# Vector Calculus Solution by DynamiteKegs

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```									                         Math 223 VECTOR CALCULUS

SOLUTION FOR MIDTERM EXAM – I (09/29)

September 29 (Mon), 2008

Instructor:              Yasuyuki Kachi

Line #: 32702.

[I] (20pts)    In R3 , with the coordinate system             x, y, z ,

(1)   the deﬁning equation of the plane Π passing through P =                  1, 0, −1 , and

having the normal vector n =           3, 6, 10 , is

3 x + 6 y + 10 z = −7.

(2)   Let Q =          0, −1, 0 . Then

−
−→
b = PQ =          0 − 1,      −1           − 0,   0−     −1

=      − 1, −1, 1 .

(3)   The distance between the plane Π and the point Q is

n ·b                    3, 6, 10 · − 1, −1, 1
projn b    =                  =               √
n                          32 + 62 + 102

1
=   √              .
145

[II] (20pts)     (1)     We may rewrite the equation which is originally written in

cylindrical coordinate system        r, θ, z       as

1
(∗)                                  z =                       ,
sin 2θ

1
using Cartesian coordinate system         x, y, z , as follows. First, recall

sin 2θ       = 2       cos θ          sin θ

double angle formula    . Accordingly, we may “artiﬁcially” rewrite the right-hand

side of the given equation (∗) as

1                                  1
=
sin 2θ                2 ·      cos θ            sin θ

r2
=                                              .
2 ·     r cos θ               r sin θ

Recall that the conversion formulas between the two coordinate systems are

x = r cos θ,

y = r sin θ,

z =        z.

Also recall that a part of the reverse conversion formulas is
x2 + y 2 =            r2 .

Hence we may rewrite the right-hand side of (∗), using x, y, z, as

x2 + y 2
.
2xy

In sum, the equation (∗) becomes
x2 + y 2
z =                              ,
2xy

or the same
2xyz = x2 + y 2                      .

2
(2)     We may prove that the surface in R3 as in (1) is ruled , as follows:

Let z0 be an arbitrary real number such that z0 > 1.                        Substitute z = z0
1
into the equation   z =               ,    thus obtain
sin 2θ

1
z0 =                         ,
sin 2θ
or, the same to say,
1
sin 2θ     =                     .
z0

1
Since              < 1,        there are four θs in          0 ≤ θ < 2 π.         We may conﬁrm
z0
this by drawing the graph of u = sin 2θ in the                   θ, u    coordinate plane, and use

the horizontal line test.        For an arbitrary choice of θ = θ1 satisfying the

highlighted equation, there is another θ = θ2 satisfying the highlighted equation,

such that

θ1 − θ2         =         π       .

In other words, we may list the complete set of solutions to the highlighted equation as

θ1 , θ2 , θ3 , θ4 ,    in such a way

θ1 − θ2 = π,          and            θ3 − θ4 = π

hold. Accordingly, the half-lines θ = θ1 and θ = θ2                         make a pair, and the

half-lines   θ = θ3      and     θ = θ4      make a pair, and each of the two pairs of

half-lines makes a ﬂat angle, thus makes a whole line.

To conclude, the intersection of the surface with the plane z = z0 is the union

of two lines. This proves that the surface is ruled.

3
[III] (20pts)    We make the triplet of vectors                   a, b, c        in R3   below an

orthonormal frame by ﬁlling in appropriate real numbers in the blanks:

25              144
a =                      , −         ,                     ,
169              169
25
b =                      ,           ,                    ,
169
√
60 2                                 119
c =     −        ,                         , −            .
169                                 169

We rely on the classiﬁcation of orthonormal frames in R3 :

a =     a2 + b2 − c2 − d2 ,          2 bc + ad ,               2 bd − ac    ,

b =     2 bc − ad ,           a2 − b2 + c2 − d2 ,              2 cd + ab    ,

c =     2 bd + ac ,           2 cd − ab ,            a2 − b2 − c2 + d2 ,

where a, b, c, d ∈ R such that a2 + b2 + c2 + d2 = 1. First, observe

the ﬁrst entry of a + the second entry of b + the third entry of c

=    3 a2 − b2 − c2 − d2

=    4 a2 −     a2 + b2 + c2 + d2

=    4 a2 − 1                    since         a2 + b2 + c2 + d2 = 1 .

This equals
25          25                119                 69
+                −               =      −       .
169         169                169                169
In short,
69
4 a2 − 1         =     −        .
169

4
From this it follows
1                69
a2 =                      1−
4                169

25
=           .
169
Next, observe

2 b2 − 2 c2     =      the ﬁrst entry of a                 −     the second entry of b
25                      25
=                     −
169                    169
=     0.

Hence b2 = c2 . Accordingly,
25
a 2 − d2        =      the ﬁrst entry of a                  =       .
169
25
Since we had a2 =            , it follows d = 0. Now we substitute d = 0,
169
and also b2 = c2 , into the above expression of the triplet a , b , c :

a =         a2 ,           2 bc,       − 2 ac      ,

b =     2 bc,                  a2 ,        2 ab     ,

c =     2 ac,          − 2 ab,        a2 − 2 b2 .

In particular,
√
60 2
the third entry of a          =       − the ﬁrst entry of c =                   ,
169

and also
144
the ﬁrst entry of b          =           the second entry of a =         −        .
169

5
144
2 bc = −                     < 0.
169

This and b2 = c2 prove c = −b.

Hence
√
60 2
the third entry of b              =       the third entry of a =                       ,
169

and
√
60 2
the second entry of c                 =   − the third entry of b =                  −      .
169

To conclude,
√
25                         144                         60 2
a =                                      −                                                  ,
169             ,          169             ,            169

√
144                        25                          60 2
b =             −                                                                       ,
169             ,         169              ,            169

√                         √
60 2                      60 2                          119
c =             −                       −                              −                .
169            ,          169             ,            169

Note       :     More generally, if

2                            2
a =                 cos α       ,           − sin α          ,                              ,

2
b =                                 ,            cos α       ,                              ,

1
c =       − √           sin 2α ,                                       ,       cos 2α
2
6
form an orthonormal frame, then they are

2                                     2               1
a =                cos α       ,                   − sin α           ,          √          sin 2α   ,
2
2                                     2                  1
b =        −       sin α       ,                         cos α       ,          √          sin 2α   ,
2

1                                     1
c =     − √              sin 2α ,              − √            sin 2α ,                  cos 2α          .
2                                     2

[IV] (20pts)   (1)       For three vectors

a =            a1 , a2 , a3 ,

b =        b1 , b2 , b3 ,

c =        c1 , c2 , c3

in R3 , their scalar triple product                       a × b              · c    coincides with the following
quantity:

a1     a2        a3
b1     b2        b3        =     a1 b2 c3 + a2 b3 c1 + a3 b1 c2
c1     c2        c3
− a3 b2 c1 − a1 b3 c2 − a2 b1 c3 .

This last quantity is called the determinant .                                 The determinant possesses the
following property:
a1        a2   a3             c1         c2    c3
b1        b2   b3        =    a1         a2    a3 ,
c1        c2   c3             b1         b2    b3

which can be rewritten in terms of the scalar triple product of vectors in R3 , as

a ×b         ·c =             c ×a          · b.

7
(2)   For three vectors a , b and c in R3 , their vector triple product                                               a ×b       ×c
coincides with
a   ·           c        b −                 b     ·    c           a.

(3)   We may prove the identity
a ·c       b ·c
a ×b            ·       c ×d             =
a ·d       b ·d

for vectors a , b , c and d in R3 , as follows:

a ×b       ·       c ×d

=          c ×d            ×a          ·b                                         by the formula in (1)

=              c       ·       a       d −                  d       ·       a      c       ·b

by the formula in (2)

=          c       ·       a               d        ·       b           −          d       ·       a        c     ·    b

=          a       ·       c               b       ·        d           −          a       ·       d        b     ·    c

a ·c            b ·c
=                           .
a ·d            b ·d

Note 1      :           We may use (3) to prove

2                    2           2                      2
a ×b                =        a       b           −     a ·b                  ,

for vectors a and b in R3 , as follows:                                   In the formula obtained in (3), substitute
c with a , and substitute d with b , and obtain

8
a ·a             b ·a
a ×b     ·    a ×b           =                                 .
a ·b             b ·b

This is the same as
2
2                 a          a ·b
a ×b           =                                     ,
2
a ·b                b

or
2               2           2                           2
a ×b        =       a           b        −          a ·b            .

Note 2       :       An instant application of the identity in “Note 1” above is the

following Cauchy–Schwarz inequality for vectors in R3 :

a ·b ≤              a        b               .

[V] (20pts)       Consider the equation

2/5                           2/5           −5/2
(#)                   r =        cos θ           +           sin θ                                   ,

written in the polar coordinate system .

(1a)   Substitute (#) into       x = r cos θ,                and         y = r sin θ                 each:

cos θ
x =                                                                5/2
,
2/5                             2/5
cos θ           +           sin θ

sin θ
y =                                                                5/2
.
2/5                             2/5
cos θ           +           sin θ

9
(1b)     x2/5 + y 2/5

2/5                                                                        2/5
cos θ                                                                     sin θ
=                      2/5                      2/5
+                                   2/5                            2/5
cos θ           +     sin θ                                        cos θ                     +    sin θ

2/5                                   2/5
cos θ                +          sin θ
=                             2/5                         2/5
=         1.
cos θ         +       sin θ

(2)     Know that the equation (#) in the previous page is modiﬁed as

1 −6/5                   −3/5                     −3/5
(##)               r             cos θ                  sin θ                        1 − r2
5
−1/5                   −1/5                                                 1/5                              1/5
=        r −2/5     cos θ                  sin θ                    −           r 2/5             cos θ                 sin θ                     .

Substitute             x = r cos θ,               and      y = r sin θ                       in (##):

1                                                                                 5                     5             1       5           1       5
1 − x2 − y 2
−1                   −1
x−3/5 y −3/5                                             =         x                    y                     − x                   y                 .
5

(3)     Take the 5-th power of the both sides of the identity in (2):

1                                                 5
x−3 y −3        1 − x2 − y 2
55
−3       5           −3       5              3       5           3       5
=        x−1 y −1 − x y                − 5            x                    y                     − x                   y
−1       5           −1        5             1       5           1       5
+ 10           x                    y                     − x                   y

=       x−1 y −1 − x y

−       5        x−2/5 y −2/5                 x−1/5 y −1/5 − x1/5 y 1/5

2
+       5        x1/5 y 1/5               x−1/5 y −1/5 − x1/5 y 1/5                                    .

10
(4)   Substitute the result of (2) into the identity you obtained in (3):
1                                            5
x−3 y −3         1 − x2 − y 2
55
=      x−1 y −1 − x y

1
− 5 x−2/5 y −2/5                   x−3/5 y −3/5        1 − x2 − y 2
5
2
1/5       1/5        1                            2         2
+ 5 x             y                x−3/5 y −3/5       1 − x       − y                   .
5
Clean up the right-hand side:
1                                            5
x−3 y −3         1 − x2 − y 2
55

=      x−1 y −1 − x y

1                                               2
− x−1 y −1        1 − x2 − y 2               +       x−1 y −1     1 − x2 − y 2                         .
5
Multiply x3 y 3 to the both sides of the identity:
1                            5
1 − x2 − y 2
55
=      x2 y 2 − x4 y 4
1                                 2
− x2 y 2     1 − x2 − y 2               +          x2 y 2    1 − x2 − y 2              .
5
This last identity can be rewritten as
5
1 − x2 − y 2             − 3125            x2 y 2 − x 4 y 4
2
+ 3125 x2 y 2            1 − x2 − y 2             − 625 x2 y 2       1 − x2 − y 2                  = 0.

The left-hand side is indeed a polynomial in x, and y.

(5)    The graph of the curve prescribed by the equation in (4) is the asteroid.

It has cusps at each of           1, 0 ,           0, 1 ,        − 1, 0 ,      0, −1 .

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[VI] (Extra 20pts)           (1)   The Jacobi Identity for vectors in R3 is as follows:

a       ×       b     ×        c

+         b    ×       c         ×       a

+         c       ×       a        ×   b        =   0.

(2)   We may calculate the Lie bracket                            A, B        = AB − BA for matrices
                                                                               
0                −a3        a2                           0                −b3    b2
A =  a3                0        −a1          and       B =  b3                0    −b1  ,
−a2                a1        0                           −b2               b1     0

as follows. First,

                                                          
0                −a3           a2      0            −b3          b2
AB =  a3                0           −a1   b3             0          −b1 
−a2               a1            0      −b2            b1          0

                                                                         
− a2 b2 − a3 b3                         a2 b1                      a3 b1
=        a1 b2                        − a1 b1 − a3 b3                 a3 b2      .
a1 b3                             a2 b3                 − a1 b1 − a2 b2

Second,

                                                          
0                −b3           b2      0            −a3          a2
BA =  b3                0           −b1   a3             0          −a1 
−b2               b1            0      −a2            a1          0

                                                                         
− a2 b2 − a3 b3                         a1 b2                      a1 b3
=        a2 b1                        − a1 b1 − a3 b3                 a2 b3      .
a3 b1                             a3 b2                 − a1 b1 − a2 b2

12
Accordingly,

A, B     = AB − BA
                                                                                          
0                       a2 b1 − a1 b2                a3 b1 − a1 b3
                                                                                              
                                                                                              
=             a1 b2 − a2 b1                               0                    a3 b2 − a2 b3    .
                                                                                              
                                                                                              
                                                                                              
a1 b3 − a3 b1                       a2 b3 − a3 b2                        0

(3)     We may paraphrase the result of (2) as
                         
0              −c3     c2
A, B             =  c3              0      −c1  ,
−c2             c1       0

where
a2          a3                              a3           a1                         a1        a2
c1 =                          ,           c2 =                                ,   c3 =                      .
b2          b3                              b3           b1                         b1        b2

Agree
c1 , c2 , c3               =        a1 , a2 , a3           ×       b1 , b2 , b3 .

(4)     By the result of (3), the cross product for vectors in R3 and the Lie bracket for
skew-symmetric 3 × 3 matrices are parallel in structure. We may use this fact to
prove the Jacobi Identity as in (1), as follows:

Clearly it suﬃces to prove

A     ,        B       ,        C

+                    B    ,     C        ,       A

+              C   ,       A       ,    B           =   O.

13
The left-hand side equals

A     B    −    B     A     C    −     C      A     B    −      B   A

+      B     C    −    C     B     A    −     A      B     C    −      C   B

+      C     A    −    A     C     B    −    B       C     A    −      A   C   .

This is simpliﬁed as

ABC − BAC − CAB + CBA

+ BCA − CBA − ABC + ACB

+ CAB − ACB − BCA + BAC

=        O,

where we have used the associativity law for matrix multiplications:

AB C = A BC .

14

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