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The homework question I have here, is:

Show that if $\overrightarrow{a} \times \overrightarrow{u} = 0$ for any unit vector $\overrightarrow{u}$, then $\overrightarrow{a}=\overrightarrow{0}$

Things I have tried:

  1. Multiplying by the unit vectors $\hat{i}, \hat{j}, \hat{k}$. Let $\overrightarrow{a}= \begin{bmatrix} a\\b\\c\end{bmatrix}$

$\begin{bmatrix} a\\b\\c\end{bmatrix} \times \begin{bmatrix} 1\\0\\0\end{bmatrix} = \begin{bmatrix} 0\\c\\-b\end{bmatrix}$

$\begin{bmatrix} a\\b\\c\end{bmatrix} \times \begin{bmatrix} 0\\1\\0\end{bmatrix} = \begin{bmatrix} -c\\0\\a\end{bmatrix}$

$\begin{bmatrix} a\\b\\c\end{bmatrix} \times \begin{bmatrix} 0\\0\\1\end{bmatrix} = \begin{bmatrix} b\\-a\\0\end{bmatrix}$

Is this what the problem meant by any unit vector $\overrightarrow{u}$? Just the three unit vectors $\hat{i}, \hat{j}, \hat{k}$? Not sure where to go next if that's the case, or if that was a wrong step taken.

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    $\begingroup$ If $\vec u = (1, 0)^T$ and $\vec a = (-1, 0 )^T$ then $ \vec u \times \vec a = 0$ but $\vec a \ne \vec 0$. $\endgroup$ – user8277998 Oct 16 '17 at 9:07
  • $\begingroup$ @123 So that is a counterexample if I understand correctly? $\endgroup$ – LovesPeanutButter Oct 16 '17 at 9:11
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    $\begingroup$ Yes. $~~~~~~~~~$. $\endgroup$ – user8277998 Oct 16 '17 at 9:13
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    $\begingroup$ I'm pretty sure that "for any unit vector" is here used in the sense of "for every unit vector". $\endgroup$ – Klaus Draeger Oct 16 '17 at 9:36
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Any unit vector implies just that, any, and arbitrary, not just $\hat{i}, \hat{j}, \hat{k}$ which are 3 specific unit vectors.

$$ \vec{a} \times \vec{u} = \vec{0} $$ tells you either $\vec{a}$ is parallel to $\vec{u}$ or $\vec{a}=\vec{0}$.

But, $\vec{u}$ is an arbitrary unit vector. $\vec{a}$ cannot be simultaneously parallel to all arbitrary $\vec{u}$.

Hence $\vec{a}=\vec{0}$.

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Don't exactly know what your problem is.

From your first equation, you have $b=c=0$.

From the second or third equation, you have $a=0$.

Therefore $\vec{a}=\vec{0}$.

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I just wanted to add a comment to your solution. Since you have $\vec a\times \hat i=\vec a \times \hat j=\vec a\times \hat k=0$, you can write any vector $\vec u$ as $\alpha \hat i+\beta \hat j+\gamma \hat k$. Now if you expand $\vec a\times \vec u$ you will get $0$. So as long as your vector is parallel to three linearly independent vectors, your vector is parallel to any $\mathbb R^3$ vector

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The question could be referring to magnitudes of vectors, since $0$ doesn't have a direction. If that's so, then I'd think that $$|a\times u|=0$$ $$\therefore |a||u|\sin{\theta}=0$$ Since $u$ is a unit vector, $|u|=1$. So $$|a|\sin\theta=0$$ $$\therefore |a|=0 \ \ \ \text{or} \ \ \ \sin\theta=0 $$

From here, we can say that, if the two vectors are not parallel or antiparallel, $|a|=0$ and so we can deduce that $a=0$. However, in the case where $sin{\theta}=0$, $|a|$ could take any value, and so $a$ does not necessarily equal zero.

I know, this is a pretty elementary and informal way of showing that it isn't correct.

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