Radius of Curvature
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(a) Find the point of intersection of the line containing the points (3; 2; 1) and (4; 3; 3), and the plane containing the points (5; 4; 2), (3; 1; 6) and (6; 5; 3).
(b) Find the radius of curvature at (1; 0; 1) on the three dimensional curve given by
r(t) = cos ti + sin tj + e^tk. Sketch this curve, and briefly describe its shape.
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Solution Summary
The equation of the line given in the parametric form is depicted.
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(i) As is known the equation of the line which contains the points ( ) and ( ) is
, (1)
Here we have = 3, = 2, = 1 , and 4, 3, 3. Hence, from (1) we obtain
, , .
Therefore, the equation of the line, given in the parametric form is
, . (2).
Let us now write the equation of the plane which contains points ( ), ( ) and . It is known that it has a form the determinant
=0, (3)
Here = 5, = 4, = 2, 3, 1, 6 and 6, =5, =3. Let us substitute these values in (3), and then evaluate the ...
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