Webpage title: How to Find a Rectangle on a Given Plane?

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Hi,

If I have the equation for a plane, Ax+By+Cz+D=0, how can I find four points that represent a rectangle width w and length l on that plane? (assuming the given point on the plane is the center of that rectangle..)

Thanks..
 
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I would find two points on the plane. Let's call them point A and point B. Then I would find vector AB. Then I would have to find another vector that is parallel to that vector with the same magnitude but also exists on the plane. Let's say that vector is vector CD. However, you would have to make sure that vector AC and vector BD are equal in magnitude and are also parallel.
 
Actually a better way would be to find three points on the plane, A, B, C, such that the dot product of vector AB and vector BC is 0, meaning that the two vectors are perpendicular. From there you have one side for the length and another side for the width. You can then find the other two sides, making sure of course that they are parallel to the opposing vectors and are of the same magnitude.
 
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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