ECE602
Homework #4 Solutions
1) Finding the vector lengths
a.)
A  3
,
B  9
b.)
A  7
,
B  15
c.)
A  15
,
C  13
,
C  11
,
B  15
C  9
,
Finding AxB:
a.)
b.)
c.)
^
^
^
ax
ay
az
2
2
1
1
8
4
^
^
 9 a y  18 az
^
^
^
ax
ay
az
2
3
6  45 ax  38 a y  34 az
10
2
^
^
^
11
^
^
^
ax
ay
az
10
10
5
5
2
14
^
^
^
 130 ax  165 a y  70 az
Finding AxC:
a.)
b.)
c.)
^
^
^
ax
ay
az
2
2
1
12
4
3
^
^
^
ax
ay
az
2
3
6
2
9
6
^
^
^
ax
ay
az
10
10
5
4
7
4
^
^
^
 10 ax  18 a y  16 az
^
^
^
 72 ax  24 a y  12 az
^
^
^
 75 ax  60 a y  30 az
1
Fall 2013
ECE602
Homework #4 Solutions
Fall 2013
The projection of B on A is equivalent to finding the A components of vector C.
b

a
Hence, we find the following:
b c o s ( ) 
a b
a
Just use the right side of the above expression to find the scalar projection of C on B:
a.)
12  32  12
 
9
b.)
9
20  18  66
 
15
c.)
8
64
15
20  14  56
15

62
15
Finally, the following scalar triple product finds the volume of a parallelepiped:
ax
ay
az
ABC   b
x


cx
by
bz
cy
cz
The volumes for parts a through c are calculated below:
2
a.)
b.)
2
1
 ABC   1


12
4
3
2
3
6
 A B C   10


2
8
2
9
4  90
11  636
6
2
ECE602
c.)
Homework #4 Solutions
10
10
 ABC   5


4
2
7
Fall 2013
5
14  915
4
2. To show that, for any three vectors A, B, and C, the following vector equation holds:
A  ( B  C )  B  (C  A )  C  ( A  B )  0
use the BAC-CAB identity to aid you in proving the relation above. The BAC-CAD identity is
the following:
A  (B  C )  B( A C )  C ( A B)
Apply the BAC-CAB identity to each of the cross product terms in the above equation to
obtain the following:
B ( A C )  C ( A B )  C ( B A )  A ( B C )  A (C B )  B (C A )  0
Remember:
a. Dot products are commutative, i.e. in general A B  B A
b. Products of a scalar and a vector is also commutative, i.e. in general
c B  Bc
3. The cross product of any two vectors, say A and B, yield a resultant vector C: A  B  C
The vector, C, is perpendicular to both vectors A and B. Check this out by taking the dot
product A and C and the dot product of B and C. Since the cosine of 90 degrees is zero,
both dot products should equal zero. As for this problem, take the cross product of both
vectors:
^
^
^
^
^
^
^
^
^
( a x  2 a y  a z )  (3 a x  a y  2 a z )  3 a x  5 a y  7 a z
Since we are looking for the unit vector divide the resultant vector by its magnitude:
^
^
^
3a x  5a y  7 a z
3 5 7
2
2
^

2
^
^
3a x  5a y  7 a z
83
4.
A vector that is orthogonal to the pane formed by the two vectors mentioned in the problem
is determined by the cross product of those two vectors:
3
ECE602
Homework #4 Solutions
^
^
^
^
^
^
Fall 2013
^
^
^
( a x  a y  2 a z )  (3 a x  2 a y  a z )   3 a x  7 a y  5 a z
^
Take the resultant vector from above and cross that vector with the vector
^
^
^
^
^
^
^
^
^
^
^
^
(  3 a x  7 a y  5 a z )  ( 2 a x  2 a y  a z )  (1 7 a x  1 3 a y  8 a z )
parallel to the plane and perpendicular to
^
2ax  2ay  az
, the new resultant vector is
^
2ax  2ay  az
.
The unit vector is the following:
^
^
^
(1 7 a x  1 3 a y  8 a z )
17  13  8
2
2
^

2
^
^
(1 7 a x  1 3 a y  8 a z )
522
5. Since Professor Chamberlin was kind enough to solve for a, we’ll solve for scalars b and c.
Take the vector equation, V, and cross both sides by vector C:
V  C  a ( A  C )  b ( B  C )  c (C  C )
, the last term cancels since the cross product
of two parallel vectors is zero. Next, dot both sides of the above equation with vector A:
(V  C ) A  a ( A  C ) A  b ( B  C ) A
, the first term in the right hand equation cancels
out since the cross product of vectors A and C is orthogonal to vector A.
Now we can rearrange the above equation and solve for b:
(V  C ) A
(B  C ) A

 AV C 


 ABC 


 b
- (5-1)
Likewise, we solve for c in the same fashion. Take the vector equation V and cross
both sides with vector A:
V  A  a ( A  A )  b ( B  A )  c (C  A ) ,
the first term cancels since the cross product
of two parallel vectors is zero. Next, dot both sides of the above equation with vector B:
(V  A ) B  c ( C  A ) B  b ( B  A ) B
, the second term in the right hand equation
cancels out since the cross product of vectors A and C is orthogonal to vector A.
4
ECE602
Homework #4 Solutions
Fall 2013
Now we can rearrange the above equation and solve for c:
(V  A ) B
(C  A ) B

 A BV 


 ABC 


 c
- (5-2)
The scalars b and c are now in terms of scalar triple products. Solve for the four triple
products and replace those values into equations (5-1) and (5-2):
2
2
 ABC   1


12
1
4  90
8
4
3
2
2
 AV C   1 0


12
;
10
2
11
V B C   1


12
8
4  1590
1
2
11  312
2
4
;
3
 A BV   1


10
4
2
3
1
8
4  216
2
11
Replace the triple scalar products back into the scalar equations for a, b, c:
(V  B ) C
(A B) C
(V  C ) A
(B  C ) A
(V  A ) B
(C  A ) B



V B C 


 ABC 


 AV C 


 ABC 


 A BV 


 ABC 


 a 


1590
90
312
90
216
90
 1 7 .6 6 7
 3 .4 6 7
  2 .4
5