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Find the maximum distance between the points (3sin , 0,0) and (4cos ,0,0).

Find the maximum distance between the points (3sin , 0,0) and (4cos ,0,0).

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Grade:12th pass

1 Answers

Nikhil
13 Points
4 years ago
Distance between/w two points is given by√[(x2-x1)^2 + (y2-y1)^2 +(z2-z1)^2]Therefore distance between the given points will be =√[(3sin-4cos)^2 +0 +0 ]=|3sin-4cos|But maximum value of (A)cos + (B)sin + (C) is given by (C) + √[(A)^2+(B)^2],Therefore the maximum/greatest distance is= 0 + √[(4)^2+(3)^2]=5

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