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18.02 Exam 1 Solutions
Problem 1.
a) P = (1, 0, 0), Q = (0, 2, 0) and R = (0, 0, 3). Therefore −−→ =ˆ
ı − 2ˆQR = −2ˆ
j + 3k
ˆ.QP j and −−→
−−→
QR·
−−→
QP �1, −2, 0� · �0, −2, 3� =4
√12 + 22√22 + 32
b) cos θ = √65
=
QP
−−→ QP
−−→
Problem 2.
a) −−→ P R = �−1, 0, 3�.P Q = �−1, 2, 0�, −→
−−→
P Q × −→
P R = ˆ
ı ˆ
j k
ˆ
−1 2 0
−1 0 3 = 6ˆ
ı + 3ˆ
j + 2k
ˆ.
1 = 1 62 + 32 + 22 = 1 √49 = 7
−−→
P Q × −→
P RThen area(Δ) = .
2 2 2 2
b) A normal to the plane is given by −
N = P Q × P R �6, 3, 2�. Hence the equation has the form
→ −−→ −→ =
6x + 3y + 2z = d. Since P is on the plane d = 6 1 + 3 1 + 2 1 = 11. In conclusion the equation of the
···
plane is
6x + 3y + 2z = 11.
c) The line is parallel to �2 − 1, 2 − 2, 0 − 3� = �1, 0, −3�. →= 6 − 6 =Since −0, the line isN · �1, 0, −3�
parallel to the plane.
Problem 3.
a) −→ ABOA = �10t, 0� and −−→ = �cos t, sin t�, hence
B
�
�
� −−→
OB = −→ AB
OA + −−→ = �10t + cos t, sin t�.
O A The rear bumper is reached at time t = π and the position of B is (10π−1, 0).
b) −
V
→= �10 − sin t, cos t�, thus
2 2
|−→
V |2 = (10 − sin t)2 + cos t = 100 − 20 sin t + sin2 t + cos t = 101 − 20 sin t.
The speed is then given by √101 − 20sin t. The speed is smallest when sin t is largest i.e. sin t = 1. It occurs
when t = π/2. At this time, the position of the bug is (5π, 1). The speed is largest when sin t is smallest;
that happens at the times t = 0 or π for which the position is then (0, 0) and (10π − 1, 0).
Problem 4.
a) |M| = −12.
b) 5, b = 7.−a =
⎡⎤⎡ x 114 0 t/12 + 1
1
3 ⎤⎡⎤⎡⎤
⎣ ⎦ =⎣ ⎣⎦ ⎦ =⎣ ⎦
c) −5 7 −8
7 −5 4 7t/12 − 2
−5t/12 + 1
ty 12
z
d�
r 1 7 5
d) dt = 12 , 12 , − 12 .
Problem 5.
a) −
N ·−
r (t) = 6, where −
N
→→→= �4, −3, −2�.
b) We differentiate −
N ·−
r (t) = 6:
→→
d d d d
−→
N −
N ·−
r (t) + −
N
→→→·−
r (t) + −
→N −
r (t)
→→and hence −
→
N ⊥ d r (t).
→
dt −−
r (t) = −
0
→→
r (t)
→
·−=0 = · ·
dt dt dt dt
