Math 1172 - Autumn 2015
Quiz 9 - In Class
Name:
Recitation Instructor:
SHOW ALL WORK!!! Unsupported answers might not receive full credit.
Problem 1 [4 pts] Suppose ~u = ı̂ − ̂ + k̂ and ~u = 2ı̂ + ̂ − 3k̂
a) [2 pts] Find ~u × ~v .
Solution: Compute the cross product using the symbolic determinant:


ı̂
̂
k̂
−1
1
1
1
1
−1
~u × ~v = det  1 −1
ı̂ − det
̂ + det
k̂
1  = det
1 −3
2 −3
2
1
2
1 −3
= 2ı̂ + 5̂ + 3k̂
where we’ve used the formula:
det
a b
c d
= ad − bc
to evaluate the 2 × 2 determinants.
b) [2 pts] Find a unit vector perpendicular to both ~u and ~v
Solution: Given any vector w,
~ a unit vector (i.e. a vector of length 1) ŵ in the same
direction is always given by:
w
~
.
ŵ =
|w|
~
Also, given any non-parallel vectors ~u and ~v , ~u × ~v is a vector that is orthogonal to both
~u and ~v , so we just need to find a unit vector ŵ in the same direction. Computing:
~u × ~v = 2ı̂ + 5̂ + 3k̂
√
|~u × ~v | = 22 + 52 + 32
√
= 38
Thus, ŵ =
i
~u × ~v
1 h
= √
2ı̂ + 5̂ + 3k̂ .
|~u × ~v |
38
Problem 2 [2 pts] Find an equation of a line parallel to ~v = 2ı̂ + 3̂ − k̂ that passes
through the point (0, 3, −1). Is the point (2, 4, 0) on this line?
Solution: A parametric description of a line parallel to a given a vector ~v that passes
through the point P0 = (x0 , y0 , z0 ) is given by:
~r (t) = ~v t + P~0 , − ∞ < t < ∞.
where P~0 = hx0 , y0 , z0 i. Thus, an equation of the desired line is given by:
~r (t) = h2, 3, −1i t + h0, 3, 1i , − ∞ < t < ∞.
To check if (2, 4, 0) is on the line, note that is this is the case, there must be a common
value for t so x(t) = 2, y(t) = 4, and z(t) = 0. We can write the equation of the line as:
~r (t) = h2t, 3t − t, −t + 1i ,

2t
 x(t) =
y(t) = 3t + 3 .
and thus:

z(t) = −t + 1
Thus, x(t) = 2 when t = 1. But, y(1) = 6, so (2, 4, 0) is NOT on the line.
Math 1172 - Autumn 2015
Quiz 9 - Take Home
Name:
Recitation Instructor:
SHOW ALL WORK!!! Unsupported answers might not receive full credit.
Problem 1 [2.5 pts] (Projecting a Vector Field Onto a Curve)
In multivariable calculus, many problems require one to find the component of a vector
(field) onto a given curve at each point along the curve.
Suppose ~r(t) = ht2 , 4t, 4 sin ti.
a) [1 pt] Calculate the unit tangent vector T̂ (t) when t = 0.
Solution: Since ~r (t) is in the direction tangent to the curve, a unit vector T̂ (t) is
~r (t)
obtained by setting T̂ (t) =
. Calculating:
|~r (t)|
~r (t) = t2 , 4t, 4 sin t
~r 0 (t) = h2t, 4, 4 cos ti
~r 0 (0) = h0 4, 4i
√
√
|~r 0 (0)| = 02 + 42 + 42 = 4 2
~r (0)
|=
Hence, T̂ (0) =
|~r (0)
1
1
0, √ , √
.
2
2
b) [.5 pts] Show that for any vector F~ and unit vector v̂ that scal v̂ F~ = F~ · v̂
Solution: For any vectors ~u and ~v , scal ~v ~u =
~u · ~v
.
|~v |
When ~v is a unit vector, |~v | = 1, so:
F~ · v̂
F~ · v̂
scal v̂ F~ =
=
= F~ · v̂
|v̂|
1
c) [1 pt] Suppose F~ = h−1, 2, 4i. Find scalT̂ (0) F~ . This is the magnitude of the
component of the vector F~ along the curve ~r(t) at t = 0.
Solution: Since T̂ (0) is a unit vector, the above formula is applicable and:
1
1
1
1
scalT̂ (0) F~ = F~ · T̂ (0) = h−1, 2, 4i · 0, √ , √
= (−1)(0) + (2) √
+ (4) √ .
2
2
2
2
6
Thus, scalT̂ (0) F~ = √ .
2
Problem 2 [1.5 pts] Suppose ~r(t) is a differentiable vector-valued function and |~r(t)| = 1.
a) [.5 pts] (True or False) Is ~r 0 (t) a unit vector for each value of t? Think about this
both conceptually and computationally!
Solution: Conceptually, this should not be the case; |~r(t)| = 1 represents a vector that
is a unit vector for each time t. There is no reason that such a vector must move at unit
speed; a constraint on the length of a vector does not inhibit how quickly it can move!
For an explicit example of this, consider the vector-valued function:
~r (t) = hcos(ωt), sin(ωt)i .
This function represents a particle moving in a circle of radius 1 with constant angular
velocity ω. It is clear that |~r (t)| = 1, but when one computes:
~r 0 (t) = h−ω sin(ωt), ω cos(ωt)i ,
one finds that |~r 0 (t)| = ω.
b) [1 pt] Show that ~r(t) and ~r 0 (t) are orthogonal for each value of t.
Hint: ~r(t) · ~r(t) = 1 for all t.
Solution: Since ~r(t) · ~r(t) = 1 for all t, differentiate both sides, using the product rule
for the dot product on the LHS:
d
d
(1) =
(~r(t) · ~r(t))
dt
dt
0 = ~r 0 (t) · ~r(t) + ~r (t) · ~r 0 (t)
0 = 2 [~r (t) · ~r 0 (t)] since the dot product is symmetric (i.e. ~u · ~v = ~v · ~u).
Hence, ~r (t) · ~r 0 (t) = 0, so ~r(t) and ~r 0 (t) are orthogonal for each value of t.