Math 2414
Homework Set 9 – Solutions
#4. (2 pts for (a) ONLY)
10 Points

⇒
 w
1
 =
u=
, −214 , 221
w
21

⇒


u=
− vv =
−
21
(a) w =
(b) v =38
3
38

i+
2
38

j−
5
38

k
#8. (2 pts)
 
x y
−7
cos θ =
−0.15453
  = =
x y
54 38
⇒
cos −1 ( −0.15453) =
1.72595 rad
θ=
The two vectors are neither parallel or orthogonal.

 a  b  1
2, −4,1 =
#9. (3 pts) proja b=  2 a=
21
a
2 −4 1
, ,
21 21 21
#12. (3 pts) First label the points P=(1, -2, 3), Q=(0, 5, -1) and R=(4, 0, -2) then the following two vectors
are in the plane,

QP = 1, −7, 4

RP = −3, −2,5
Then the cross product of these two vectors will be orthogonal to the plane containing the vectors and
hence the points.

 
i
j k
 
QP × RP = 1 −7 4
−3 −2 5
#1.
( −2, 4 ) is a point and


i
j





 


1 −7 =−35i − 12 j − 2k − 5 j + 8i − 21k =−27i − 17 j − 23k
−3 −2
Not Graded
−2, 4 is a vector. Here’s a sketch of the point and several possible
representation of the vector.
Math 2414
Homework Set 9 – Solutions
#2.

a = 4 + 36 =
#3.

a=

8b = −16, −8
40 = 2 10
10 Points
 
4b − 3a = −8, −4 − 6, −18 = −14,14




81 + 1=
82
8b= 16 i + 80 j − 24 k
 




 


4b − 3a =
8 i + 40 j − 12 k − ( 27i − 3 j ) =−19i + 43 j − 12k
 
a b
#5. =
 
#6. a  b
=
−4 )
( 2 )( 0 ) + ( 5)( 3) + ( −6 )(=
 
a=
b cos ( 25π )
39
=
( 2 )( 31
)( 0.30917 )
 
−456
pq
=
−1
#7. cos θ =
  =
p q
342 608
19.16854
θ=
π The two vectors are parallel.
⇒

 a  b  1
−1 1 7
, ,
#10. proja b = 2 a = −1,1, 7 =
51
51 51 51
a
 
 
#11. Remember that w × v =− ( v × w ) .
  
i j k
 
v × w= 2 0 −4
3 1 −6


 
w × v =−4i − 2k
 
i j








2 0= 0i − 12 j + 2k − ( −12 j ) − ( −4 ) i − ( 0 ) k = 4i + 2k
3 1

=
a
#13. First label the vectors

0, 7, −1 =
,b

1,3, −2 and c = 9, 0,3 then compute the volume of
the parallelepiped that is determined by these vectors. If the volume is zero then they all lie in the same
plane.
0 7 −1 0 7
  
a b × c =
−120 ≠ 0
1 3 −2 1 3 =
9 0 3 9 0
(
)
So, the vectors do NOT all lie in the same plane.