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Question. My goal is to determine all the vectors of $\mathbb R^3$ that make an angle of $\frac{\pi}{3}$ with the vector $(1,0,0)$.

My attempt. We want to determine all the vectors $x = (x_1,x_2,x_3) \in \mathbb R^3$ such that

$$ \cos\left(\frac{\pi}{3}\right) = \frac{x \cdot (1,0,0)}{\| x \| \, \| (1,0,0) \|} = \frac{x_1}{\| x \|}. $$

Taking into account that $\cos(\pi/3) = 1/2$ and that $\| x \| = \sqrt{x_1^2 + x_2^2 + x_3^2}$ , the precedent equation can be rewritten as

$$ \tag{1} \frac{1}{2} \sqrt{x_1^2 + x_2^2 + x_3^2} = x_1 \Leftrightarrow \frac{1}{4} (x_1^2 + x_2^2 + x_3^2) = x_1^2 \Leftrightarrow 3x_1^2 = x_2^2 + x_3^2. $$

Solving for $x_1$, we obtain that

$$ x_1 = \pm \sqrt{\frac{1}{3}(x_2^2 + x_3^2)} = \pm \frac{1}{3} \sqrt{3x_2^2 + 3x_3^2} $$

Now, there is one slight problem with my solution. I believe that this problem comes from assuming the minus sign when I remove the square from $x_1$. For example, if I take $x_2 = 1$ and $x_3 = \sqrt{2}$, one of the possible solutions according to my reasoning is $(x_1,x_2,x_3) = (-1,1,\sqrt{2})$. Simple calculations yield that this vector makes an angle of $2\pi/3$ with $(1,0,0)$, which is not $\pi/3$.

In other words, I am looking for a justification that allows me to not consider the minus sign when taking the square roots from the equation above. I believe this should be related to the first equation in $(1)$, where it's clear that $x_1$ must be non-negative. Either way, I would like to confirm my reasoning.

Thanks for any help in advance.

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    $\begingroup$ Check your first step: $x=1 \implies x^2=1$ while $x^2=1\implies x=\pm1$ $\endgroup$ Commented Nov 14, 2024 at 0:29
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    $\begingroup$ Your solution is fine except that you can only take the positive square root for $x_1$. That is obvious if you consider that a vector with a negative $x_1$ component would make an angle of greater than.$90^{\circ}$ with $(1,0,0)$. You could have also deduced that $x_1\gt 0$ from your first equation that stated that $x_1=\frac{1}{2}\left\lVert\vec v\right\rVert$. $\endgroup$ Commented Nov 14, 2024 at 2:15

1 Answer 1

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One of your steps is incorrect:

$$\frac{1}{2} \sqrt{x_1^2 + x_2^2 + x_3^2} = x_1 \iff \frac{1}{4} (x_1^2 + x_2^2 + x_3^2) = x_1^2$$

This is not an equivalence. $\frac{1}{2} \sqrt{x_1^2 + x_2^2 + x_3^2} = x_1$ implies the $\frac{1}{4} (x_1^2 + x_2^2 + x_3^2) = x_1^2$ but not the other way around because when you take square root you need to consider the absolute value.

You can remedy this like this:

$$\frac{1}{2} \sqrt{x_1^2 + x_2^2 + x_3^2} = x_1 \iff \frac{1}{4} (x_1^2 + x_2^2 + x_3^2) = x_1^2\ \text{ and }\ x_1\ge 0$$

Also, you should exclude the possibility $x_1=x_2=x_3=0$ which satisfy the equation you got but is obviously not a solution to your problem.

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