I'm confused by this exercise here :

Using the cross product, for which value(s) of t the vectors w(1,t,-2) and r(-3,1,6) will be parallel

I know that if I use the cross product of two vectors, I will get a resulting perpenticular vector. However, how to you find a parallel vector?

Thanks for your help

  • 1
    $\begingroup$ Hint consider when the magnitude of cross product is zero. When does this happen? $\endgroup$
    – Karl
    Aug 23 '15 at 17:42
  • $\begingroup$ (Notice that using the cross product is not the easiest way to do this.) $\endgroup$
    – user84413
    Aug 23 '15 at 20:25

You can use the fact that for two vectors $u,v$, $$ | u \times v | = |u| \ |v| \ \sin\theta,$$ where $\theta$ is the angle between $u$ and $v$.

Since you want $u$ and $v$ to be parallel, you want $\sin\theta = 0$, so $|u\times v|=0$.

This means you can solve your problem by finding the cross product and then setting its magnitude equal to $0$ and solving for $t$.

So the cross product is $(6t+2, 0, 3t+1)$.

So first we set its magnitude equal to $0$: $$0=\sqrt{(6t+2)^2+(3t+1)^2}$$

And squaring this we get \begin{align} 0&=(6t+2)^2+(3t+1)^2 \\ &=(36t^2+24t+4)+(9t^2+6t+1) \\ &= 45t^2+30t+5 \\ &= 9t^2+6t+1 \end{align}

Then we can use the quadratic formula:

\begin{align}t &= \dfrac{-6\pm\sqrt{6^2-4\cdot 9}}{2\cdot9} \\ &= \dfrac{-6\pm\sqrt{36-36}}{18} \\ &= \frac {-1}3 \end{align}

  • $\begingroup$ ok, so I find the cross product which gives me coords (6t +2 , 0, 3t + 1). Now solving for t when the magnitude is equal to 0 I get $$ t = \pm \sqrt{\frac{-5}{9}} $$ is this correct? $\endgroup$
    – Machinegon
    Aug 23 '15 at 18:14
  • $\begingroup$ I got the same cross product, but I got $t=-1/3$. I'll edit the steps into my answer. $\endgroup$
    – coldnumber
    Aug 23 '15 at 18:18
  • 1
    $\begingroup$ Shit, that's not what I did at all to solve for t. I need to review these basic factoring stuff before I get myself owned again. Thanks! $\endgroup$
    – Machinegon
    Aug 23 '15 at 18:28

v1(1,t,-2); v2(-3,1,6); 3*v1 is parallel to v2 if t=-1/3

Also: Perform vector Cross Product v1Xv2=v3 to get components:

v3((6*t+2),(6-6),(1+3*y)); the cross product is 0 when t=-1/3

  • $\begingroup$ sorry; (1+3*y)should be (1+3*t) $\endgroup$ Aug 24 '15 at 14:08

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