Find < SPT in degrees
2 answers:
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I assume there are some plus signs that aren't rendering for some reason, so that the plane should be 
.
You're minimizing
subject to the constraint 
. Note that 
and 
attain their extrema at the same values of 
, so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.
The Lagrangian is
Take your partial derivatives and set them equal to 0:
Adding the first three equations together yields
and plugging this into the first three equations, you find a critical point at
.
The squared distance is then
, which means the shortest distance must be 
.
You're minimizing
The Lagrangian is
Take your partial derivatives and set them equal to 0:
Adding the first three equations together yields
and plugging this into the first three equations, you find a critical point at
The squared distance is then