Can you give a more specific question?

kskiraly
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Homework Statement



Find 2 unit vectors u1, u2, lying in the plane, x-y+z=0, which are not parallel to each other.

The Attempt at a Solution



I've tried taking the unit vector of the normal vector <1,-1,1> which is <1/rad(3),-1/rad(3),1/rad(3)>, but my teacher has told me it is not in the plane.

Then for my second vector, I created a parallel vector, but then I reread the question and found out they are not supposed to be parallel.
 
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kskiraly said:

Homework Statement



Find 2 unit vectors u1, u2, lying in the plane, x-y+z=0, which are not parallel to each other.

The Attempt at a Solution



I've tried taking the unit vector of the normal vector <1,-1,1> which is <1/rad(3),-1/rad(3),1/rad(3)>, but my teacher has told me it is not in the plane.
Yes, obviously. x- y+ z= (1-(-1)+ 1)/rad(3)= rad(3) not 0! You are aware, I am sure, that the normal vector to a plane is not in the plane! (If you are not, reread the definition of "normal vector".)

Then for my second vector, I created a parallel vector, but then I reread the question and found out they are not supposed to be parallel.
A very good case for always rereading the question (several times)! Yes, it says "not parallel".

To find a vector in the plane x- y+ z= 0, look for values of x, y, z that satisfy that equation. There are, of course, an infinite number of such choices. You can choose any values you like for x and y, for example, and then solve the equation for z. 1 and 0 are good, simple, choices.
If x= 1 and y= 0, then z must satisfy 1- 0+ z= 0.
If x= 0 and y= 1, then z must satisfy 0- 1+ z= 0.

Now form unit vectors from <x, y, z> and show that they are not parallel.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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