Solutions
Block I : Vector A r i t h m e t i c
U n i t 4 : The Dot Product
comment
While w e w a i t u n t i l t h e f i n a l p a r t of t h i s e x e r c i s e t o have you
supply t h e v a r i o u s d e t a i l s , t h e main aim of t h i s e x e r c i s e i s t o
emphasize t h a t , s i n c e c e r t a i n r u l e s f o r numerical p r o d u c t s a r e n o t
obeyed by d o t p r o d u c t s , t h e r e i s no r e a s o n t o s u s p e c t t h a t t h e two
t y p e s of m u l t i p l i c a t i o n w i l l behave e x a c t l y a l i k e . I n p a r t i c u l a r ,
i n numerical a r i t h m e t i c w e may conclude t h a t i f a b = a c and a # 0
t h e n b = c . What w e a r e having you show i n p a r t ( a ) i s t h a t i f
2 # d t h e r e a r e a number of ( i n f a c t , i n f i n i t e l y many) v e c t o r s +C
+ +
+ +
such t h a t A - B = A - C . T h i s shows t h a t t h e c a n c e l l a t i o n law i s
d i f f e r e n t f o r d o t p r o d u c t s and w e cannot go around b l i t h e l y can+
+ +
+ +
c e l l i n g A ' s i n a n e x p r e s s i o n such a s A * B = A - C , s i n c e i t need n o t
+
+
be t r u e t h a t B = C.
A t any r a t e , t a c k l i n g t h i s e x e r c i s e w i t h t h e p a r t s i n t h e g i v e n
o r d e r , w e have:
a.
Introducing t h e abbreviation (a,b,c) t o denote t h e vector
+
+
+
a i + b j + ck, we have
We desire that
Since we a r e dealing i n C a r t e s i a n coordinates, our r e c i p e f o r
finding dot products yields
+ +
A - C = (1,1,1)
(x,y,Z) = ( 1 )+~ ( 1 ) +~ ( 1 1 2 = x + y +
Z
.,
Solutions
Block 1: Vector A r i t h m e t i c
Unit 4: The Dot Product
,. .. :
,
,
1 . 4 . 1 (L) continued
+ +
-++
From (1) and ( 2 ) we s e e t h a t A - B = AoC i f and o n l y i f
Since ( 3 ) i s a l i n e a r e q u a t i o n i n t h r e e unknowns, we have two
+
d e g r e e s of freedom i n determining C.
I n other.words, w e can pick
any two of i t s components a t random and s o l v e uniquely f o r i t s
t h i r d component with e q u a t i o n ( 3 ) .
For example, i f w e wish t o have x = 5 and y = 7 , t h e n e q u a t i o n (3)
t e l l s u s t h a t z = -3.
+ -+
A o B = (1,1,1)
+ +
A - C = (1,1,1)
+-+
+
I n t h i s c a s e , C = (5,7 ,-3).
(2,3,4) = 9
(5,7,-3)
+ +
= 5
+
7
-
A s a check
3 = 9
Therefore, A - B = A * C
More g e n e r a l l y , we could use (3) t o show t h a t z = 9
+
l e t t i n g C = (x,y,9-x-y), w e have t h a t
- x - y.
Then
+ +
+-+
and comparing t h i s with e q u a t i o n (1) w e see t h a t A - B = A 0 C a s
b.
asserted.
+ +
+ +
We have A * B = A - C , whence
+-+
A*B
- +A - C+
= 0
Hence, by t h e d i s t r i b u t i v e p r o p e r t y of t h e d o t p r o d u c t , e q u a t i o n
(4 )
becomes
*For t h e r i g o r - o r i e n t e d r e a d e r , what we r e a l l y a r e d o i n g i s
+ +
+ +
+ +
+ +
A . B - A'C
A * B + (-A)*C
+ -+
+ -+
+ (-1)A-C
+ +
+
+
= A-B + A*(-l)C
+ - + -+
+
= A - B + A*(-C)
+ +
+
= A a ( B + [-C])
+ +
+
= A-B
= A-(B
-
C)
+
-+
( s i n c e -A = ( - l ) A )
+ +
+ +
( s i n c e m A 0 C = A 0 m C f o r any s c a l a r m)
+
+
( s i n c e ( - l ) C = -C)
(by t h e " u s u a l " d i s t r i b u t i v e r u l e )
+
(by t h e d e f i n i t i o n of B
- +C )
Solutions
Block 1: Vector A r i t h m e t i c
U n i t 4: The Dot Product
1 . 4 . 1 (L) c o n t i n u e d
S i n c e t h e d o t p r o d u c t of two v e c t o r s i m p l i e s t h a t e i t h e r one of
-+
t h e v e c t o r s i s 0 o r t h e two v e c t o r s a r e p e r p e n d i c u l a r , w e may conc l u d e from (5) t h a t
+
+
and w e r e q u i r e d t h a t B # C.
+
+
C is perpendicular t o A.
A s a check,
+
+
I n p a r t ( a ) w e chose A s o t h a t A #
-+ Hence, i t must be t h a t B
-
d
notice t h a t
Therefore,
Therefore,
c. S i n c e w e have a l r e a d y a g r e e d t h a t v e c t o r s do n o t depend on their
p o i n t of o r i g i n , l e t u s assume, w i t h o u t l o s s of g e n e r a l i t y , t h a t
-+
and B o r i g i n a t e a t a common p o i n t , t h e o r i g i n .
-+
+ -+
Now, t h e l o c u s of a l l v e c t o r s V such t h a t A - V = 0 i s t h e p l a n e t o
-+
which A i s p e r p e n d i c u l a r .
(Again, f o r t h e s a k e of u n i f o r m i t y , w e
-f
a r e assuming t h a t a l l v e c t o r s V s t a r t a t t h e o r i g i n . )
W e then
have, p i c t o r i a l l y ,
Solutions
Block 1: Vector A r i t h m e t i c
Unit 4: The Dot Product
1.4.1(L)
continued
+
I f we denote t h i s plane by M I C may be any v e c t o r which o r i g i n a t e s
i n t h e plane M and t e r m i n a t e s a t t h e head of B.
Again, p i c t o r i a l l y ,
(P is any p o i n t i n M
+
Therefore, OP l i e s i n M
+
Theref o r e , OP I A
+
+
+
+
But, OP + C = B + OP =
+
+
B C.
Therefore,
Theref o r e ,
1(
)
+ +
+ +
-
-
A - B = A*C)
+
+ +
+ +
I n summary, t h e n , t o f i n d a l l C such t h a t A - B = A - C , we c o n s t r u c t
+
+
t h e plane t o which A i s p e r p e n d i c u l a r . Then, any v e c t o r C which
+
o r i g i n a t e s i n t h i s plane and t e r m i n a t e s a t t h e head of B s a t i s f i e s
t h e requirement.
+
I n terms of p a r t ( a ) , i n t h i s p a r t i c u l a r problem, i f each C which
works i s placed a t ( 0 , 0 , 0 ) it w i l l t e r m i n a t e i n t h e plane, whose
e q u a t i o n , i n f a c t , i s x + y + z = 9.
To s e e t h i s more v i s u a l l y t h e diagram below shows t h e plane
+
+
z = 9 i n t h e f i r s t o c t a n t ( x , y , z 3 0 ) . Notice t h a t t o
f i n d where t h e plane i n t e r s e c t s , s a y , t h e x-axis, w e r e c a l l t h a t
t h i s p o i n t i s of t h e form (x,O,O) and hence w e s o l v e x + y + z = 9
x
y
with y = z = 0.
W e o b t a i n x = 9.
Solutions
Block 1: Vector ~ r i t h m e t i c
U n i t 4: The Dot Product
1.4.1 (L) c o n t i n u e d
(1) ASQR r e p r e s e n t s t h e p o r t i o n of t h e p l a n e x
i s i n the f i r s t octant.
(2)
+
y
+
z = 9 which
Summarizing o u r p r e c e d i n g remarks, p i c k any p o i n t i n
+
+
+ +
+ +
y + z = g I s a y , Po. Then OPo i s a C which s a t i s f i e s A * B = A - C
x +
A
with
A
and % a s g i v e n i n t h e e x e r c i s e .
A s , a more c o m p u t a t i o n a l check, w e have:
+
I f S ( X ~ , ~ ~i .s Zi n~ t)h e p l a n e p e r p e n d i c u l a r t o A = ( 1 1 1 ) t h e n
-P
+
+
A - O S = 0.
S i n c e OS = (xl, ylr z ) w e have
Equation ( 6 ) t e l l s us t h a t (xl, yl, zl) b e l o n g s t o M i f x
1
+
y1
+
z1 = 0 , o r z 1 =
- (xl
+ yl). So suppose ( ~ ~ . y ~ does
, z ~b )e l o n g t o M.
Hence, S h a s t h e form (xl,ylr-x1
-
yl).
and t e r m i n a t e s a t ( 2 , 3 , 4 ) we have t h a t
Then zl = - f x l + y l ) . Then s i n c e B s t a r t s a t 0
Solutions
Block 1: Vector Arithmetic
Unit 4: Tne Dot Product
1 . 4 . 1 (L) continued
+
From ( 7 ) we s e e t h a t t h e sum of t h e components of C i s
and t h i s i s i n agreement with equation (3) with x = 2-xl,
y = 3-yl,
and z = 4+x1 + yl.
Hopefully, w e now have a c o r r e l a t i o n between t h e a r i t h m e t i c answer
and t h e geometric c o n s t r u c t i o n which y i e l d s t h i s answer.
d. One f r i n g e b e n e f i t of t h e f i r s t t h r e e p a r t s of t h i s e x e r c i s e i s
t h a t n o t only h a s i t s u p p l i e d more d r i l l i n v e c t o r a r i t h m e t i c b u t
i t h a s a l s o made it seem q u i t e n a t u r a l i n v e c t o r a r i t h m e t i c t h a t
+ +
+ + +
+
+
+
we can have A - B = A * C without e i t h e r A = 0 o r B = C.
The purpose of t h i s p a r t of t h e e x e r c i s e i s t o once again emphas i z e t h e c r u c i a l concept of s t r u c t u r e .
Recall t h a t , i n o r d i n a r y
a r i t h m e t i c , i f a , b , and c a r e numbers and i f ab = a c , t h e n e i t h e r
a = 0 o r b = c.
This r e s u l t f o l l o w s inescapably from t h e proper-
t i e s of m u l t i p l i c a t i o n of numbers. The d o t product has very d i f f e r e n t p r o p e r t i e s from t h e "ordinary" product. To be s u r e t h e d o t
+ +
+ +
product i s commutative_ ( A - B = B - A ) and it obeys t h e d i s t r i b u t i v e
+ +
+
p r o p e r t y of m u l t i p l i c a t i o n "overH a d d i t i o n ; t h a t i s , A * ( B + C ) =
+ +
+ +
A - B + A - C , b u t h e r e t h e resemblance ends.
For one t h i n g , t h e d o t
product i s n ' t even a b i n a r y o p e r a t i o n ( i . e . , it i s n o t c l o s e d )
s i n c e by use of i t we combine two v e c t o r s t o form a s c a l a r . For
another t h i n g , t h e d o t product i s n o t a s s o c i a t i v e . Indeed not
+ + +
+ + +
o n l y i s it n o t t r u e t h a t A * ( B * C ) = ( A * B ) * C b u t even more c r u c i a l l y ,
+ + +
+ + +
t h e e x p r e s s i o n s A * ( B * C ) and ( A * B ) * C a r e n o t even d e f i n e d , s i n c e ,
+ +
f o r example, A - B i s a number and we do not " d o t H a number ( s c a l a r )
-
with a v e c t o r .
T h i r d l y , t h e r e i s no analog of a m u l t i p l i c a t i v e
identity.
Namely, we cannot d o t a v e c t o r with any v e c t o r t o o b t a i n
t h e o r i g i n a l v e c t o r a s t h e product s i n c e t h e d o t product of two
vectors i s n o t a vector.
F i n a l l y , t h e concept of m u l t i p l i c a t i v e
i n v e r s e s i s a l s o l a c k i n g f o r d o t products.
Namely, not only i s
t h e r e no i d e n t i t y v e c t o r with r e s p e c t t o d o t products, b u t even i f
t h e r e were, i t would n o t be p o s s i b l e , a s w e have a l r e a d y s a i d
s e v e r a l times i n t h i s paragraph, t o d o t two v e c t o r s t o o b t a i n a
vector.
Solutions
Block 1: Vector A r i t h m e t i c
U n i t 4 : The Dot Product
1 . 4 . 1 (L) c o n t i n u e d
Among o t i l e r t h i n g s w e now s e e how c o n v e n i e n t it i s when we d e a l
w i t h b i n a r y o p e r a t i o n s , f o r a s soon a s we combine " l i k e t h i n g s " t o
form " u n l i k e t h i n g s " it o f t e n becomes i m p o s s i b l e t o combine more
t h a n two l i k e t h i n g s . For example, i n t h e p r e s e n t i l l u s t r a t i o n
we could n o t t a l k m e a n i n g f u l l y a b o u t t h e d o t p r o d u c t of t h r e e ( o r
more) v e c t o r s even though t h e c o n c e p t was unambiguously d e f i n e d
f o r two v e c t o r s .
I n any e v e n t , t h e n , t h e key t h o u g h t i s t h a t t h e " c a n c e l l a t i o n law"
f o r d o t p r o d u c t s does n o t c o n t r a d i c t t h e " c a n c e l l a t i o n law" f o r
" o r d i n a r y " p r o d u c t s s i n c e t h e two t y p e s of p r o d u c t s have d i f f e r e n t
-
properties
and hence p o s s e s s d i f f e r e n t i n e s c a p a b l e c o n c l u s i o n s
based on t h e s e p r o p e r t i e s . * Another way of s a y i n g t h i s i s t h a t w e
do n o t c a l l d i f f e r e n t c o n c l u s i o n s c o n t r a d i c t o r y u n l e s s t h e y b o t h
f o l l o w i n e s c a p a b l y from t h e same s e t of assumptions ( p r o p e r t i e s ) .
*A Note on F r a c t i o n s I n v o l v i n g Dot P r o d u c t s
-- I n o r d i n a r y a r i t h m e t i c i f a f 0 t h e n ab
T h i s r e s u l t was
ac
c'
dependent on s e v e r a l p r o p e r t i e s of numerical a r i t h m e t i c . For
example, l e t
Then, ab = (ac)x But, by t h e a s s o c i a t i v e p r o p e r t y ( a c ) x = a ( c x ) .
ab = a (cx),
and s i n c e a f 0 i t f o l l o w s t h a t Therefore, Solutions
Block 1: Vector Arithmetic
Unit 4: The Dot Product
Note continued
Comparing (1) and (2) w e s e e t h a t
have
=
% if
a # 0.
Suppose now w e
-+
C e r t a i n l y w e might f e e l tempted t o c a n c e l on A from both t h e
numerator and denominator i n ( 3 ) .
I f we d i d t h i s w e would o b t a i n
Well, t o begin w i t h , i n terms of t h e p r o p e r t i e s of v e c t o r arithmet i c , i t i s n o t v a l i d t o c a n c e l s i n c e we have seen i n t h e p r e s e n t
e x e r c i s e t h a t t h e c a n c e l l a t i o n law, i n i t s usual form, does n o t
B
Secondly, w e would l i k e t o observe t h a t t h e e x p r e s s i o n , - - i s not
C defined!
g o r example, w i t h i n t h e ,proper c o n t e x t of o u r remarks how should
B
,--be i n t e r p r e t e d ? To c a p t u r e t h e f l a v o r of d i v i s i o n being t h e
n
L
i n v e r s e of m u l t i p l i c a t i o n and t h a t h e r e m u l t i p l i c a t i o g r e f e r s t o a
B
d o t product, w e would r e q u i r e t h a t t h e d e f i n i t i o n of ,--be
C i f t h e meaning of d i v i s i o n h e r e i s t o be analogous with i t s meaning
i n numerical a r i t h m e t i c .
But (5) h a r d l y makes sense s i n c e t h e l e f t s i d e of t h e e q u a l i t y i s
a number and t h e r i g h t s i d e i s a v e c t o r .
Notice,that t h i s problem d i d n o t occur with s c a l a r m u l t i p l i c a t i o n
A
since m
- could be d e f i n e d by
Solutions
Block 1: Vector A r i t h m e t i c
Unit 4: The Dot Product
Note c o n t i n u e d
and e q u a t i o n ( 6 ) i s meaningful s i n c e a s c a l a r m u l t i p l i e d by a
i s t h a t v e c t o r which
Indeed ( 6 ) s a y s t h a t
vector i s a vector.
-+
when s c a l e d t o m t i m e s i t s l e n g t h (if* > 0) y i e l d s t h e v e c t o r A.
+ +
Mechanically, t h e key p o i n t i s t o view A - B a s a s i n g l e number and
-h
+
thus t h a t t h e A and B a r e i n s e p a r a b l e i n t h e e x p r e s g i g n . I n t h i s
$
A *B
way w e w i l l n o t ( h o p e f u l l y ) be tempted t o "reduce" +--7
(Perhaps
A*C
t h i s i s more analogous t o t h e f a c t t h a t i n o r d i n a r y f r a c t i o n s we
to
cannot reduce a+b
a +c
denominator.)
a.
$ even
though a- i s i n b o t h numerator and
+
W e a r e g i v e n t h a t Po ( 2 , 3 , 4 ) i s i n o u r p l a n e M and t h a t V = ( 5 , 8 , 6 )
i s perpendicular t o our plane.
a t Po.
Thus,
+
W e p l a c e V s o t h a t it o r i g i n a t e s
+
Now t h e p o i n t P ( x , y , z ) i s i n t h e p l a n e i f and o n l y i f V
+ -*
t h i s i n t u r n r e q u i r e s t h a t V-POP = 0. That i s
(5,8,6)
(x-2, y-3,
2-4)
= 0
or
5 (x-2)
+
8 (y-3)
+
6 (2-4)
= 0
-4
I POP,
and
Solutions
Block 1: Vector Arithmetic
Unit 4: The Dot Product
1 . 4 . 2 continued Therefore, 5x
b.
+
8y
+
62 = 58
When x = y = 1 , equation ( 1 ) y i e l d s
Therefore,
i s i n M.
plane.
I
Since 72
< 8 , it f o l l o w s t h a t ( 1 , 1 , 8 ) i s above t h e
Solutions
Block 1: Vector Arithmetic
Unit 4: The Dot Product
1.4.3 continued
Now
A% = (3-1, 3-2, 5-3)
=
(2,1,2)
. . , '.
Therefore,
and
+
AC = (4-1, 4-2, 9-31 = (3,2,61
Therefore,
Putting this into (1) yields
(3,2,6) = (3)(7) cos 8
(2,1,2)
6
+
2
+
12 = 21 cos 8
20 = 21 cos 0
cos 0 =
20
Therefore,
20
-1 20
21 - cos
0 = arc cos -
21
(the angle whose cosine is
20
21).
We may look this up in the tables if we wish to conclude
cos 0 q 0.952
Therefore,
i
Solutions
Block 1: V e c t o r Arithmetic
Unit 4: The Dot Product
1.4.4(L1
S t r i c t l y s p e a k i n g , i t i s o u r b e l i e f t h a t from a c o m p u t a t i o n a l
p o i n t of view t h e r e i s no need f o r t h i s t o be l a b e l e d a l e a r n i n g
e x e r c i s e . Y e t , w e would l i k e t o c r e a t e a n o p p o r t u n i t y t o emphas i z e once a g a i n how v e c t o r methods can be used t o s o l v e a nonv e c t o r problem.
I n t h i s c a s e , we a r e given a l i n e i n t h e xy-plane
and a p o i n t n o t i n t h e xy-plane
( i n particular, then, t h e point i s
n o t on t h e g i v e n l i n e ) .
From o u r knowledge of Euclidean geometry
w e know t h a t t h e r e i's one and o n l y one l i n e t h a t c a n be drawn from
t h e given p o i n t perpendicular t o t h e given l i n e . This e x e r c i s e
a s k s u s t o f i n d t h e c o o r d i n a t e s of t h e p o i n t a t which t h e perpend i c u l a r from t h e p o i n t i n t e r s e c t s t h e given l i n e .
Pictorially,
One p l a n of a t t a c k i s a s f o l l o w s . From t h e e q u a t i o n y = -3x + 2
we can f i n d any number of p o i n t s which l i e on t h e given l i n e .
In
f a c t , w e may p i c k x a t random and t h e n l e t y = -3x
+
2.
One such
p o i n t (and you may, i f you wish, p i c k any o t h e r p o i n t and see what
happens) i s g i v e n by l e t t i n g x = 0 i n which c a s e y = 2 . T h e r e f o r e ,
we.may choose P
0
= ( 0 , 2 ) t o l i e on o u r l i n e .
To i n d i c a t e t h a t w e
a r e involved i n three-space w e w i l l w r i t e t h i s a s (0,2,0) r a t h e r
t h a n a s ( 0 , 2 ) . C l e a r l y t h i s i n v o l v e s no l o s s of g e n e r a l i t y . The
+
p o i n t P w e seek i s c h a r a c t e r i z e d by t h e f a c t t h a t PA must be p e r +
p e n d i c u l a r t o PPo. I f w e now e l e c t t o i n t r o d u c e t h e language of
-+ +
v e c t o r s , we a r e s a y i n g t h a t PA-PP = 0. The computation now
0
Solutions
Block 1: Vector A r i t h m e t i c
U n i t 4 : The Dot Product
1 . 4 . 4 (L) c o n t i n u e d
becomes q u i t e simple.
We f i r s t r e c o g n i z e t h a t t h e g e n e r a l p o i n t P
on o u r l i n e i s g i v e n by ( x , -3x
+
2, O ) , whence
while
Hence,
Hence, from (1),
PA-h0=
0 i f and o n l y i f x = 0 o r l o x
+
1 = 0.
I f .x = 0 t h e n y = - 3 x + 2 = 2, and w e o b t a i n P = (0',2) which i s t h e
same a s p and i n t h i s c a s e we do n o t have t h r e e d i s t i n c t p o i n t s ,
0-+ +
( I n o t h e r words t h e f a c t t h a t PA*PPo = 0 does n o t
A , P, and Po.
fro
-L
preclude t h a t
= 0. TO be c o n s i s t e n t w i t h t h e f a c t t h a t
+ +
-+
-b
+
A * B = 0 means A and B a r e o r t h o g o n a l , one o f t e n d e f i n e s t h e 0-vector
t o be o r t h o g o n a l t o e v e r y v e c t o r . )
I f l o x + 1 = 0 t h e n x = -1/10, whereupon y = -3x + 2 y i e l d s t h a t
y = 23/10.
I n o t h e r words, t h e l i n e from A p e r p e n d i c u l a r t o t h e
g i v e h l i n e meets' t h a t l i n e a t t h e p o i n t (-1/10, 23/10).
( A s a f i n a l n o t e on t h i s t e c h n i q u e , o b s e r v e t h a t w e might have
been lucky and p i c k e d o u r p o i n t Po i n such a way t h a t i t t u r n e d
o u t t o be t h e p o i n t a t which t h e p e r p e n d i c u l a r from A m e t o u r l i n e .
Had t h i s happened, a l g e b r a i c a l l y , t h e q u a d r a t i c e q u a t i o n r e s u l t i n g
+ +
from PA-PPo = 0 would have had t h e q u a d r a t i c t e r m v a n i s h and t h e r e
would have been o n l y t h e one s o l u t i o n t o e q u a t i o n (1). Anyone who
i s i n t e r e s t e d c a n work t h i s e x e r c i s e choosing Po t o be
(-1/10,
23/10) and see t h a t t h i s d o e s indeed happen.)
Solutions
Block 1: Vector Arithmetic
Unit 4: The Dot Product
1.4.4 (L) continued
Pictorial Summary
Our immediate aim in this exercise is to show structurally that
carrying out the operation
is the same as in "ordinaryn numerical multiplication. That is,
using our rules for vector arithmetic, we can show that
(where we are using
+
to abbreviate vl
+
+ 2 +
vl and v2 = v2
-P
v2).
Without the details of a formal mathematical proof, we proceed as
follows:
Solutions
Block 1: Vector Arithmetic
Unit 4: The Dot Product
1.4.5 (L) continued
The point is that (1) always holds,
we're used to, namely
The more convenient result that
+ -b
' t - t.t
was valid because i - j = 0 and i - 1 - J 3
-
1.
f
More generally, a glance at equation (1) shows that this "short
-+
+
-+2
cit* will prevail whenever vl
v2 = 0 and vl = ;22 = 1, for
clearly, in this case, equation (1) becomes
.
What we want to "hammer home" in this exercise is that while equation (1) always holds, equation (2) is a special case, in particu-+
-+
lar, equation (2) will yield an incorrect answer if vl
v2 # 0 or
-+
2
v12 # 1 or v2
1.
-+
+
In our present exercise
-f
- + 2 -+
= (3.2.5)
v1
(1) v1 = v1
(2)
(3)
+
V2
2
G1
-
-
+
=V2
-+
-+
v2 = (2,7,3)
v2 = (3.2.5)
2
(3.2,5) = 3
+
22
+ 52
(2.7.3) = 22
+
72
+
2(7)
+
(2,7.3) = 3(2)
+
= 38
32 = 62
5(3) = 35
(The answer (3G1 + 4G2)
-+
-+
+ 2
vl
v2 = 0 and vl = :22
+
(4vl
-
+
3v2) = 0 would have been t r u e had
= 1, b u t t h i s was n o t t h e c a s e h e r e . )
An a l t e r n a t i v e method would have been t o w r i t e a l l v e c t o r s i n terms
+
+
of i , j, and k. Then
+
-+
B = 4vl
-
3G2 = 4 ( 3 , 2 , 5 )
I n t h i s form ( s i n c e
-
3(2,7,3) = (6,-13,ll)
l2= j2 = g2 =
+
= 6i
-
+
13j
+
+
Ilk
+ +
+ +
+ +
1 and i * j = i - k = k = j = 0 ) w e
have
and t h i s a g r e e s w i t h o u r answer o b t a i n e d by t h e former method.
One o f t e n chooses v e c t o r s which have t h i s s o - c a l l e d orthonormal
p r o p e r t y where t h e magnitude of each v e c t o r i s 1 and t h e d o t prod u c t of a l l p a i r s of d i f f e r e n t v e c t o r s i s 0. Such a c h o i c e a l l o w s
u s t o compute u s i n g e q u a t i o n ( 2 ) r a t h e r t h a n e q u a t i o n (1).
+ +
A s a f i n a l n o t e , n o t i c e t h a t whether (1) o r (2) h o l d s , A - B = 0
+
+
means t h a t A and B a r e o r t h o g o n a l r e g a r d l e s s of how w e e l e c t t o
*See n o t e a t t h e end o f t h i s e x e r c i s e .
Solutions
Block 1: Vector ~ r i t h m e t i c
Unit 4: The Dot Product
1.4.5 (L) c o n t i n u e d
-+
-+
r e p r e s e n t A and B.
-+ -+
A-B,
The r e p r e s e n t a t i o n o n l y a f f e c t s how w e compute
f o r example, whether e q u a t i o n (1) must be used o r whether w e
can u s e ( 2 )
.
A Note on E x e r c i s e 1.4.5(LI
:
'I
P <-
,
-
I n high school algebra w e learned t h e so-called
" r u l e of f o i l "
o u t e r , knner, l a s t ) which a s s e r t e d
(first, -
One r a t h e r n e a t way of d e m o n s t r a t i n g (1) was by t h e geometric ob-
+ b)(c +
( a + b ) by
servation t h a t (a
d ) was t h e a r e a of a r e c t a n g l e whose
dimensions were
(c
+
d)
.
That i s ,
On t h e one hand, t h e r e c t a n g l e h a s
area (a + b) (c
hand, i t i s a c
+
+
d ) ; on t h e o t h e r
ad + bc + bd.
There i s , however, a way t o d e m o n s t r a t e (1) s o t h a t w e i l l u s t r a t e
t h e s t r u c t u r e of our game.
Namely, s i n c e a
+
b i s a number, s a y e l we have
But, by t h e d i s t r i b u t i v e p r o p e r t y e ( c
By t h e commutative p r o p e r t y ( a
d(a
+
b).
+
+ d)
b)c = c(a
= ec
+
+
ed.
b ) and ( a
Hence,
+
b)d.,=
Solutions
Block 1: Vector Arithmetic
Unit 4: The Dot Product
Note continued
Therefore,
= ac
+
ad
+
bc
+
bd
( s i n c e bc
+
ad = ad
+
bc)
The c r u c i a l p o i n t i s t h a t every p r o p e r t y of numerical a r i t h m e t i c
which was needed t o prove t h i s r e s u l t a l s o happens t o be t r u e i n
+
+
+
+ + +
+ +
our v e c t o r c a s e . Namely, A + B = B + A , A - B = BaA, and
+
+
A0(B
+
+ +
+ +
C) = A W B+ A.C.
-f
w e l l t o d o t products.
Hence, t h e H r u l e of f o i l " a p p l i e s e q u a l l y
Indeed, merely by recopying o u r e a r l i e r
proof we have
(where
+ +
= c-a
a.
+
+ +
c-b
+ 3.; + 3.G
We have
+ +
Since A - C = 0, we may d o t both s i d e s of
+
(1) with A t o o b t a i n
Solutions
Block 1: Vector A r i t h m e t i c
U n i t 4: The Dot Product
-
-
1.4.6 (L) c o n t i n u e d
i . 5 = ii.( x i + E ) + +
= if* (d)
+ A-C
= x +2
A + O
2
+
= xA
+ -+
+2
Since A - B , A
, and
x a r e s c a l a r s , e q u a t i o n ( 2 ) may be s o l v e d t o
yield
+
-+
+
R e f e r r i n g t o o u r diagram t h e v e c t o r p r o j e c t i o n of B o n t o A i s xA;
hence, from , ( 3 ) , t h i s i s
+
+
F i n a l l y , from ( l ) ,C = B
- xA-+
( n o t i c e how w e c o n t i n u e t o u s e v e c t o r
a l g e b r a i n a r a t h e r n a t u r a l way), whereupon ( 3 ) y i e l d s
b. I n p a r t ( a ) we emphasized a l g e b r a i c v e c t o r t e c h n i q u e s i n o r d e r t o
show a g a i n t h e s t r u c t u r e of v e c t o r a r i t h m e t i c .
In t h i s particular
e x e r c i s e , t h i s was e q u i v a l e n t t o s o l v i n g a geometric problem nongeometrically.
The purpose of p a r t ( b ) i s t o emphasize t h e geometric f a c t t h a t i f
+
+ +
uA d e n o t e s t h e u n i t v e c t o r i n t h e d i r e c t i o n of A t h e n B-uA d e n o t e s
+ +
( I f B-u i s p o s i t i v e
t h e l e n g t h of t h e p r o j e c t i o n of 3 o n t o d.
t h e n t h e a n g l e between
d
t i v e , t h e angle i s obtuse
and
-
8
A -+
i s a c u t e , w h i l e i f % - u A i s riega-
a s we s h a l l s e e below.)
To s e e t h i s r e s u l t w i t h o u t r e f e r e n c e t o d o t p r o d u c t s , w e have
Solutions
Block 1: Vector A r i t h m e t i c
Unit 4: The D o t Product
1.4.6 (L) c o n t i n u e d
+
I f 0 > 90° t h e n l ~ l c o s0 i s n e g a t i v e o r t h e p r o j e c t i o n i s measured
-+
i n t h e o p p o s i t e s e n s e of A.
That i s
NOW
0 i s c l e a r l y t h e a n g l e between
A
and
8.
Thus
-+ +
Equation ( 6 ) s t a r t s t o look l i k e
(= B * A ) e x c e p t t h a t t h e f a c t o r
-+
i s missing. W e t h e n n o t i c e t h a t i f
= 1 t h e missing f a c f o r
+
c a u s e s no change.
But
= 1 i s t h e same a s s a y i n g t h a t A i s a
-+
u n i t v e c t o r and t h i s c e r t a i n l y a p p l i e s t o u,.
Moreover, t h e f a c t
it-8
IA~
121
121
-b
t h a t u,
n
-+
i s + d e f i n e d t o have t h e same s e n s e a s A ( t h a t i s , s i n c e
. A 7i s a p o s i t i v e s c a l a r m u l t i p l e of A) means t h a t t h e
1AI
+
a n g l e between A and
Pictorially,
5
i s t h e same a n g l e a s between
A
and
5.
Solutions
lock 1: Vector A r i t h m e t i c
U n i t 4: The Dot Product
1.4.6 ( L ) c o n t i n u e d
I n o t h e r words,
But, by d e f i n i t i o n of t h e d o t p r o d u c t ,
Comparing ( 7 ) and ( 8 ) y i e l d s t h e r e s u l t t h a t t h e ( d i r e c t e d ) l e n g t h
+ +
of . t h e p r o j e c t i o n of 5 o n t o 2 i s B*uA
i s simply t h e d i r e c t e d
Thus, t h e v e c t o r p r o j e c t i o n of 5 o n t o
l e n g t h of t h e p r o j e c t i o n of 5 o n t o % m u l t i p l i e d by t h e u n i t v e c t o r
i n t h e d i r e c t i o n of A.
+
S i n c e uA =
-,
I ti I
That i s ,
(9) becomes
Solutions
Block 1: Vector A r i t h m e t i c
Unit 4: The Dot Product
1 . 4 . 6 ( L ) continued
Notice t h a t (10) and ( 4 ) are i d e n t i c a l , a s they should b e .
1.4.7 P i c t o r i a l l y , w e have But
Therefore,
. .
?
.
.,
.
Solutions
Block 1: Vector A r i t h m e t i c
U n i t 4 : The Dot Product
c
The s l o p e of R i s - a
- s i n c e a x + by + c = 0 , b f 0
y = a
b
-Ex 5.
b
Therefore, t h e
Hence, t h e l i n e p e r p e n d i c u l a r t o R h a s s l o p e
-+
-b
-+
v e c t o r V = a i + bj i s p e r p e n d i c u l a r t o R.
-+
a.
-
The d i s t a n c e we s e e k i s e q u a l t o t h e l e n g t h of t h e p r o j e c t i o n of
o n t o $, and t h i s i s merely t h e d o t p r o d u c t of A:P
-+
u n i t v e c t o r i n t h e d i r e c t i o n of N.
A
P:
-+
ue = u n i t v e c t o r i n d i r e c t i o n of
8
=
ai
+
with t h e
bj
m
while
A:P
=
(yo
-
0. yo
-
- 1 )
= (xo. yo
+
C
Therefore,
+
* T h i s p o i n t was o b t a i n e d b y l e t t i n g x = 0 i n a x
by + c = 0 ( o t h e r
p o i n t s would do a s w e l l ) .
T h e o n l y t i m e we a r e i n t r o u b l e i s when
b = 0, b u t i n t h i s c a s e a x + by + c = 0 r e d u c e s t o a x + c = 0
which i s t h e l i n e x =
--a p a r a l l e l
C
t o the y-axis.
In t h i s case,
is t r i v i a l t o find the required distance.
S o , w i t h o u t l o s s of g e n e r a l i t y , we may a s w e l l a s s u m e t h a t b # 0. it
Solutions Block 1: Vector Arithmetic Unit 4: The Dot Product 1.4.8 (L) continued Finally, since distance is defined to be non-negative and
axo + byo + c might be negative, we put our final answer in the
form
(An interesting aside occurs if
becomes
a
= 1, for in this case (1)
which means that we find the distance from (xo,yo) to ax + by + c = 0
simply by replacing x and y by xo and yo in the left side of the
equation for the line.)
Solutions Block 1: Vector Arithmetic Unit 4: The Dot Product 1.4.9
continued Hence, by the Pythagorean Theorem, MIT OpenCourseWare
http://ocw.mit.edu
Resource: Calculus Revisited: Multivariable Calculus
Prof. Herbert Gross
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