# Practice Exam 1 Solutions - Cerro Coso Community College

advertisement ```Math 251
Practice Exam 1
Cerro Coso Community College
Solutions
1. Express 9i − 2 j + 6k as a product of its length and direction.
Length = Magnitude =
92 + ( −2 ) + 62 = 121 = 11.
2
2
6 
9
Product of length and direction = 11 i − j + k  .
 11 11 11 
2. Given P1(1, –4, 7) and P2(3, –3, 2).
a) Find the distance between P1 and P2.
Distance =
( 3 − 1)2 + ( −3 + 4 )2 + ( 2 − 7 )2
= 30.
b) Find the direction of P1P2.
( 3 − 1) i + ( −3 + 4 ) j + ( 2 − 7 ) k =
30
30
30
2
1
5
i+
j−
k.
30
30
30
c) Find the midpoint of the line segment P1P2.
 ( 3 + 1) ( −3 − 4 ) ( 2 + 7 )  
7 9
,
,

 =  2, − ,  .
2
2  
2 2
 2
3. Find the center and radius of the sphere x2 + y2 + z2 + 4x – 6y –2z –1 = 0.
x2 + 4x + y2 – 6y + z2 – 2z = 1
(x2 + 4x + 4) + (y2 – 6y + 9) + (z2 – 2z + 1) = 1 + 4 + 9 + 1
(x + 2)2 + (y – 3)2 + (z – 1)2 = 15
Center: (–2, 3, 1) Radius: 15 .
Continued on the next page.
4. Let v = −i + j and u = 2i + 3j + 2k.
a) Find v ⋅ u, v , and u .
v ⋅u = − 2 + 3
v =
( −1)2 + 12
u =
( 2) + ( 3)
= 2.
2
2
+ 22 = 9 = 3.
b) Find the cosine of the angle between v and u.
cos θ =
u⋅v − 2 + 3
=
.
u v
3 2
c) Find the scalar component of u in the direction of v.
Scalar component = u cos θ =
u⋅v − 2 + 3
=
.
v
2
d) Find the vector projv u.
 u⋅v 
− 2+ 3
projv u =  2  v =
( −i + j ) .
 v 
2


4. Given P(1, 1, 1), Q(2, 1, 3), and R(3, –1, 1).
a) Find the area of the triangle created by points P, Q, and R.
PQ = 1, 0, 2 , PR = 2, −2, 0 .
Area =
i j k
1 2
1 0
1 uuuv uuuv 1
1 0 2
1
PQ &times; PR = 1 0 2 =
i−
j+
k = 4i + 4 j − 2k
2 0
2 −2
2
2
2 −2 0
2
2 −2 0
=
1 2
2
4 + 42 + ( −2 ) = 3.
2
b) Find a unit vector perpendicular to plane PQR.
2 2 1
i + j − k.
3 3 3
Continued on the next page.
5. Find the volume of the parallelepiped formed by the vectors
u = 2,1, 0 , v = 2, −1,1 , and w = 1, 0, 2 .
2 1 0
 −1 1
2 1
2 −1 
Volume = abs ( ( u &times; v ) ⋅ w ) = abs 2 −1 1 = abs  2
−1
+0

0 2
1 2
1 0 

1 0 2
= abs ( 2 ( −2 ) − 1( 3) + 0 ) = 7
6. Consider the following diagram.
F
45o
Q
P
Find the magnitude of the torque exerted by F on the bolt at P if │PQ│= 6 inches and
│F│= 50 pounds. Answer in foot-pounds.
Magnitude of torque = │PQ&times;F│=│PQ││F│sin 45o =
 2
.5 ( 50 ) 
 = 17.68 foot-pounds.
 2 
7. Find the vector and parametric equations for a line through (0, -7, 0) and
perpendicular to the plane 3x + 7y – 5z = 21.
The vector perpendicular to the plane is 3i + 7j – 5k.
Vector: 3ti + (-7 + 7t)j + (-5t)k
Parametric: x = 3t, y = -7 + 7t, z = -5t
8. Find the an equation for the plane through ( 2, 4, 5), (1, 5, 7), and (-1, 6, 8).
P(2, 4, 5), Q(1, 5, 7), and R(-1, 6, 8). PQ = -i + j +2k, PR = -3i +2j +3k.
i
j k
PQ &times; PR = −1 1
2 = −i − 3 j + k. Choose P(2, 4, 5).
−3 2 3
-(x – 2) – 3(y – 4) + (z – 5) = 0; -x + 2 – 3y +12 + z – 5 = 0; x + 3y – z = 9.
Continued on the next page.
9. Find the plane determined by the intersecting lines.
L1: x = t , y = 3 − 3t , z = −2 − t , − ∞ &lt; t &lt; ∞.
L2 : x = 1 + s, y = 4 + s, z = −1 + s, − ∞ &lt; s &lt; ∞.
Find the point of intersection first.
4t = 0
t = 1+ s
t = 0, s = −1.
− s +t =1
3 − 3t = 4 + s
− s − 3t = 1
Substituting either t = 0 or s = −1 into L1 or L2 yields x = 0, y = 3, z = −2.
Working backwards from the parametric form of both lines gives us the two vectors:
i − 3 j − k and i + j + k. Crossing these two vectors yields
i j k
−3 −1 1 −2
1 −3
i−
j+
k = −2i − 2 j + 4k
1 −3 −1 =
1 1
1 1
1 1
1 1 1
−2 ( x − 0 ) − 2 ( y − 3) + 4 ( z + 2 ) = 0; − 2 x − 2 y + 6 + 4 z + 8 = 0; − 2 x − 2 y + 4 z = −14; x + y − 2 z = 7.
x2 y2 z
12. Consider the following quadric surface.
+
= . A diagram follows.
4
9 6
z
x
y
a.
Which conic section results when we slice the above quadric surface with
xy-plane? Ellipse
b.
Which conic section results when we slice the above quadric surface with
yz-plane? Parabola
Continued on the next page.
( )
11. Let r(t) = e − t i + ( 2 cos 3t ) j + ( 2sin 3t ) k be the position of a particle in space.
a.
Find the particle’s velocity vector.
Velocity =
b.
dr
= −e −t i − ( 6sin 3t ) j + ( 6 cos 3t ) k.
dt
Find the particle’s acceleration vector.
Acceleration =
c.
d 2r
dt
2
= e −t i − (18cos 3t ) j − (18sin 3t ) k.
Find the particle’s speed and direction of motion at time t = 0.
dr
= −e −0i − ( 6sin 3 ( 0 ) ) j + ( 6 cos 3 ( 0 ) ) k = −i + 6k.
dt t =0
Speed is the magnitude of the velocity vector, hence speed =
( −1)2 + 62
= 37.
−1
6
i+
k.
37
37
Velocity is the product of speed and direction, hence at t = 0 the velocity of the
Direction is the unit vector of the velocity vector, hence direction is
6
 −1

particle is 37 
i+
k .
37 
 37
12. Let r ( t ) = ( sin t ) i + tj + ( cos t ) k , t ≥ 0, be a position vector of a particle in space at
time t. Find the time or times in the given time interval when velocity and
acceleration vectors are orthogonal.
r ' ( t ) = ( cos t ) i + j + ( − sin t ) k
r '' ( t ) = ( − sin t ) i + ( − cos t ) k
Now set r ' ( t ) ⋅ r '' ( t ) = 0 and solve for t.
− sin t cos t + sin t cos t = 0; This statement is true for all t ≥ 0.
13. Evaluate
(
 2
3 
i+
k  dt.

2
2
0
1
t
+
1
t
−


∫
1
) (
 2sin −1 t i +

1
π 3
π 
π 
3 tan −1 t  = 2   i + 3   j = π i +
j.
0
4
2
4
)
Continued on the next page.
14. Solve the following initial value problem for r as a vector function of t.
Differential Equation:
Initial Conditions:
d 2r
= − (i + j + k ).
dt 2
r ( 0 ) = 10i + 10 j + 10k and
dr
= 0.
dt t =0
dr
= − ( i + j + k ) dt = − it − jt − kt + C1 = − ( it + jt + kt ) + C1.
dt ∫
dr
Since
= 0, − i ( 0 ) − j ( 0 ) − k ( 0 ) + C1 = 0. Therefore, C1 = 0
dt t =0
and
dr
= −it − jt − kt.
dt
Now, r ( t ) = ∫ − ( it + jt + k )dt = −
t2
t2
t2
i − j − k + C2 .
2
2
2
02
02
02
i − j − k + C2 = 10i + 10 j + 10k.
2
2
2
 t2
  t2
  t2

t2
t2
t2
Therefore, r ( t ) = − i − j − k + 10i + 10 j + 10k =  − + 10  i +  − + 10  j +  − + 10  k.
 2
  2
  2

2
2
2

 
 

Since r ( 0 ) = 10i + 10 j + 10k , −
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