Minimum distance between a line and an ellipsoid

Calculus Level 5

Given the ellipsoid

x 2 5 2 + y 2 7 2 + z 2 1 0 2 = 1 \dfrac{x^2}{5^2} + \dfrac{y^2}{7^2} + \dfrac{z^2}{10^2} = 1

and a spatial line whose parametric equation is given by p ( t ) = p 0 + t d \mathbf{p}(t) = \mathbf{p}_0 + t \mathbf{d} , where p 0 = ( 5 , 7 , 10 ) \mathbf{p}_0 = (5, 7, 10) , d = ( 1 , 1 , 1 ) \mathbf{d} = (1, -1, 1) , and t R t \in \mathbb{R} is an arbitrary parameter, we want to find the minimum distance d d^* , between a point P = ( x 1 , y 1 , z 1 ) P =(x_1, y_1, z_1 ) on the line and a point Q = ( x 2 , y 2 , z 2 ) Q = (x_2, y_2, z_2 ) on the ellipsoid. If the minimum occurs with P = ( x 1 , y 1 , z 1 ) P^* = ({x_1}^*,{y_1}^*,{z_1}^*) , and Q = ( x 2 , y 2 , z 2 ) Q^* = ( {x_2}^*,{y_2}^*,{z_2}^*) , then report the value of 100 ( x 1 + y 1 + z 1 + x 2 + y 2 + z 2 + d ) \lfloor 100({x_1}^*+{y_1}^*+{z_1}^* + {x_2}^*+{y_2}^*+{z_2}^* + d^* ) \rfloor , where \lfloor \cdot \rfloor is the floor function, for example, 7.6 = 7 \lfloor 7.6 \rfloor = 7 .


The answer is 3702.

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