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
Best Answer
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}