3
$\begingroup$

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?

$\endgroup$
3
  • 1
    $\begingroup$ Which two vectors? $\langle b,a,b \rangle$ and $\langle a,b,a \rangle$ ? If so, what does "cross" mean? $\endgroup$ – GEdgar Jan 28 at 22:46
  • 1
    $\begingroup$ Welcome to Mathematics Stack Exchange. To be parallel, $b/a=a/b\implies b^2=a^2\implies b=\pm a$ $\endgroup$ – J. W. Tanner Jan 28 at 22:46
  • $\begingroup$ See my edits to get a hint on how we format mathematics. $\endgroup$ – John Hughes Jan 28 at 23:03
8
$\begingroup$

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 $b=\pm a$.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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
$\lambda = \pm 1$

$\endgroup$
0
$\begingroup$

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 $x=\pm1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.