Calc III: Vector parallel to the line of intersection

  • Thread starter Seri
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Homework Statement


Hi there,

Two points on the sphere S of radius 1 have spherical coordinates P: \phi = 0.35 \pi, \theta = 0.8 \pi and Q: \phi = 0.7 \pi, \theta = 0.75 \pi. Find a vector parallel to the line of intersection of the tangent planes to S at the points P and Q.


2. The attempt at a solution
First, I converted spherical to cartesian coordinates.
Cartesian:
P (-.72083942, .26684892, -.80901699)
Q (-.5720614, -.41562694, -.70710678)

Then, I found two tangent planes at P and Q. To do this, I found the gradient of the general sphere equation x^2 + y^2 +z^2 = 1. And then the general equation of the tangent plane for each point.

Tangent P = -1.44167x + .533698y - 1.161803z -2.4906521

Tangent Q = -1.1441228x - .83125388y - 1.4142136z -2



3. Relevant equations
I'm lost on what to do after that though. I talked to a tutor about this question and he said to set the two equation so that they equal z. Then add them together to get a single equation of a line. And then use parametrics since it's asking for a vector parallel. Does this sound right?
 

Answers and Replies

  • #2
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/nudges to the top
Wake up your monday morning with some hearty, delicious calculus! ;)
 
  • #3
tiny-tim
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Tangent P = -1.44167x + .533698y - 1.161803z -2.4906521

Tangent Q = -1.1441228x - .83125388y - 1.4142136z -2

That looks very complicated. :frown:

The tangent planes at P and Q are perpendicular to OP and to OQ.

So they're simply r.P = 0 and r.Q = 0, and they intersect in the line r.P = r.Q = 0.

Try that! :smile:
 

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