A vector perpendicular on OZ axis and makes 45 degree angle with pozitive OX

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Determine the vector w such that |w|= 2, w is perpendicular on the axis Oz and makes a 45◦angle with the positive direction of Ox. Could anybody enlighten me a bit ? I don't understand that well 3D vectors.

This is what have I done but it seems more or less pointless for what I have to do.

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I don't understand that well 3D vectors.

That's fine. That's probably why you were given this exercise in the first place: To get to know 3D vectors better.

That being said, here are some pointers: Most people are capable of imagining the three axes in front of them, and doing that gives you a great starting point. Use your hands if that helps. Use pens, or knitting needles, or any other long, straight and narrow tool if that helps. Draw it, or parts of it, if that helps. Anything that helps visualisation is allowed (as long as it's legal, and doesn't disturb others too much).

Usually it's imagined with the positive $x$-axis to the right, the positive $y$-axis away from you, and the positive $z$-axis pointing up, but you're free to choose whichever orientation suits you best.

From here you start asking questions. What does the collection of all vectors with $|w| = 2$ look like? What does the collection of all vectors perpendicular to the $z$-axis look like? (These are extremely standard questions to ask about collections of vectors, so memorizing the answers if you haven't already will probably be a good idea.)

Your vector has both of these properties, so it must be part of both of these collections simultaneously. What does that collection look like? Then use the final piece of information, and you're done (note that there are two vectors which fulfill all the criteria).

Edit after seing your work: Well, ok, then. That looks very reasonable. You've found that $x_1 = y_1 = \sqrt 2$ (note that $y_1 = -\sqrt 2$ is also a solution; it should be $\sin\alpha = \frac{|y_1|}2$). And while I can't see it written down, it seems from the drawing that you've figured out that $z_1 = 0$. So you have found $(\sqrt 2, \sqrt 2, 0)$. Well done!