Explanation Given information: Consider the given lines: r 1 ( t ) = 〈 2 , 1 , 0 〉 + 〈 t , 2 t , 3 t 〉 r 2 ( t ) = 〈 2 , 1 , 0 〉 + 〈 3 t , t , 8 t 〉 Concept used: Plane is passing through P 0 ( x 0 , y 0 , z 0 ) with normal vector n = 〈 a , b , c 〉 So, equation of the plane is given by, Vector form: n . 〈 x , y , z 〉 = d Scalar form: a ( x − x 0 ) + b ( y − y 0 ) + c ( z − z 0 ) = 0 a x + b y + c z = d where d = a x 0 + b y 0 + c z 0 Calculation: The direction vectors of the given lines are v = 〈 3 , 1 , 8 〉 and w = 〈 1 , 2 , 3 〉 . Since, these vectors contained in the plane, so the cross product of these direction vectors gives the normal vector of the plane. So, vector cross product can find out using the formula v × w . n = v × w n = [ i j k 3 1 8 1 2 3 ] n = ( 3 − 16 ) i − ( 9 − 8 ) j + ( 6 − 1 ) k n = − 13 i − j + 5 k Now, the line is passing through the point (2, 1, 0), since r 1 ( t ) = 〈 2 , 1 , 0 〉 + 〈 t , 2 t , 3 t 〉

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Question

Chapter 13.5, Problem 27E

To determine

To evaluate:

The equation of the plane containing the lines r1(t)=〈2,1,0〉+〈t,2t,3t〉 and r2(t)=〈2,1,0〉+〈3t,t,8t〉.

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Chapter 13.5, Problem 26E

Chapter 13.5, Problem 28E

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Chapter 13 Solutions

Calculus - Standalone book

Ch. 13.1 - Prob. 1PQ Ch. 13.1 - Prob. 2PQ Ch. 13.1 - Prob. 3PQ Ch. 13.1 - Prob. 4PQ Ch. 13.1 - Prob. 5PQ Ch. 13.1 - Prob. 6PQ Ch. 13.1 - Prob. 1E Ch. 13.1 - Prob. 2E Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E

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The equation of the plane containing the lines r 1 ( t ) = 〈 2 , 1 , 0 〉 + 〈 t , 2 t , 3 t 〉 and r 2 ( t ) = 〈 2 , 1 , 0 〉 + 〈 3 t , t , 8 t 〉 .

Solution Summary: The author evaluates the equation of the plane containing the lines: lr_1(t) =.

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To evaluate:

The equation of the plane containing the lines r1(t)=〈2,1,0〉+〈t,2t,3t〉 and r2(t)=〈2,1,0〉+〈3t,t,8t〉.

Want to see the full answer?

Chapter 13 Solutions

The equation of the plane containing the lines r 1 ( t ) = 〈 2 , 1 , 0 〉 + 〈 t , 2 t , 3 t 〉 and r 2 ( t ) = 〈 2 , 1 , 0 〉 + 〈 3 t , t , 8 t 〉 .

Solution Summary: The author evaluates the equation of the plane containing the lines: lr_1(t) =.

Concept explainers

Videos

To evaluate: The equation of the plane containing the lines r1(t)=〈2,1,0〉+〈t,2t,3t〉 and r2(t)=〈2,1,0〉+〈3t,t,8t〉.

Want to see the full answer?

Chapter 13 Solutions

To evaluate:

The equation of the plane containing the lines r1(t)=〈2,1,0〉+〈t,2t,3t〉 and r2(t)=〈2,1,0〉+〈3t,t,8t〉.