Explanation Given information: The equation of line is x − 3 2 = y + 1 1 = z − 2 − 1 and point is P ( 3 , − 2 , 1 ) Concept and Formula Used: Distance of a point ( x 0 , y 0 , z 0 ) from the line x − x 1 a = y − y 1 b = z − z 1 c is d = ‖ v × Q P ‖ ‖ v ‖ Calculation: It is given that equation of line is x − 3 2 = y + 1 1 = z − 2 − 1 The point is P ( 3 , − 2 , 1 ) = ( x 0 , y 0 , z 0 ) Distance of a point from line is given by d = ‖ v × Q P ‖ ‖ v ‖ On comparing the equation line x − 3 2 = y + 1 1 = z − 2 − 1 with x − x 1 a = y − y 1 b = z − z 1 c a = 2 , b = 1 & c = − 1 and line contain the point Q ( x 1 , y 1 , z 1 ) = ( 3 , − 1 , 2 ) Therefore Q P = ( 3 − 3 ) i + ( − 2 − ( − 1 ) ) j + ( 1 − 2 ) k Q P = 0 i − j − k = − j − k & v is the vector parallel to given line given by v = 2 i + 1 j − 1 k Distance between point P ( 3 , − 2 , 1 ) = ( x 0 , y 0 , z 0 ) and l

6th Edition

ISBN: 9781465208880

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 9, Problem 37SP

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To Find: Distance of a point P(3,−2,1) from a line x−32=y+11=z−2−1 .

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Chapter 9, Problem 36SP

Chapter 9, Problem 38SP

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Chapter 9 Solutions

Calculus

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Distance of a point P ( 3 , − 2 , 1 ) from a line x − 3 2 = y + 1 1 = z − 2 − 1 .

Solution Summary: The author calculates the distance of a point P(3,-2,1) from the equation x-32.

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To Find: Distance of a point P(3,−2,1) from a line x−32=y+11=z−2−1 .

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Chapter 9 Solutions