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MATH 2242: Graded Homework # 1 – KEY
- Let a = (1, 1), b = (2, −1), α = −2, β = 3.
(a) Calculate αa + βb algebraically and geometrically.
(b) Describe in words the vector b − a.
(c) Normalize the vector b (i.e find the vector
b).
(d) Find the orthogonal projection of a along b.
Solution.
(a) αa+βb = −2(1, 1)+3(2, −1) = (− 2 , −2)+(6, −3) = (4, −5). See Figure 1 for the graphical representation.
(b) b − a is the vector point from the tip of a to the tip of b when the tails of a and b coincide.
(c) ||b|| =
2
5, so
b =
b ||b||
√^1 5
(d) ab =
a·b b·b
b =
(1,1)·(2,−1) 5
1 5
1
Figure 1: The vectors a (red), b (blue), − 2 a (magneta), 3b (cyan), and − 2 a + 3b (black)
- Let L be the line containing the points (1, 1 , 1) and (2, 2 , 3). Let L
′ be the line with direction vector v =
(1, − 1 , −1) and containing the point (0, 2 , 2).
(a) Find an equation L(t) for the line L.
(b) Find an equation L
′ (t) for the line L
′ .
(c) Do L and L
′ intersect? (Hint: Can you find a value of t for which L(t) = L
′ (t)?)
Solution.
(a) Put v = (2, 2 , 3) − (1, 1 , 1) = (1, 1 , 2). So L(t) = (1, 1 , 1) + t(1, 1 , 2).
(b) L ′ (t) = (0, 2 , 2) + t(1, − 1 , −1).
(c) The problem is ambiguous as stated. Ignoring the hint, the question is asking if the two lines intersect
anywhere (as point sets) in the plane. Taking the hint into account, I’m asking if there exists a value of t
for which the two lines intersect the same point simultaneously. To put it in other words, suppose Alice
walks along line L and Bob along line L ′
. In the first interpretation, I’m asking if Alice and Bob every
cross the same point (possibly at different times). In the second interpretation, I’m asking if they ever
cross the same point simultaneously (i.e. bump into each other). I wanted the second interpretation, so
I’ll work that one out (the answer is negative). I’ll leave it to you to check, however, that the answer is
affirmative if you take the first interpretation.
Indeed suppose there is a t such that L(t) = L
′ (t). Then the components of the two lines must be the
same at time t. This gives the system of three equations
1 + t = t
1 + t = 2 − t
1 + 2t = 2 − t
The first equation says 1 + t = t, which is impossible, so the lines L(t) and L
′ (t) do not simultaneously
intersect. Nevertheless, you can check that L(0) = (1, 1 , 1) = L
′ (1). I hope this wasn’t too confusing.
- (a) Find the equation of the plane P with normal vector n = (− 1 , 4 , 2) and containing the point p 0 = (1, 1 , 1).
(b) For what value of z is the point p = (0, 1 , z) in the plane P?
Solution.
(a) Use the equation (r − r 0 ) · n = 0. This gives (x − 1 , y − 1 , z − 1) · (− 1 , 4 , 2) = 0. That is, −(x − 1) + 4(y −
- 2(z − 1) = 0, which simplifies to −x + 4y + 2z = 5.
(b) From (a), if the point (0, 1 , z) is in P, then −0 + 4 + 2z = 5, which implies z =
1 2
- Let a = (− 1 , 0 , 2), b = (3, 4 , 5), c = (1, 1 , 1), α = 2, and β = −3. Calculate the following:
(a) a · b. Are a and b orthogonal (i.e. perpendicular)? Explain.
(b) a × b and b × a (Hint: for the second one, use a property of cross products - don’t rework it).
(c) a × (b × c) (Hint: Use the BAC-CAB rule).
(d) a · (b × c) (Hint: Use the scalar triple product rule).
Solution.
(a) a · b = (− 1 , 0 , 2) · (3, 4 , 5) = −3 + 0 + 10 = 7. No, a and b are not orthogonal because a · b 6 = 0.
(b)
a × b =
ˆi ˆj ˆk
Next, b × a = −a × b = (− 8 , − 11 , 4).
(c)
a × (b × c) = b(a · c) − c(a · b) = (3, 4 , 5)(1) − (1, 1 , 1)(7) = (− 4 , − 3 , −2).
(d)
a · (b × c) =
- Recall from physics that the work W done by a force F acting over a displacement d is given by W = F · d.
Suppose an electric field exerts the constant force F = 2
i − 3
j +
k while moving a charged particle from the
point a = (− 1 , 3 , 2) to the point b = (0, 7 , 3). Find the work W done by the field on the particle. (Note:
Normally all physics equations carry units – but nevermind that in this class.)
Solution. The displacement is d = b − a = (0, 7 , 3) − (− 1 , 3 , 2) = (1, 4 , 1). Next, W = F · d = (2, − 3 , 1) ·
(1, 4 , 1) = 2 − 12 + 1 = −9. The minus sign indicates the field took energy out of the particle.
- Let a = 2
i −
j and b =
k +
i.
(a) What is the angle between a and b?
(b) What is the distance between a and b (i.e. ||b − a||)?
Solution.
(a) ||a|| =
5, ||b|| =
2, and a · b = (2, − 1 , 0) · (1, 0 , 1) = 2. So
θ = cos
− 1 (
a · b
||a||||b||
) = cos
− 1 (
) ≈ 0 .89 rad ≈ 51
◦ .
(b) ||b − a|| =
- The plane P contains the points (1, 2 , −1), (3, 3 , 0), and (− 2 , − 5 , 1). Find an equation of the form ax + bz = c
describing the intersection of the plane P with the xz-coordinate plane.
In (b) we encountered the fact that we cannot use the formula θ = tan
− 1 (|y/x|) + fixup when x = 0. When
that happens, determine if θ =
π 2 or
3 π 2 according as to whether y > 0 or y < 0. If y = 0 as well (so x = y = 0)
then θ really is undefined. The same goes for part (d).
- Convert the following cylindrical or spherical coordinates to Cartesian coordinates.
(a) (r, θ, z) = (2,
π 4
(b) (r, θ, z) = (3, 1000 , −4)
(c) (ρ, θ, ϕ) = (10,
π 2 , π).
(d) (ρ, θ, ϕ) = ( √^1 2
π 2
π 4
Solution.
(a) (x, y, z) = (r cos θ, r sin θ, z) = (2 cos
π 4
, 2 sin
π 4
√ 2 2
√ 2 2
(b) (x, y, z) = (r cos θ, r sin θ, z) = (3 cos(1000), 3 sin(1000), −4) ≈ (1. 7 , 2. 5 , −4).
(c) (x, y, z) = (ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ) = (10 sin π cos
π 2 , 10 sin π sin
π 2 , 10 cos π) = (0, 0 , −10).
(d) (x, y, z) = (ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ) = (
1 √ 2
sin
π 4 cos
π 2
1 √ 2
sin
π 4 sin
π 2
1 √ 2
cos
π 4
1 2
1 2
- (a) Which coordinate system, cylindrical or spherical, is best suited to describe the surface x 2
(Hint: Think about what kind of symmetry this equation has).
(b) Rewrite this equation in the coordinate system you chose above.
Solution.
(a) The equation has spherical symmetry (we can replace x with −x, y with −y, and z with −z and nothing
changes). So spherical coordinates make the most sense here.
(b) The left hand side x
2
2
2 is just ρ
2 in spherical coordinates, so the equation becomes ρ
2 = 9, i.e.
ρ = 3 in spherical coordinates. This describes a shell of radius 3 about the origin.
- Describe in words the geometric meaning of the following mappings.
(a) (r, θ, z) 7 → (r,
θ 2 , z) (in cylindrical coordinates).
(b) (ρ, θ, ϕ) 7 → (ρ, θ + π, 2 ϕ) (in spherical coordinates). Assume ϕ is between 0 and
π 2
Solution.
(a) This transformation maps cylinders to half-cylinders (think of a Japanese folding fan).
(b) This transformation rotates everything around the z-axis by π radians (i.e. 180 degrees) and then maps
the upper half space (z ≥ 0) onto all of space.
