Explanation Given: The parametric equations of a line are: x = 4 t − 2 , y = 3 and z = − t + 1 And the provided point Q is ( 1 , 5 , − 2 ) . Formula used: The distance between a point Q and a line is: D = ‖ P Q ⇀ × u ‖ ‖ u ‖ Here, D is the distance between point and the line, u is a direction vector for the line and P is a point on the line. Calculation: In the parametric equations of the line the direction numbers of a line are the coefficients of parameter t Equations of the line are: x = 4 t − 2 , y = 3 and z = − t + 1 Direction vector is, u = 〈 a , b , c 〉 = 〈 4 , 0 , − 1 〉 Substituting t = 0 , to find a point on a line, then obtain the point as: For the x coordinate of the point: x = 4 t − 24 = 4 ( 0 ) − 2 = − 2 For the y coordinate of the point: y = 3 For the z coordinate of the point: z = − t + 1 = 0 + 1 = 1 Now from the above calculation the point P on the line is: P = ( − 2 , 3 , 1 ) Now, find the vector P Q → So, the vector P Q → extending from point P to point Q will be P Q → = 〈 ( 1 − ( − 2 ) ) , ( 5 − 3 ) , ( − 2 − 1 ) 〉 = 〈 3 , 2 , − 3 〉 . Cross product of P Q → and u is, P Q → × u = | i j k 3 2 − 3 4 0 − 1 | = | 2 − 3 0 − 1 | i − |

11th Edition

ISBN: 9781337275576

Author: Larson, Ron, Edwards, Bruce

Publisher: Cengage Learning,

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Chapter 11.5, Problem 95E

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To calculate: For the given line and point, Find the distance between the provided point (1,5,−2) and the provided line x=4t−2,y=3 and z=−t+1.

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Chapter 11.5, Problem 94E

Chapter 11.5, Problem 96E

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For the given line and point, Find the distance between the provided point ( 1 , 5 , − 2 ) and the provided line x = 4 t − 2 , y = 3 and z = − t + 1 .

Solution Summary: The author calculates the distance between the provided point and the parametric equations of the line.

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To calculate: For the given line and point, Find the distance between the provided point (1,5,−2) and the provided line x=4t−2,y=3 and z=−t+1.

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Chapter 11 Solutions