Explanation Given information : The distance from point P ( x , y , z ) to the line x − 1 4 = y + 1 − 1 = z − 0 3 is 5 units. Formula/s used : The cross product of two vectors A → and B → is given by A → × B → = | A → | | B → | sin θ n ^ . Calculations: The distance from the point P ( x , y , z ) to the given line L : x − 1 4 = y + 1 − 1 = z − 0 3 is 5. Clearly, a point on the line is Q ( 1 , − 1 , 0 ) . Since the direction ratio of the line are 4, -1, and 0, hence, the line L is parallel to vector Q A → =4i-j+3k where PA is perpendicular to the line from point Pas shown in figure here. The vector Q P → = ( x − 1 ) i+ ( y + 1 ) j+ ( z − 0 ) k . Let d be the distance from point P to the line L , then d = P A = Q P sin θ where θ angle between QA and QP . The unit vector along Q A → =4i-j+3k is a ^ = Q A → ‖ Q A → ‖ = 4i-j+3k ‖ 4i-j+3k ‖ = 4i-j+3k 26 . Now the vector a ^ × Q P → = ‖ a ^ ‖ ‖ Q P → ‖ sin θ n ^ = Q P sin θ n ^ = d n ^ ∵ ‖ a ^ ‖ =1 . Thus, we have d = ‖ a ^ × Q P → ‖

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ISBN: 9781465208880

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

Publisher: Kendall Hunt Publishing

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Chapter 9.6, Problem 40PS

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Tofind:Anequation for the set of all points P(x,y,z) such that distance from this point to the given line is fixed (5 units).

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Tofind: Anequation for the set of all points P ( x , y , z ) such that distance from this point to the given line is fixed (5 units).

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