Explanation Given Information: It is provided equation of lines x − 3 4 = z − 1 2 ; y = − 2 and x − 3 3 = y + 2 1 = z − 1 − 2 Formula Used: Cartesian equation of the straight line passing through a fixed point ( x 1 , y 1 , z 1 ) having direction ratios a , b , c is given by x − x 1 a = y − y 1 b = z − z 1 c Equation of a plane passing through a given point whose position vector is a → and perpendicular to the given vector n → , is r → . n → = a → . n → and the Cartesian equation of the plane passing through a given point ( x 1 , y 1 , z 1 ) and perpendicular to the normal whose directional ratios are a , b , c respectively is given by: a ( x − x 1 ) + b ( y − y 1 ) + c ( z − z 1 ) = 0 Calculation: The general equation of the plane is given by: a x + b y + c z = d The direction ratios of the given lines are ( 4 , 0 , 2 ) , ( 0 , 1 , 0 ) and ( 3 , 1 , − 2 ) Therefore p → = 4 i ^ + 2 k ^ ; q → = j ^ and s → = 3 i ^ + j ^ − 2 k ^ hence r

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 16SP

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To find: The equation of the plane containing the lines x−34=z−12;y=−2 and x−33=y+21=z−1−2 .

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Chapter 9, Problem 15SP

Chapter 9, Problem 17SP

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Chapter 9 Solutions

Calculus

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The equation of the plane containing the lines x − 3 4 = z − 1 2 ; y = − 2 and x − 3 3 = y + 2 1 = z − 1 − 2 .

Solution Summary: The author explains the Cartesian equation of the straight line passing through a fixed point (x_1,y

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To find: The equation of the plane containing the lines x−34=z−12;y=−2 and x−33=y+21=z−1−2 .

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Chapter 9 Solutions