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Solutions to the Practice Problems
Math 2500 Feb. 6, 2008
- Given the following pairs of vectors ~u and ~v, find the angle θ between them and compute the cross product ~u × ~v.
(a) ~u = (1, 2 , 3), ~v = (− 2 , 1 , 0) The angle θ is given by cos θ =
~u · ~v ‖~u‖‖~v‖
so the angle is θ = π/2. Next we compute the cross product:
~u × ~v = (0 − 3 , − 6 − 0 , 1 + 4) = (− 3 , − 6 , 5).
(b) ~u = (1, 0 , 2), ~v = (2, 1 , 0) The angle θ is given by cos θ =
~u · ~v ‖~u‖‖~v‖
so the angle is θ = cos−^1 (2/5) ' 1 .16. Next we compute the cross product:
~u × ~v = (0 − 2 , 4 − 0 , 1 − 0) = (− 2 , 4 , 1).
(c) ~u = (1, 1 , 0), ~v = (1, 0 , 1) The angle θ is given by cos θ =
~u · ~v ‖~u‖‖~v‖
so the angle is θ = cos−^1 (1/5) ' 1 .37. Next we compute the cross product:
~u × ~v = (1 − 0 , 0 − 1 , 0 − 1) = (1, − 1 , −1).
- Consider the plane Π 1 containing p = (2, 1 , 3), with normal vector ~n = (− 1 , 2 , 0).
(a) Write down the linear equation any point (x, y, z) in this plane must satisfy. The linear equation of Π 1 is given by
0 = ~n · ((x, y, z) − p) = (− 1 , 2 , 0) · (x − 2 , y − 1 , z − 3) = −x + 2y.
(b) Find the angle between the plane Π 1 and the plane Π 2 determined by x − y = 2. The angle between Π 1 and Π 2 is the same as the angle between the two normal vectors ~n and m~ = (1, − 1 , 0). This angle θ is given by cos θ =
~n · m~ ‖~n‖‖ m~‖
so θ = cos−^1 (− 3 /
(c) Parameterize the line l which is the intersection of Π 1 and Π 2. We can parameterize l as l(t) = q + ~vt, where q is a point common to both planes Π 1 and Π 2 and ~v is lies in the same direction as l. We could compute ~v by computing the cross product ~n × m~, but in this case there’s an easy trick: observe that both ~n and m~ lie in the x − y plane, so they are both perpendicular to the vector ~v = (0, 0 , 1). Now we have to find a point q in both Π 1 and Π 2. The coordinates (x, y, z) of q will satisfy the system
x − y = 2 − x + 2y = 0.
Simultaneously solving for x and y we see x = 4, y = 2. What is z? It can be anything. One can see that from the fact that l is parallel to the z-axis. We choose z = 0, so q = (4, 2 , 0) and
l(t) = (4, 2 , 0) + t(0, 0 , 1) = (4, 2 , t).
(d) Find the distance between the plane Π 1 and the point q = (3, 3 , 3). We can compute the distance by computing the orthogonal projection of q−p = (3, 3 , 3)−(2, 1 , 3) = (1, 2 , 0) onto ~n = (− 1 , 2 , 0). This orthogonal projection is
~a = (cos θ)|q − p|
~n |~n|
(q − p) · ~n |~n|
~n |~n|
The distance is the length of this orthogonal projection
d = |~a| =
- Consider the vectors ~u = (1, 2 , 1) and ~v = (0, 1 , −1).
(a) Explain why all the planes parallel to both ~u and ~v will have the same normal vectors (up to scaling). All parallel planes have the same normal vector (up to scaling), because any plane is given by ~n·((x, y, z)− (x 0 , y 0 , z 0 )), for some choice of of normal vector ~n and point (x 0 , y 0 , z 0 ) on the plane. Changing which plane you pick in a family of parallel planes amounts to varying (x 0 , y 0 , z 0 ) (in the direction of ~n). Thus they all have the same normal vector. Another way to think of this is: the angle between two planes is the angle between the two normal vectors. So two planes are parallel precisely when their normal vectors are parallel, which is another way of saying the normals are the same up the scaling. (b) Are these planes parallel to the plane given by x + y + z = 2? Explain your answer. The normal vector to the plane spanned by ~u and ~v is
~n = ~u × ~v = (− 3 , 1 , 1),
while the plane x + y + z = 3 has the normal vector m~ = (1, 1 , 1). These normal vectors are not parallel, so the two planes are not parallel.
- Consider the plane curve given by c(t) = (cos(t), sin(2t)), for 0 ≤ t ≤ 2 π.
(a) Sketch this curve. The curve c is a sideways figure 8, with the twist in the 8 at the origin. Here is a sketch.
(b) Set up, but do not evaluate, the integral to compute the arclength of c. length =
∫ (^2) π 0 |c
′(t)|dt = ∫^2 π 0
sin^2 (t) + 4 cos^2 (2t)dt. (c) Notice c is periodic (c(0) = c(2π)). Is c a simple closed curve? In othre words, are the t parameters 0 and 2 π the only times c crosses itself? No, c crosses itself at the origin, which corresponds to t = π/2 and t = 3π/2.
- (a) Consider the right circular cone C, with vertex at (0, 0 , 0), and slope 1. In other words, the cone C is what you get when you rotate the line y = z in the y − z plane about the z-axis. Write C in cylindrical coordinates. For a given height z, the distance from the z-axis is also z and the angle can be anything. So in cylindrical coordinates, this is r = z, with θ unrestricted. (b) Write the part of the shell 1 ≤ x^2 + y^2 + z^2 ≤ 4 lying in the x < 0 , y > 0 , z < 0 octant in spherical coordinates. First, x^2 + y^2 + z^2 = ρ^2 , so we have 1 ≤ ρ ≤ 2. Next we have to find bounds on the angles. The quadrant x < 0, y > 0 in the x − y plane corresponds to π/ 2 ≤ θ ≤ π. Also, z < 0 means we restrict π/ 2 ≤ φ ≤ π. So this region is 1 ≤ ρ ≤ 2 , π/ 2 ≤ θ ≤ π, π/ 2 ≤ φ ≤ π.
(a) Find the velocity and acceleration vectors of this curve. The velocity is c′(t) = d dt
(t cos t, t sin t, t) = (cos t − t sin t, sin t + t cos t, 1),
and the acceleration is
c′′(t) =
d dt
(c′) =
d dt
(cos t − t sin t, sin t + t cos t, 1) = (−2 sin t − t cos t, 2 cos t − t sin t, 0).
(b) Is the tangent line to c ever parallel to the x − y plane? Be sure to explain your answer. In order for the tangent line to c to be parallel to the x−y plane, we would need c′^ to be a horizontal vector. In other words, we need the third component of c′^ to be 0. This never happens; the third component of c′^ is 1. So the tangent line to c is never parallel to the x − y plane. (c) Set up, but do not evaluate, the integral to compute the arclength of c for 0 ≤ t ≤ π.
length =
∫ (^) π
0
|c′(t)|dt =
∫ (^) π
0
(cos t − t sin t)^2 + (sin t + t cos t)^2 + 1dt
∫ (^) π
0
cos^2 t − 2 t cos t sin t + t^2 sin^2 t + sin^2 t + 2t cos t sin t + t^2 cos^2 t + 1dt =
∫ (^) π
0
2 + t^2 dt.
- Consider the planes Π 1 and Π 2 , given as follows. The first plane Π 1 passes through p = (1, 2 , 3) and has the normal vector ~n = (1, 0 , −1). The second plane Π 2 is given by the linear equation x + y + z = 1.
(a) Explain how one can tell that Π 1 and Π 2 are not parallel, and compute the cosine of the angle θ between them. The normal vector to Π 1 is ~n = (1, 0 , −1) and the normal vector to Π 2 is m~ = (1, 1 , 1). These two vectors are not parallel, so the planes are not parallel If θ is the angle between them, then
cos θ =
~n · m~ |~n|| m~|
(b) The two planes Π 1 and Π 2 intersect in a line l. Find a parameterization for l. First observe that the equation of Π 1 is
x − z = (1, 0 , −1) · (1, 2 , 3) = − 2.
To find the line, we need a point and a direction. Let’s find the point first, by looking for simultaneous solutions of x − z = − 2 x + y + z = 1. First we try setting y = 0, which gives us
x − z = − 2 x + z = 1.
Now we add the two equations, which gives us 2x = −1, or x = − 1 /2, and so z = 3/2. Thus a point on the line l is q = (− 1 / 2 , 0 , 3 /2). Now we can find a direction by taking the cross product of the two normal vectors: ~v = ~n × m~ = (1, 0 , −1) × (1, 1 , 1) = (1, − 2 , 1). Finally, we can parameterize the line by
l(t) = q + t~v = (− 1 / 2 , 0 , 3 /2) + t(1, − 2 , 1).
- Consider the two vectors ~v = (3, − 2 , 1) ~u = (2, 1 , 4).
(a) Compute ~u × ~v.
~u × ~v = (2, 1 , 4) × (3, − 2 , 1) = (1 − (−8), 12 − 2 , − 4 − 3) = (9, 10 , −7)
(b) Find a vector w~ which is perpendicular to ~v, and explain why the answer you give is correct. There are many possible choices. You could choose w~ = ~u × ~v = (9, 10 , −7), which you just computed. Or you could choose something to make the dot product zero, like w~ = (2, 3 , 0) or w~ = (0, 1 , 2). All these answers are equally correct.
- Consider the planes Π 1 and Π 2 , where Π 1 contains the point p 1 = (3, 2 , 1) and has the normal vector n~ 1 = (2, 0 , 1), while Π 2 is given by the equation x + y − z = 3.
(a) Find the cosine of the angle θ between Π 1 and Π 2. The angle θ between the planes is the same as the angle between the two normals ~n 1 and n~ 2. The cosine is given by the dot product:
cos θ =
~n 1 · n~ 2 |~n 1 || n~ 2 |
(b) Write down the equation for Π 1. The equation of the plane is given by
~n 1 · (x, y, z) = ~n 1 · (3, 2 , 1) ⇒ 2 x + z = 7.
(c) Find a vector ~v which is parallel to both Π 1 and Π 2. We want ~v to be parallel to both Π 1 and Π 2 , so we must have ~v perpendicular to both ~n 1 and n~ 2. One such vector is the cross product:
~v = ~n 1 × ~n 2 = (2, 0 , 1) × (1, 1 , −1) = (− 1 , 3 , 2).
(d) Parameterize the line l of intersection between Π 1 and Π 2. To parameterize a line, we need to find two things: a direction vector and a basepoint. We just found the direction vector ~v = (− 1 , 3 , 2) is the previous part (it’s parallel to both planes), so all that remains is to find a basepoint, which is a simultaneous solution to the two equations
2 x + z = 7 x + y − z = 3.
Notice y doesn’t appear in the first equation, so let’s try setting y = 0. Then we get
2 x + z = 7, x − z = 3.
Adding these two equations together, we get 3x = 10, or x = 10/3. Plugging this back into either equation, we get z = 1/3. So a basepoint is (10/ 3 , 0 , 1 /3) and a parameterization of the line of intersection is
l(t) = (10/ 3 , 0 , 1 /3) + t(− 1 , 3 , 2).
- Consider the parameterized curve ~r(t) = (et, t, e−t).
(a) Find the velocity vector of ~r. Take a derivative: d~r dt
d(et) dt
d(t) dt
d(e−t) dt
) = (et, 1 , −e−t).
(b) Is the tangent line to ~r ever parallel to the yz–plane? Be sure to explain your answer. No. The normal to the yz–plane is the vector ~n = (1, 0 , 0), so we’d need
0 = ~n ·
d~r dt = (1, 0 , 0) · (et, 1 , −e−t) = et.
However, exponentials are never zero, so this doesn’t happen. (c) Set up, but do not evaluate, the integral to compute to arclength of the section of ~r for 0 ≤ t ≤ 1. The length is given by the integral of the speed:
L =
0
~r dt
|dt =
0
|(et, 1 , −e−t)|dt =
0
e^2 t^ + 1 + e−^2 tdt.
- Consider the two vectors ~v = (− 1 , 2 , 1), ~u = (2, 0 , 1).
(a) Find the cosine of the angle θ between ~v and ~u.
cos(θ) =
~v · ~u |~v||~u|
