can solve for $$a$$ in terms of $$b$$.

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
$$\lambda = \pm 1$$

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 $$x=\pm1$$.