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I am rather new to calculus, and was asked to answer the following question. I think I have the right answer, but I would love some feedback, particularly because some answers I derived empirically, without really understanding the reasoning behind it… Also, of course, there might be mistakes I did not pick up on.

$\mathbf u=(u_1,u_2,u_3)$, a vector in space.

Which of the following statements are correct and which ones are incorrect?

  1. there is a vector $\mathbf u$ that creates identical $45^\text{o}$ angles with the unit vectors $\mathbf i,\mathbf j$.
  2. there is a vector $\mathbf u$ that creates identical $60^\text{o}$ angles with the unit vectors $\mathbf i,\mathbf j$.
  3. there is a vector $\mathbf u$ that creates identical $30^\text{o}$ angles with the unit vectors $\mathbf i,\mathbf j$.
  4. All vectors $\mathbf u$ that are perpendicular to $\mathbf i+\mathbf j+ \mathbf k$ are located on any straight line in space.
  5. The projection of vector $\mathbf u$ on the z-axis is the vector (0,0,$u_3$)

My answers are:

1 and 2 are correct; I found an example for both (the first was obvious, the second took a tiny amount of work)

3 is incorrect, but I do not understand the reason in theoretical terms (I would love to know more, also what distinguishes this one from number 2)

4 is incorrect, because I found all vectors to fit this definition to be on a plane, and not a straight line (namely the plane $x+y+z=0$).

5 is correct

Thank you!

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All of your answers are correct. To see why 3 is impossible, set $\mathbf u =(u_1,u_2,u_3)$. A general property of $\mathbb R^3$ is that the dot product of 2 vectors can be expressed geometrically:

$$\mathbf u \mathbf v = \lvert\mathbf u\rvert\lvert\mathbf v\rvert\cos(\alpha),$$

where $\alpha$ is the angle between $\mathbf u$ and $\mathbf v$.

If we apply this to $\mathbf u$ and $\mathbf i$ (angle $\alpha_1$) we get

$$\mathbf u \mathbf i = u_1=\lvert\mathbf u\rvert\cdot 1\cdot \cos(\alpha_1),$$

and similiarly for $\mathbf u$ and $\mathbf j$ (angle $\alpha_2$):

$$\mathbf u \mathbf j = u_2=\lvert\mathbf u\rvert\cdot 1\cdot \cos(\alpha_2).$$

This means

$$u_1^2+u_2^2=\lvert\mathbf u\rvert^2\cos^2(\alpha_1) + \lvert\mathbf u\rvert^2\cos^2(\alpha_2) = (u_1^2+u_2^2+u_3^2)(\cos^2(\alpha_1) + \cos^2(\alpha_2))$$

and finally

$$\cos^2(\alpha_1) + \cos^2(\alpha_2) = \frac{u_1^2+u_2^2}{u_1^2+u_2^2+u_3^2}\le 1.$$

As you can see that inequality is an equality for $\alpha_1=\alpha_2=45°$, is true for $\alpha_1=\alpha_2=60°$ but not true for $\alpha_1=\alpha_2=30°.$

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Angle is equivalent to distance on the unit sphere and thus also follows the triangle inequality: if $\angle\mathbf u\mathbf v =\theta$, then for any $\mathbf w$, $\angle \mathbf u \mathbf w +\angle \mathbf v \mathbf w \ge \theta$. This should be reasonably obvious: if you walked between two points a thousand miles (0.252 rad) apart on the Earth, you have to walk 500 (0.126 rad) miles and then 500 more. Not merely 300 (0.0757 rad) miles and 300 more.

As for the others: constructing an explicit example is not just a perfectly valid existence proof, it's often the most sought-after! (Similarly, constructing a counterexample is a great method of disproving an "all x do y" statement)

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  • $\begingroup$ Argh, what'd I do $\endgroup$ – Dan Uznanski Mar 15 at 14:20

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