from the "advanced example" in the lecture clip on lagrange multipliers. Why are u1, u2 and u3 perpendicular to the sides of the base triangle?? http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/part-c-lagrange-multipliers-and-constrained-differentials/session-41-advanced-example/

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from the "advanced example" in the lecture clip on lagrange multipliers. Why are u1, u2 and u3 perpendicular to the sides of the base triangle?? http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/part-c-lagrange-multipliers-and-constrained-differentials/session-41-advanced-example/

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  • phi
I guess it is hard to visualize. But consider the "slant height" of a face from point P down to its base leg. That slant height makes a 90 degree angle with the leg |dw:1442951690471:dw| now rotate that slant height until it lies in the x-y plane |dw:1442951759766:dw| it will remain 90 degrees to the leg Also, point P when projected down to the x-y plane will lie on this rotated "slant height" Thus we have (looking down on the base) |dw:1442951912332:dw|
yep... got it!! you mean to say that the "slant height" and the "projected slant height" both lie on the plane perpendicular to the base leg right?
  • phi
yes, that is good way to describe it.

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