Part A) The two vectors 2 i + 5 j 3 k and 2 i 3 j + 4 k define a plane (it is th
ID: 1589294 • Letter: P
Question
Part A)
The two vectors 2 i + 5 j 3 k and 2 i 3 j + 4 k define a plane (it is the plane of the triangle with both tails at one vertex
and each head at one of the other vertices). Which of the following vectors is perpendicular to the plane?
a.
A.B = 5.0 i + 1.0 j + 3.0 k
b.
12
c.
20
d.
6.0
e.
A.B = [minus] 14 i + 4.0 j + 22 k
Part C)
Two displacement vectors are: a = (4.0 m)i + (5.0 m)j + (3.0 m)k and b = (3.0 m)i + (6.0 m)j - (2.0 m)k. What is the angle between these two vectors?
a.
22°
b.
35°
c.
67°
d.
43°
e.
61°
Part D)
Consider the position of a particle moving in three dimensional space to be given by the position vector
r = 1.5 t 3 i - 6.0 t 2 j + 4.00 t k., where r is meters and t in seconds. What is the magnitude of the acceleration
of the particle at t = 1.0 s?
a.
15 m/s2
b.
25 m/s2
c.
12 m/s2
d.
18 m/s2
e.
10 m/s2
Explanation / Answer
A) Vector perpendicular to plane =a x b = 2 i + 5 j 3 k x 2 i 3 j + 4 k = 11i -14j-16k
B) if A = 3.0 i + 5.0 j + 2.0 k and B = 3.0 i + 5.0 j 2.0 k.
A.B = -9+25-4 =12
C) a = (4.0 m)i + (5.0 m)j + (3.0 m)k and b = (3.0 m)i + (6.0 m)j - (2.0 m)k
cos (theta) = a . b / |a| .|b| = 12+30-6 /7.071 +7 =36 /14.071 =2.558
theta = arccos(2.558)
D) r = 1.5 t 3 i - 6.0 t 2 j + 4.00 t k
a = d^2r/dt^2 = 9t-12
acceleration at t=1 sec , a =9*1-12 = -3m/s^2 (answer doesn't match any options)