Vectors, Planes, and Spheres, OH MY (Help mep lease)

jdj0202
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Homework Statement



Find an equation of the plane that contains both the center and "south pole" of the sphere

(x-3)^2+(y-2)^2+(z-3)^2 = 400

and is parallel to the line given in vector form by r(t) = [-43*t-82, -44*t+39, -84*t+94]. Write your answer in the form ax + by + cz = d

Homework Equations


The Attempt at a Solution



All of my attempts are so long it would be absurd to type them up; this is my last ditch effort at finding an answer before giving up.

Please help me with this, thanks.
 
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If I'm not going to get help because I didn't type up all of my work, will someone let me know so I can put it in here? Thanks.
 
Is my question too hard? If so, I agree.
 
You don't have to put in all of your efforts. Pick one that seems halfway reasonable, and show us what you have tried to do.
 
I will ask you few questions.

Can you determine the center of the sphere with its equation given?

Can you determine the directional vector of the line?

And finally, what is the south pole of the sphere?
 
To find the equation of a plane, you have to find a vector normal to the plane. And that will be the cross product of two vectors in or parallel to the plane. One you are given and the other is the vector from the center of the sphere to it "south pole" (by which I guess you mean the point on the sphere with lowest z).
 
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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