Intersecting x_1x_2 Plane and Curve x at t=π/2

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Homework Statement


Find the intersection of the x_1x_2 plane and the normal plane to the curve
x= (cos(t)e_1 + (sin(t))e_2 + (t)e_3

At the point t = pi/2


Homework Equations




I have looked everywhere for a formula or an example for this and cannot find one? Can anyone help me as to what I should be looking up, if there is a formula, or a hint on the method I should try.

Thanks
 
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First, determine the planes! The tangent vector to cos(t)e_1 + (sin(t))e_2 + (t)e_3
is -sin(t)e_1+ cos(t)e_2+ e_3 and at pi/2 that is -e_1+ e_3. Of course, at pi/2, the curve goes through the point e_2+(pi/2)e_3.

The equation of a plane with normal vector -e_1+ e_3 containing point (0,1,pi/2) is, of course, -x_1+ x_3- pi/2= 0 or x_3- x_1= pi/2.

I assume you know that the equation of the x_1x_2plane is x_3= 0.

Find all points that satisfy x_3- x_1= pi/2 and x_3= 0. The intersection of two planes is, of course, a line.
 
Why did you look for the formula? Such things are easy to derive for yourself.
 
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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