MA261-A Calculus III 2006 Fall Homework 3 Solutions Due 9/22/2006 8:00AM
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1 MA6-A Calculus III 6 Fall Homework Solutions Due 9//6 :AM 9. # Find the parametric euation and smmetric euation for the line of intersection of the planes + + z = and + z =. To write down a line euation, we need a directional vector and a point. To get a point, rst, we assume that z =. So, we have + = and =. This implies that = and =. Thus, the point (; ; is in the intersection of our two planes, which is a point we need. We can read the normal vector of the plane + + z = to be (; ; and the normal vector of the plane +z = to be (; ;. Since the intersection line lies on both planes, the line must be perpendicular to both normal vectors. So, the directional vector of the line can be (; ; (; ; = (; ;. Therefore, the parametric euation of the line is < : = + ( t = t = + ( t = z = + ( t = t and the smmetric euation of the line should look like = = z. But, we cannot divide b. Thus, we should write a euation for individuall. So, the smmetric euation of the line is = z and =. 9. # Is the line through (; ; and (; ; perpendicular to the line through ( ; ; and (; ;? The directional vector for the rst line is ( ; ; ( = ( ; ;. The directional vector for the second line is ( ( ; ; = (; ;. B dot product, since ( ; ; (; ; = 6=, these two lines are not perpendicular. 9. # (a Find parametric euations for the line through (; ; that is perpendicular to the plane + z =. (b In what points does this line intersect the coordinate planes?
2 (a To write down a line euation, we need a directional vector and a point. We have a point (; ;. Note that the normal vector (; ; is perpendicular to the plane. So, it can serve a directional vector of the line. Thus, the parametric euation is < = + ( t = + t = + ( t = t. : z = + ( t = t (b When the line intersect -plane, the intersection point looks like (; ;. Thus, b the parametric euation, we have = z = t. This implies that = + ( = and = ( =. Hence, the line intersects the -plane at (; ;. When the line intersect z-plane, the intersection point looks like (; ; z. Thus, b the parametric euation, we have = = t, that is t =. This implies that = + ( = 7 and z = ( =. Hence, the line intersects the z-plane at (7; ;. When the line intersect z-plane, the intersection point looks like (; ; z. Thus, b the parametric euation, we have = = + t, that is t =.. This implies that = = 7 and z = =. Hence, the line intersects the -plane at ; 7;. 9. # Find the euation of the plane that passes through the line of intersection of the planes z = and + z = and is perpendicular to the plane + z =. We can read the directional vector of the plane z = is (; ; and the directional vector of the plane + z = is (; ;. The directional vector of the line of intersection of both planes is perpendicular to both. So, it can be (; ; (; ; = (; ;. Also, b letting z = for both planes, we can have a point in the line of intersection (; ;. The normal vector of the plane + z = is (; ;. Since our plane is perpendicular to the plane + z =, we know that (; ; is on our plane. Now, we have two vectors (; ; and (; ; on our plane and a point (; ;. Thus we have the normal vector of our plane is (; ; (; ; = (; ;. Hence, the euation of the plane is which is ( ; ; z (; ; =, + + z =. 9. # Find an euation for the plane consisting of all points that are euidistant from the points ( ; ; and (; ;. Assume that (; ; z is the point that are euidistant from the points ( ; ; and (; ;. The distance between (; ; z and ( ; ; is ( ( + ( + (z and the distance between (; ; z and (; ; is ( + ( ( + (z.
3 Since the are euidistant, we have ( ( + ( + (z = ( + ( ( + (z. It turns out that which is ( ( + ( + (z = ( + ( ( + (z, + z =. 9. # Find parametric euations for the line through the point (; ; that is parallel to the plane + + z = and perpendicular to the line = + t, = t, z = t. To write down a parametric euation for a line, we need a directional vector and a point. We have a point (; ;. Assume that our directional vector is (a; b; c. Since our line is parallel to the plane ++z =, we know that (a; b; c is perpendicular to the normal vector of ++z =, which is (; ;. This gives us that (a; b; c(; ; =, or, a + b + c =. Also, since our line is perpendicular to the line = + t, = t, z = t, we know that (a; b; c is perpendicular to the direction vector of this line, which is (; ;. This gives us that (a; b; c (; ; =, or, a b + c =. Now, we can solve a; b; c from the two euations we just found to get (a; b; c = ( ; ; (This is just one possible solution b letting c =. We can have our own one. Thus, the parametric euation is < : = + ( t = t = + ( t = + t z = + ( t = + t [Alternative Solution] As above, we found that our directional vector (a; b; c is perpendicular to both (; ; and (; ;. Thus, we can use the cross product to get (a; b; c = (; ; (; ; = ( ; ;. 9. # Find parametric euations for the line through the point (; ; that is perpendicular to the line = + t, = t, z = t and intersects this line. To write down a parametric euation for a line, we need a directional vector and a point. We have a point (; ;. Since our line is perpendicular to the line = + t, = t, z = t, we know that the driectional vector is perpendicular to the direction vector of this line, which is (; ;. Now, draw a gure with the given line = + t, = t, z = t with an point of this line (the easiest point of this time to nd is (; ;. Plot (; ;. Draw a vector ~v from (; ; to (; ;. Also, draw a line which is perpendicular to the given line and passes through (; ;. Mark the intersection point as P. Does our gure look like a right-angle triangle we use to derive the projection formula?.
4 In this gure, the last line (perpendicular one is the line we want. The vector ~w from P to (; ; plas the role of the directional vector. B projection formula discussion, ~w is the orthoginal projection of ~v onto (; ; (see Eercise 9.. #7. So, ( ; ; (; ; ~w = ~v proj (; ; ~v = ( ; ; j(; ; j (; ; = ( ; ; (; ; = ; ;. With a directional vector and a point, we know that the line euation is < = + t = t = + : t = + t. z = + ( t = + t 9. # Find euations of the planes that are parallel to the plane + z = and two units awa from it. To write down a plane euation, we need a normal vector and a point. Since the planes are parallel to the plane + z =, the normal vector is (; ;. Also, the unit normal vector is (; ; = ; ; k(;; k. B setting = and z =, we have a point (; ; on the plane + z =. There are two points in the direction of the normal vector (; ; which are units awa from it. The can be found b starting from (; ; and walking along the direction (or opposite direction of the normal vector (; ; (; ; + ; ; = and. Thus, the are ; ; (; ; ; ; = ; ;. So, we know that for these two points the diatance to the point (; ; are all. Thus, we have our points. For ; ;, we have the euation of the plane is which is, ; ; z + z = 7. (; ; =, For ; ;, we have the euation of the plane is ; ; z (; ; =, which is, + z =.
5 9. # Find the distance between the skew lines with parametric euations = + t, = + 6t, z = t and = + s, = + s, z = + 6s. (This is similar to Eample in the tetbook page 67. First, the normal vector is (; 6; (; ; 6 = (6; ;. Second, b putting s = in the euations of the second line, we get the point (; ;. Thus, the euation of the plane that our second line lies on it is ( ; ; z + (6; ; =, that is, 6 + z + =. Now, b putting t = in the euations of the rst line, we get the point (; ;. So the distance from (; ; to the plane 6 + z + = is D = j6 ( ( + ( + j = 6 + ( + 7 =. 9.6 # Let f (; = ln ( +. (a Evaluate f (;. (b Evaluate f (e;. (c Find and sketch the domain of f. (d Find the range of f. (a f (; = ln ( + = ln =. (b f (e; = ln (e + = ln e =. (c To use the natural logarithm function, we have to have + >. Thus, the domain is f(; j + > g. (d It is eas to see that + can be eual to an positive real number. Thus, the range of f is R. 9.6 # Sketch the graph of f (; = cos Thr graph is
6 6. z # Use traces to sketch the graph of the function f (; =. For a = k, we have traces z = k. z For a = k, we have traces z = k. z For a z = k, we have traces k =.
7 7 The graph is z 9.6 # Classif the surface + z 6 z + = b comparing with one of the standard forms in Table. Then sketch its graph. + z 6 z + = = + z 6 z + = z z + = ( + (z
8 = ( (z + B comparing to the Table, the surface is a Elliptic Paraboloid at the top (; ;. The graph is # Show that the curve of intersection of the surfaces + z + = and + z = lies in a plane. The rst surface can be written as z = z = + + z =. So, it is a hperboloid of one sheet. The second surface can be written as + z = + + z = + z = 6 + z = 6.
9 So, it is also a hperboloid of one sheet. For an intersection point (; ; z, it satis es both + z + = and + z =. Thus, we substract the second euation from the rst euation and get + =, that is, 6 + =. So, an intersection point must lie on the plane 6 + =. 9.6 # Find an euation for the surface consisting of all points P for which the distance from P to the -ais is twice the distance from P to the z-plane. Identif the surface. 9
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