Determine whether the following lines intersect, are parallel, or skew. L 1 : x = 6t y = 1 + 9t z = 3t. x = 1 + 2s y = 4 3s z = s

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1 Homework Solutions 5/ Determine whether the following lines intersect, are parallel, or skew. L 1 : L 2 : x = 6t y = 1 + 9t z = 3t x = 1 + 2s y = 4 3s z = s A vector parallel to L 1 is 6, 9, 3. A vector parallel to L 2 is 2, 3, 1. If these vectors are parallel, then the lines are parallel too. So check the ratios of the components: The ratio of the first components is 6 = 3. 2 The ratio of the second components is 9 = 3. 3 The ratio of the third components is 3 = 3. 1 So 6, 9, 3 = 3 2, 3, 1, and these vectors are parallel. Therefore L 1 is parallel to L 2. We could obviously stop here, but I will test for intersection and skewness just for demonstration purposes. If the two lines intersect, there must be a point (x, y, z) that satisfies the equations for L 1 for some value of t, and the equations for L 2 for some value of s. For this value of s and t, we must have 6t = x = 1 + s 1 + 9t = y = 4 3s 3t = z = s From the first equation, we see that t = 1 1 s. The second equation becomes s = 4 3s 3 2 s = 9 2 s = 3 1

2 and thus t = 1 1 = 2. Plugging this into the left side of the third equation gives t = 2 s So the third equation cannot be satisfied. Therefore there is no point where L 1 and L 2 intersect, which is expected because they re parallel. If the test for the lines being parallel had failed, then we would conclude the lines are skew Find the equation of the plane through the origin and parallel to 2x y + 3z = 1. Let P 1 be the plane we re finding the equation for, and P 0 be the plane with equation 2x y + 3z = 1. Then we know P 1 is parallel to P 0 and (0, 0, 0) P 1. To find the equation for P 1, we need a vector normal to P 1 and a point in P 1. We already have the point, so we just need the normal vector. Since P 1 is parallel to P 0, every line L 1 in P 1 must be parallel to some line L 0 in P 0. That means any vector parallel to L 1 is also parallel to L 0. But by definition, any vector normal to P 0 is orthogonal to every vector parallel to L 0. So a vector normal to P 0 must also be normal to P 1. Thus 2, 1, 3 is normal to P 1. So an equation for P 1 is 2(x 0) (y 0) + 3(z 0) = 0 or more compactly 2x y + 3z = Find an equation of the plane containing ( 1, 2, 1) and the line of intersection of the planes x + y z = 2 and 2x y + 3z = 1. We are given a point on the plane, so we need only to find a normal vector. If the point ( 1, 2, 1) is on the line of intersection of the other two planes, then we don t have enough information to get the equation of the plane. Otherwise, we can find a normal vector by taking the cross product of a vector parallel to that line of intersection, and a vector going from ( 1, 2, 1) to some point on that line. (Both of those vectors are parallel to lines in the 2

3 plane we want the equation of, so a vector orthogonal to both of them must be normal to the plane). First, we need to find the line of intersection of the planes x + y z = 2 and 2x y + 3z = 1. This is done by solving for two of the variables in terms of the other one. A point (x, y, z) in the first plane must have z = x + y 2. If this point is also in the second plane, then 2x y + 3x + 3y 6 = 1 5x + 2y = 7 y = x So if we let x = t, then y = 7 5t and z = x + y 2 = 3 3 t. A vector parallel to this line is then 1, 5, The vector 2, 5, 3 is also parallel to this line (it s a scalar multiple of the other one) and is going to be easier to deal with. A point on the line of intersection is ( 0, 7, 3 2 2) which is reached at t = 0. The vector from ( 1, 2, 1) to this point is 1, 3 2, 1 2. This means the vector 2, 3, 1 is also in the plane as it is a scalar multiple. So we take the cross product to get a normal vector: 2, 3, 1 2, 5, 3 = 9 + 5, 2 + 6, 10 6 = 4, 8, 16 Thus, an equation of the plane is 4(x + 1) + 8(y 2) 16(z 1) = 0, or x + 2y 4z = 1 if we distribute and divide through by Find the point at which the line x = 3 t y = 2 + t z = 5t intersects the plane x y + 2z = 9. Just substitute for each variable in the equation of the plane with the expressions from the equation of the line. 3

4 (3 t) (2 + t) + 2(5t) = t = 9 t = 1 So the point is (2, 3, 5) Determine whether the planes P 1 : x = 4y 2z P 2 : 8y = 1 + 2x + 4z are parallel, perpendicular, or neither. If neither, find the angle between them. As in a problem above, we see that planes are parallel if and only if their normal vectors are parallel. An equation in standard form for P 1 is x 4y + 2z = 0, and one for P 2 is 2x + 4z 8y = 1. So a normal vector for P 1 is n := 1, 4, 2 and a normal vector for P 2 is m := 2, 4, 8. Check the ratios of the components: The first component ratio is 1, while the second is 1. So they are not parallel. 2 It can be shown using elementary geometry that the angle between planes is equal to the angle between their normal vectors. So if θ is the angle between P 1 and P 2, then ( ) n m θ = cos 1 n m ( ) = cos ( ) 34 = cos ( ) = cos ( ) 34 = cos 1 42 ( = cos 1 17 )

5 Find parametric equations for the line L 1 through the point (0, 1, 2) that is parallel to the plane P 1 with equation x + y + z = 2 and is perpendicular to the line L 2 with parametric equations x = 1 + t y = 1 t z = 2t For the equation of a line we need a point on the line and a vector parallel to it. So all we need is a vector a parallel to L 1. Since L 1 is parallel to P 1, it must be parallel to some line L 3 in P 1. Any vector parallel to L 1 must be parallel to L 3, and so must be orthogonal to any normal vector of P 1, one of which is 1, 1, 1. So a is orthogonal to 1, 1, 1. Since L 1 is perpendicular to L 2, any vector parallel to L 1 must be orthogonal to any vector parallel to L 2. The vector 1, 1, 2 is parallel to L 2, so we must have a orthogonal to 1, 1, 2 also. Thus a is some multiple of i j k 1, 1, 1 1, 1, 2 = = 2 + 1, 1 + 1, 1 1 = 3, 2, So an equation for the line L 1 is x = 3t y = 1 + 2t z = 2 2t Describe and sketch the surface z = cos(x). This is a cosine cylinder, consisting of lines parallel to the y-axis and having an intersection with the xz-plane that is the cosine curve. 5

6 Find the traces of the surface y 2 = x 2 + z 2 in the planes x = k, y = k, z = k and sketch it. With x = k, we have y 2 z 2 = k 2. This is a hyperbola opening up along the line parallel to the y-axis for k 0. For k = 0 it is the two lines x = 0, y = t, z = t and x = 0, y = t, z = t. With y = k, we have x 2 + z 2 = k 2. This is a circle of radius k for k 0, just the origin for k = 0. With z = k, we have y 2 x 2 = k 2. This is a hyperbola opening up along the line parallel to the y-axis for k 0. For k = 0 it is the two lines x = t, y = t, z = 0 and x = t, y = t, z = 0. 6

7 Since the traces are hyperbolas or pairs of lines in both x = k and z = k planes, and ellipses (or circles) in the y = k planes, this surface is a cone opening along the y-axis Find traces for the surface y = z 2 x 2 and sketch it. With x = k, we have y = z 2 k 2, which is a parabola opening up in the positive-y direction along the line parallel to the y-axis. 7

8 With y = k, we have z 2 x 2 = k, which is a hyperbola opening up along the z-axis for k > 0. For k = 0 it is the pair of lines x = t, y = 0, z = t and x = t, y = 0, z = t. For k < 0 it is a hyperbola opening up along the x-axis. With z = k, we have y = x 2 +k 2, which is a parabola opening up in the negative-y direction along the line parallel to the y-axis. Since the traces are hyperbolas in y = k and parabolas in x = k and z = k, it is a hyperbolic paraboloid opening up along the y-axis, and along the x-axis to the right of the xz-plane, along the z-axis to the left of the xz-plane

9 Reduce the equation x 2 y 2 + z 2 4x 2y 2z + 4 = 0 to one of the standard forms, classify the surface, and sketch it. Completing the square thrice gives ( x 2 4x + 4 ) 4 ( y 2 + 2y + 1 ) ( z 2 2z + 1 ) = 0 (x 2) 2 (y + 1) 2 + (z 1) 2 = 0 (x 2) 2 + (z 1) 2 = (y + 1) 2 This is a cone with center (2, 1, 1), that is, where the tips of the two cones meet, and which opens up around the line x = 2, y = t, z = 1 which is parallel to the y-axis Find an equation for the surface consisting of all points equidistant from ( 1, 0, 0) and the plane x = 1. Classify this surface. In 10.5 we have the distance from a point (x 1, y 1, z 1 ) to the plane ax + by + cz + d = 0 given by 9

10 D = ax 1 + by 1 + cz 1 + d a2 + b 2 + c 2 So if the distance from ( 1, 0, 0) and the plane x = 1 are equal, then x 1 (x + 1) 2 + y 2 + z 2 = = x Squaring both sides gives (x 1) 2 + y 2 + z 2 = (x 1) 2 (x + 1) 2 + y 2 + z 2 = (x 1) 2 y 2 + z 2 = (x 1) 2 (x + 1) 2 y 2 + z 2 = 4x 1 4 y2 1 4 z2 = x which is the equation of an elliptic paraboloid opening in the negative x-direction around the x-axis, with vertex at (0, 0, 0). The point ( 1, 0, 0) is said to be the focus of the paraboloid, and the plane x = 1 is the directrix. 10