Explanation Formula used: Write the expression to find equation of the plane through the point P 0 ( x 0 , y 0 , z 0 ) with normal vector n = 〈 a , b , c 〉 as follows. a ( x − x 0 ) + b ( y − y 0 ) + c ( z − z 0 ) = 0 (1) The normal vector ( n ) to the plane is the cross product of two direction vectors in the plane. Write the normal vector from the plane x − z = 1 . Rearrange the plane x − z = 1 and write the normal vector. x + ( 0 ) y − z = 1 n 1 = 〈 1 , 0 , − 1 〉 Write the normal vector from the plane y + 2 z = 3 . Rearrange the plane y + 2 z = 3 and write the normal vector. ( 0 ) x + y + 2 z = 3 n 2 = 〈 0 , 1 , 2 〉 The cross product of the normal vectors n 1 and n 2 is the direction vector ( a ) and it is parallel to the required plane. Find the direction vector a from the normal vectors of two intersected planes. a = n 1 × n 2 Substitute 〈 1 , 0 , − 1 〉 for n 1 and 〈 0 , 1 , 2 〉 for n 2 , a = 〈 1 , 0 , − 1 〉 × 〈 0 , 1 , 2 〉 Calculate the cross product as follows. a = | i j k 1 0 − 1 0 1 2 | = i [ ( 0 ) ( 2 ) − ( 1 ) ( − 1 ) ] − j [ ( 1 ) ( 2 ) − ( 0 ) ( − 1 ) ] + k [ ( 1 ) ( 1 ) − ( 0 ) ( 0 ) ] = i ( 0 + 1 ) − j ( 2 − 0 ) + k ( 1 − 0 ) = i − 2 j + k a = 〈 1 , − 2 , 1 〉 The normal vector of the perpendicular plane is the parallel direction vector to the required plane. Write the normal vector from the perpendicular plane x + y − 2 z = 1 and consider it as vector b . b = 〈 1 , 1 , − 2 〉 As both the vectors a and b lie on the plane, the cross product of a and b is the orthogonal of the plane and it is considered as normal vector. Find the normal vector ( n ) . n = a × b Substitute 〈 1 , − 2 , 1 〉 for a and 〈 1 , 1 , − 2 〉 for b , n = 〈 1 , − 2 , 1 〉 × 〈 1 , 1 , − 2 〉 Calculate the cross product as follows

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ISBN: 9781305266643

Author: James Stewart

Publisher: Cengage Learning

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Chapter 12, Problem 25RE

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To find: An equation of the plane that passes through the line of intersection of the planes x−z=1 and y+2z=3, and perpendicular to the plane x+y−2z=1.

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Chapter 12, Problem 24RE

Chapter 12, Problem 26RE

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An equation of the plane that passes through the line of intersection of the planes x − z = 1 and y + 2 z = 3 , and perpendicular to the plane x + y − 2 z = 1 .

Solution Summary: The author explains how to find an equation of the plane that passes through the line of intersection of two planes.

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To find: An equation of the plane that passes through the line of intersection of the planes x−z=1 and y+2z=3, and perpendicular to the plane x+y−2z=1.

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Chapter 12 Solutions