Explanation Given information: The equation of lines are x + 1 1 = y − 3 2 = z + 2 3 and line passing through the point ( 1 , 3 , − 2 ) & ( 0 , 1 , − 1 ) Concept and Formula Used: Distance between two lines is equal to the distance between the parallel planes containing them. To find the distance‘d’ between two planes find the distance between a point on one plane to the other plane. d = ‖ A x 0 + By 0 +Cz 0 + D ‖ A 2 + B 2 + C 2 Equation of plane A ( x − x 0 ) + B ( y − y 0 ) + C ( z − z 0 ) = 0 General equation of line x − x 1 a = y − y 1 b = z − z 1 c Equation line passing from two points ( x 1 , y 1 , z 1 ) & ( x 2 , y 2 , z 2 ) is x − x 1 x 2 − x 1 = y − y 1 y 2 − y 1 = z − z 1 z 2 − z 1 Calculation: It is given that equation of lines are L 1 : x + 1 1 = y − 3 2 = z + 2 3 & L 2 : x − 1 − 1 = y − 3 − 2 = z + 2 1 ( using x − x 1 x 2 − x 1 = y − y 1 y 2 − y 1 = z − z 1 z 2 − z 1 ) Let line L 1 lie on plane p 1 , therefore any point ‘Q’ on plane p 1 on comparing with general equation of line is Q ( − 1 , 3 − 2 ) Now, vectors parallel to line L 1 & L 2 are v 1 = 〈 1 , 2 , 3 〉 & v 2 = 〈 − 1 , − 2 , 1 〉 respectively. Also, cross product of v 1 = 〈 1 , 2 , 3 〉 & v 2 = 〈 − 1 , − 2 , 1 〉 i .e

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

6th Edition

ISBN: 9781524908102

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

Publisher: Kendall Hunt Publishing

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Chapter 9.6, Problem 36PS

To determine

To Find: Distance between lines x+11=y−32=z+23 & line passing through the point (1,3,−2)&(0,1,−1)

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Chapter 9.6, Problem 35PS

Chapter 9.6, Problem 37PS

Chapter 9 Solutions

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

Ch. 9.1 - Prob. 1PS Ch. 9.1 - Prob. 2PS Ch. 9.1 - Prob. 3PS Ch. 9.1 - Prob. 4PS Ch. 9.1 - Prob. 5PS Ch. 9.1 - Prob. 6PS Ch. 9.1 - Prob. 7PS Ch. 9.1 - Prob. 8PS Ch. 9.1 - Prob. 9PS Ch. 9.1 - Prob. 10PS

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Chain Rule dy:dx = dy:du*du:dx; Author: Robert Cappetta;https://www.youtube.com/watch?v=IUYniALwbHs;License: Standard YouTube License, CC-BY

CHAIN RULE Part 1; Author: Btech Maths Hub;https://www.youtube.com/watch?v=TIAw6AJ_5Po;License: Standard YouTube License, CC-BY

Distance between lines x + 1 1 = y − 3 2 = z + 2 3 & line passing through the point ( 1 , 3 , − 2 ) & ( 0 , 1 , − 1 )

Solution Summary: The author calculates the distance between two parallel planes by using the equation of lines d=Vert.

Videos

To Find: Distance between lines x+11=y−32=z+23 & line passing through the point (1,3,−2)&(0,1,−1)

Want to see the full answer?

Chapter 9 Solutions

Distance between lines x + 1 1 = y − 3 2 = z + 2 3 &amp; line passing through the point ( 1 , 3 , − 2 ) &amp; ( 0 , 1 , − 1 )

Solution Summary: The author calculates the distance between two parallel planes by using the equation of lines d=Vert.

Videos

To Find: Distance between lines x+11=y−32=z+23 & line passing through the point (1,3,−2)&(0,1,−1)

Want to see the full answer?

Chapter 9 Solutions

Distance between lines x + 1 1 = y − 3 2 = z + 2 3 & line passing through the point ( 1 , 3 , − 2 ) & ( 0 , 1 , − 1 )