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

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Author: Larson

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Question

Chapter 11.5, Problem 98E

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To calculate: The distance between the provided point (4,−1,5) and the provided line.

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Chapter 11.5, Problem 97E

Chapter 11.5, Problem 99E

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Chapter 11 Solutions

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The distance between the provided point ( 4 , − 1 , 5 ) and the provided line.

Solution Summary: The author calculates the distance between a point (4,-1,5) and the provided line.

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To calculate: The distance between the provided point (4,−1,5) and the provided line.

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Chapter 11 Solutions