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Unformatted text preview: 378 C H A P T E R 13 VECTOR GEOMETRY (ET CHAPTER 12) 3. Which of the following planes are not parallel to the plane x + y + z = 1? (a) 2 x + 2 y + 2 z = 1 (b) x + y + z = 3 (c) x y + z = SOLUTION The two planes are parallel if vectors that are normal to the planes are parallel. The vector n = 1 , 1 , 1 is normal to the plane x + y + z = 1. We identify the following normals: • v = 2 , 2 , 2 is normal to plane (a) • u = 1 , 1 , 1 is normal to plane (b) • w = 1 , 1 , 1 is normal to plane (c) The vectors v and u are parallel to n , whereas w is not. (These vectors are not constant multiples of each other). Therefore, only plane (c) is not parallel to the plane x + y + z = 1. 4. To which coordinate plane is the plane y = 1 parallel? SOLUTION The plane y = 1 is parallel to the x zplane. y x z 1 5. Which of the following planes contains the zaxis? (a) z = 1 (b) x + y = 1 (c) x + y = SOLUTION The points on the zaxis are the points with zero x and y coordinates. A plane contains the zaxis if and only if the points ( , , c ) satisfy the equation of the plane for all values of c . (a) Plane (a) does not contain the zaxis, rather it is orthogonal to this axis. Only the point ( , , 1 ) is on the plane. (b) x = 0 and y = 0 do not satisfy the equation of the plane, since 0 + = 1. Therefore the plane does not contain the zaxis. (c) The plane x + y = 0 contains the zaxis since x = 0 and y = 0 satisfy the equation of the plane. 6. Suppose that a plane P with normal vector n and a line L with direction vector v both pass through the origin and that n · v = 0. Which of the following statements is correct? (a) L is contained in the P . (b) L is orthogonal to P . SOLUTION The direction vector of the line L is orthogonal to the vector n that is normal to the plane. Therefore, L is either parallel or contained in the plane. Since the origin is common to L and P , the line is contained in the plane. That is, statement (a) is correct. P O v n Exercises In Exercises 1–8, write an equation of the plane with normal vector n passing through the given point in vector form and the scalar forms (3) and (4). 1. n = 1 , 3 , 2 , ( 4 , 1 , 1 ) S E C T I O N 13.5 Planes in ThreeSpace (ET Section 12.5) 379 SOLUTION The vector equation is 1 , 3 , 2 · x , y , z = 1 , 3 , 2 · 4 , 1 , 1 = 4 3 + 2 = 3 To obtain the scalar forms we compute the dot product on the lefthand side of the previous equation: x + 3 y + 2 z = 3 or in the other scalar form: ( x 4 ) + 3 ( y + 1 ) + 2 ( z 1 ) + 4 3 + 2 = 3 ( x 4 ) + 3 ( y + 1 ) + 2 ( z 1 ) = 2. n = 1 , 2 , 1 , ( 3 , 1 , 9 ) SOLUTION The vector equation is 1 , 2 , 1 · x , y , z = 1 , 2 , 1 · 3 , 1 , 9 = 3 + 2 + 9 = 8 To obtain the scalar form we compute the dot product on the lefthand side above: x + 2 y + z = 8 or in the other scalar form: ( x 3 ) + 2 ( y 1 ) + ( z 9 ) = 8 + 3 2 9 = ( x 3 ) + 2 ( y 1 ) + ( z 9 ) = 3. n = 1 , 2 , 1 , ( 4 , 1 , 5 ) SOLUTION The vector form is 1 , 2 , 1 · x , y , z = 1 ,...
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 Spring '08
 Hubscher
 Vectors, ThreeSpace

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